Random Normal Matrices: Eigenvalue Correlations Near a Hard Wall

Abstract

We study pair correlation functions for planar Coulomb systems in the pushed phase, near a ring-shaped impenetrable wall. We assume coupling constant $\Gamma =2$ and that the number n of particles is large. We find that the correlation functions decay slowly along the edges of the wall, in a narrow interface stretching a distance of order 1/n from the hard edge. At distances much larger than $1/\sqrt{n}$, the effect of the hard wall is negligible and pair correlation functions decay very quickly, and in between sits an interpolating interface that we call the “semi-hard edge”. More precisely, we provide asymptotics for the correlation kernel $K_{n}(z,w)$ as $n\rightarrow \infty$ in two microscopic regimes (with either $|z-w| = \mathcal{O}(1/\sqrt{n})$ or $|z-w| = \mathcal{O}(1/n)$), as well as in three macroscopic regimes (with $|z-w| \asymp 1$). For some of these regimes, the asymptotics involve oscillatory theta functions and weighted Szegő kernels.

1 Introduction: The Pushed Phase Coulomb Gas

Hard edge boundary conditions are well-known in the theory of Hermitian random matrices, where they are, for instance, associated with the Bessel kernel, see [32, 41, 65]. In dimension two, the study of one-component plasmas near a hard wall is likewise of interest and variants appear in for example [1, 11, 12, 15, 16, 20, 30, 31, 33,34,35,36,37, 41, 47,48,49, 58, 60,61,62]. As in the majority of these works, we take the coupling constant to be $\Gamma =2$, or equivalently, we consider eigenvalues of random normal matrices (see [28, 66]).

In [47], Jancovici considers two versions of the Ginibre ensemble,¹ the first being the usual “soft edge” ensemble. In the second model a wall is placed along the boundary of the droplet (nowadays this is called a “soft/hard edge”). One of Jancovici’s motivations for studying a Coulomb system near a hard wall is its interest for describing an electrolyte near a colloidal wall or an electrode plate.

In both cases it was shown that the pair correlation functions along the edge decay very slowly compared with the situation in the bulk. More precisely, they decay only as an inverse power of the distance along the edge. The heuristic is that the screening cloud that surrounds a particle sitting near the edge is prevented by the external field and/or the wall from being rotationally symmetric, which gives the particle plus cloud system a nonvanishing electrical dipole moment. This dipole moment is not strongly localized and causes long-range correlations along the edge [47].

In the soft edge-case, Jancovici’s results were first generalized to the elliptic Ginibre ensemble in [43] and later to quite general potentials in the paper [14], where a leading term for macroscopic correlations near the outer boundary component of the droplet is found and expressed in terms of a Szegő kernel. The paper [10] provides corresponding results near the boundary of a ring-shaped spectral gap, where some additional oscillations (depending on the number of particles) enter the picture. Microscopically, the oscillations are expressed in terms of the Jacobi theta function, which enters the subleading term of the microscopic density near the soft edge; macroscopically, the pair correlations along the boundary involve an oscillatory Szegő kernel. The heuristic is that there is an additional uncertainty concerning the number of particles that fall near each of the two boundary components of the spectral gap, and this manifests itself in terms of some oscillations. It is observed in [10] that the subleading term of the microscopic density near the edge is closely related to fluctuations of linear statistics.

The goal of the present work is to adapt the above results to the setting where a hard wall is placed inside the bulk of a plasma. This is known as a “pushed phase” [35,36,37] or as a “hard edge” [11, 61], and is quite different from the “critical phase”, i.e., the soft/hard edge in Jancovici’s original work. In the pushed phase, the plasma undergoes two transitions, at distances $\mathcal{O}(1/\sqrt{n})$ and $\mathcal{O}(1/n)$ respectively, from the hard edge.

Before we continue, it is expedient to briefly recall the Coulomb gas model. We consider Coulomb systems $\{z_j\}_{j=1}^n$ in the complex plane $\mathbb {C}$ subjected to an appropriate confining potential Q, where the Hamiltonian is

$$H_n=\sum _{i\ne j} \log \frac{1}{|z_i-z_j|}+n\sum _{j=1}^n Q(z_j),$$

i.e. it is the sum of the logarithmic interaction energy and the energy of interaction with the external field. The probability law of the system is given by the Gibbs measure

$$\begin{aligned} d \mathbb {P}_n(z_1,\ldots ,z_n)=\frac{1}{n!Z_n}e^{-H_n(z_1,\ldots z_n)}\, \prod _{j=1}^n \frac{d^2 z_j}{\pi }\end{aligned}$$

(1.1)

where $Z_n$ is the normalizing constant.

Classically, one approximates the random measure $\frac{1}{n} \sum _{j=1}^n \delta _{z_j}$ by a continuous unit charge distribution $\sigma =\sigma [Q]$. This distribution is precisely given by Frostman’s equilibrium measure associated with the potential Q, which is the unique minimizer $\sigma$ of the “Q-energy”

$$\begin{aligned} \nu \mapsto I_Q[\nu ]=\iint _{\mathbb {C}^2}\log \frac{1}{|z-w|}\, d\nu (z)\, d\nu (w)+\int _\mathbb {C}Q\, d\nu \end{aligned}$$

among all compactly supported unit measures $\nu$. The support $S={\text {supp}}\sigma$ is called the droplet in potential Q. (The measures $\frac{1}{n}\sum _{j=1}^n \delta _{z_j}$ converge, in a probabilistic and weak sense, to $\sigma$ as $n\rightarrow \infty$, see e.g. [7, 39].)

If Q is smooth in a neighbourhood of the droplet, then by Frostman’s theorem (see [59]), $\sigma$ is absolutely continuous with respect to the area measure $d^2 z$ and takes the form $d\sigma (z)=\Delta Q(z)\chi _S(z)\,\frac{d^2 z}{\pi }$ where $\Delta Q:=\frac{1}{4} (Q_{xx}+Q_{yy})$ is the usual Laplacian divided by four and $\chi _S$ is the indicator function of S. For example, if $Q(z)=|z|^2$ we obtain the Ginibre ensemble for which $S=\{z:\,|z|\le 1\}$ and $d\sigma (z)=\chi _S(z)\, \frac{d^2 z}{\pi }$.

We now fix a suitable (smooth) potential. In this paper we assume the rotational symmetry $Q(z)=Q(|z|)$ and that the droplet is a disc S. We also fix an open subset G of the interior of S which we take to be an annulus $r_1<|z|<r_2$ and modify the potential inside G to $+\infty$, i.e. we put

$$\begin{aligned} \tilde{Q}(z)={\left\{ \begin{array}{ll}Q(z),&{} z\in \mathbb {C}\setminus G,\\ +\infty ,&{} z\in G.\\ \end{array}\right. } \end{aligned}$$

(1.2)

Replacing Q by $\tilde{Q}$ has a drastic effect even at the level of the equilibrium measure. Indeed, the equilibrium measure $\tilde{\sigma }$ associated with the potential $\tilde{Q}$ puts zero mass in the gap G, and the restriction $\sigma |_G$ is swept to a measure $\nu _G$ supported on the boundary of G via a construction known as balayage [59]. In short, the measure $\nu _G$ has the same total mass as $\sigma |_G$, is supported on the boundary of G and takes the form $\nu _G=c_1\nu _1+c_2\nu _2$ where $\nu _j$ denotes the arclength measure along the circle $|z|=r_j$ and $c_j$ are certain constants, given in Eq. (2.4) below. The balayage measure $\nu _G$ and the restriction $\sigma |_G$ have furthermore identical logarithmic potentials in the complement $\mathbb {C}\setminus \overline{G}$.

The equilibrium measure $\tilde{\sigma }$ can now be written

$$\tilde{\sigma }=\sigma |_{\tilde{S}}+\nu _G,$$

where $d\sigma =\Delta Q(z)\frac{d^2 z}{\pi }$ and $\tilde{S}=S\setminus G$ is the droplet associated with the potential $\tilde{Q}$.

On the level of the Coulomb system $\{z_j\}_{j=1}^n$, the picture is that most of the particles that originally occupied the gap get “swept” to a thin interface in $\tilde{S}$ near the boundary $\partial G$, at a distance of $\mathcal{O}(1/n)$; following [11] we call this the “hard edge regime”. If we denote this interface by $E_n$ then the corresponding random measure $\frac{1}{n}\sum _{z_j\in E_n}\delta _{z_j}$ approximates the singular part $\nu _G$ of $\tilde{\sigma }$.

The 1-particle density of the pushed system is of order of magnitude $n^2$ in the hard edge regime. Further inside the bulk of $\tilde{S}$, at distances much larger than $1/\sqrt{n}$ from $\partial G$, the effect of the hard wall becomes negligible and the 1-particle density is, to a first order approximation, given by n times the equilibrium density $n \Delta Q(z)\frac{d^2 z}{\pi }$ of the unconstrained ensemble (associated with the potential Q). In between these regimes sits a transitional regime which we call the “semi-hard edge”, following [11].

As we already mentioned, a different kind of wall, known as a soft/hard wall or a critical phase, is obtained by placing the hard wall along the boundary of the droplet S. This corresponds to setting $G=\mathbb {C}\setminus S$ in Eq. (1.2). In this case the local statistics near the wall, in a $\mathcal{O}(1/\sqrt{n})$-interface, is affected, but the equilibrium measure and the droplet are unchanged, and no hard-edge regime is present.

In studying hard walls, we must face the difficulty that the wall might cause long-range correlations merely due to its “symmetry breaking”. A main insight in the forthcoming work [38] is that, for “general” potentials, symmetry breaking is avoided precisely if the hard wall is placed along the boundary of the droplet associated with the potential $Q/\tau$ where $0<\tau <1$ is a fixed suitable constant. In other words, the hard wall must be placed along the boundary of a mass-$\tau$ droplet associated with Q (cf. [53] for more about $\tau$-droplets). If this is done, some first order universality results can be proven; however, we emphasize that much more refined asymptotic results, such as the ones we obtain below, remain currently out of reach in this generality. Hard walls which break the symmetry are so far studied mostly at the level of the equilibrium measure, see the works [1, 31, 33].

For rotationally symmetric pushed phase models (including Coulomb gases in $\mathbb {R}^d$ and Yukawa gases) more work has been done. In [35] it is proven that the weighted logarithmic energy $I_Q[\tilde{\sigma }]$ of the equilibrium measure exhibits a third-order phase transition as the wall crosses the critical phase and enters the pushed phase. In the planar case ($d=2$), a large n-expansion of the free-energy is proved in [30], allowing for the case of annular spectral gaps. The third order phase transition is reflected in the leading coefficient of that expansion. In the work [61], Seo finds the leading order microscopic one-point density near a hard wall and proves it to be universal for a class of rotationally symmetric ensembles (which may have an additional logarithmic singularity along the hard edge). Partition functions with hard edges are related with so-called large gap (or hole) probabilities; this topic has a long history, see e.g. [5, 25, 40], and has found important recent applications to physics, see e.g. [52]. Disc counting statistics are considered in [11, 12, 20] gives a functional limit theorem for certain smooth radially symmetric linear statistics in the hard edge regime, near the outer boundary of the droplet.

It is worth noting that a different type of hard-edge ensemble, corresponding to the external potential $Q=0$ and a hard wall outside some compact set $\Sigma$ bounded by a Jordan curve has attracted some recent interest (e.g. [49]). In this case the entire system will tend to occupy the hard wall regime, i.e. the portion of $\Sigma$ which has distance $\mathcal{O}(1/n)$ to the boundary of $\Sigma$. In a way, this case is simpler than the kind of hard walls studied in the present paper, since there is essentially just one regime (no “bulk” to interact with). In the case when $\Sigma$ is an elliptic disc (possibly with an added logarithmic singularity along the hard edge) correlations are studied in [3, 56], following the earlier work [67] on truncated unitary matrices.

The present work gives a comprehensive study of correlations near a ring-shaped hard wall in the hard and semi-hard regimes on the microscopic and macroscopic levels. We shall consider the class of underlying rotationally symmetric potentials of the form

$$\begin{aligned} Q(z)=|z|^{2b}+\frac{2\alpha }{n}\log \frac{1}{|z|} \end{aligned}$$

(1.3)

where $b>0$ and $\alpha >-1$. These are sometimes called model Mittag-Leffler potentials, and they give rise to the droplets $S=\{z:\,|z|\le b^{-1/2b}\}$. The corresponding point-processes Eq. (1.1) are well studied and are sometimes called model Mittag-Leffler ensembles, see [6, 18, 22, 29] and the references there.

For the potentials Eq. (1.3) and an annular spectral gap $G=\{r_1<|z|<r_2\}$ with $0<r_1<r_2<b^{-1/2b}$ we provide a full asymptotic picture of the correlations associated with the potential $\tilde{Q}$, cf. Figs. 1 and 2 for illustrations.

In the rest of this paper, we will use the symbol Q to denote the pushed-phase Mittag-Leffler potential (above denoted $\tilde{Q}$) which is $+\infty$ in G, and we will write $\mu$ for the equilibrium measure of Q and $S={\text {supp}}\mu =\{|z|\le b^{-1/2b}\}{\setminus } G$ for the pushed-phase droplet, see Fig. 2.

2 Description of the Model

The k-point correlation functions $\{R_{n,k}:\mathbb {C}^{k}\rightarrow [0,+\infty )\}_{k= 1}^{n}$ associated with the point process $\{z_j\}_{j=1}^n$ in Eq. (1.1) are defined such that

$$\begin{aligned} \mathbb {E}_n[f(z_1,\ldots ,z_k)]=\frac{(n-k)!}{n!}\int _{\mathbb {C}^k}f(z_1,\ldots ,z_k)\, R_{n,k}(z_1,\ldots ,z_k)\, \prod _{j=1}^{k}\frac{d^{2}z_{j}}{\pi } \end{aligned}$$

holds for all continuous and compactly supported functions f on $\mathbb {C}^{k}$. The point process Eq. (1.1) is determinantal, meaning that all correlation functions exist and that there exists a correlation kernel $K_{n}(z,w)$ such that

$$\begin{aligned} R_{n,k}(z_{1},\ldots ,z_{n}) = \det (K_{n}(z_{i},z_{j}))_{i,j=1}^{k}, \qquad \hbox {for all } n \in \mathbb {N}_{>0}, \; k\in \{1,2,\ldots ,n\}, \; z_{1},\ldots ,z_{k}\in \mathbb {C}. \end{aligned}$$

It should be noted that while the correlation functions are uniquely defined, a correlation kernel $K_n(z,w)$ is only defined up to a multiplicative “cocycle”, i.e., a function of the form $c_n(z,w)=g_n(z)\overline{g_n(w)}$ where $g_n$ is a unimodular function. We fix $K_n$ uniquely by taking $K_n(z,w)$ to be the reproducing kernel of the n-dimensional subspace of $L^2$ consisting of weighted polynomials $p(z)e^{-nQ(z)/2}$ where p is a holomorphic polynomial of degree at most $n-1$. This $K_n$ is called the canonical correlation kernel and is used without exception in the following; see Eq. (2.5) for an explicit formula.

We shall study large n asymptotics for pair correlations $K_n(z,w)$ when z, w are close to a hard wall. However, before specializing to our setting, it is convenient to recall a few well known asymptotic results in other regimes, such as the bulk and soft edge regimes. Our discussion is far from exhaustive and we refer to the recent survey [23] for more background.

If $z_{0}$ is a regular point in the bulk (i.e. the interior of S) then [19] the rescaled kernels

$$\begin{aligned} \textrm{K}_n(\textrm{z},\textrm{w})=\frac{1}{n\Delta Q(z_{0})} K_{n}\Big ( z_{0} + \frac{\textrm{z}}{(n\Delta Q(z_{0}))^{1/2}} , z_{0} + \frac{\textrm{w}}{(n\Delta Q(z_{0}))^{1/2}} \Big ), \qquad \textrm{z},\textrm{w}\in \mathbb {C}, \end{aligned}$$

converge as $n\rightarrow + \infty$ (after multiplication by suitable cocycles $c_n(\textrm{z},\textrm{w})$) to the Ginibre kernel $e^{\textrm{z} \overline{\textrm{w}}-|\textrm{z}|^{2}/2-|\textrm{w}|^{2}/2}$, and where we recall that $\Delta Q$ denotes the standard Laplacian divided by four, i.e. $\Delta Q=(Q_{xx}+Q_{yy})/4$.

In the soft edge case, the large n behavior of $K_{n}(z,w)$ is well understood when z, w are close to the “outer boundary” of S, provided that this is an everywhere regular Jordan curve and that $\Delta Q>0$ along the curve. By the outer boundary, we mean the component $\partial U$ of $\partial S$ where U (throughout) denotes the unbounded component of $\mathbb {C}{\setminus } S$. In this case, the leading order asymptotics of $K_{n}(z,w)$ is found in e.g. [15, 42, 46, 50, 63] for the case when z, w are at a distance of order $n^{-1/2}$ from the outer boundary $\partial U$ and such that $|z-w| = \mathcal{O}(n^{-1/2})$.² The papers [10, 54] provide subleading corrections to the microscopic kernel near an outer boundary, and [10] also gives such correction terms near the boundary of a ring-shaped spectral gap and relates them to fluctuations of linear statistics.

As we already noted in Sect. 1, two points z, w near a (smooth) outer soft edge $\partial U$ are strongly correlated even if $|z-w|\asymp 1$,³ i.e. even if z, w lie at a macroscopic distance of each other. (By contrast, if z, w are in the bulk, $K_{n}(z, w)$ gets exponentially small as $n\rightarrow +\infty$ with $|z-w|\asymp 1$, and $K_{n}(z, w)$ has also a Gaussian decay in the distance $|z-w|$ for fixed n). A formula for long-range correlations along the outer boundary $\partial U$ of the droplet is computed in [14]. As noticed in [10], the situation is more subtle when z, w are in a $n^{-1/2}$ neighborhood of the boundary of a (soft) spectral gap that is not simply connected: in this case, the asymptotics of $K_{n}(z,w)$ are oscillatory and described in terms of Jacobi $\theta$-functions (if $|z-w|=\mathcal{O}(n^{-1/2})$) or weighted Szegő kernels (if $|z-w|\asymp 1$).

Some further related results for correlations along the boundary are found in [4] (higher dimensional elliptic Ginibre ensemble) and [27] (lemniscate ensembles).

Scaling limits have also been investigated in other situations than the bulk and soft edge regimes, see e.g. [2, 18] near root-type bulk singularities, [15, 16, 24, 51, 60] near singular boundary points (such as cusp-like singularities and so-called local droplets), [3, 9, 26, 44] for bandlimited point processes, and [15,16,17, 34, 47, 58] near soft/hard edges and other sorts of soft/hard boundary conditions.

Much less is known about pair correlations near the hard edge for pushed-phase ensembles (i.e. with a non-constant potential). Indeed, to our knowledge, the only hitherto recorded results appear in the paper [61], where large n asymptotics for $K_{n}(z,w)$ is studied the microscopic regime where z, w satisfy $|z-w|=\mathcal{O}(n^{-1})$ and are in a $n^{-1}$ neighborhood of $\partial U$ (with $\partial U$ being a hard edge), in the case of rotation-invariant Q (which may have a weak logarithmic singularity along the hard edge).

Recall that the droplet associated with the Mittag-Leffler potential Eq. (1.3) is given by $\{|z|\le b^{-1/2b}\}$. We now fix numbers $r_1,r_2$ with $0<r_1<r_2<b^{-1/2b}$ and redefine that potential to be $+\infty$ in the gap (or hard wall) $G:=\{z:|z|\in (r_{1},r_{2})\}$, i.e. we set

$$\begin{aligned} Q(z) = {\left\{ \begin{array}{ll} |z|^{2b} - \frac{2\alpha }{n}\log |z|, &{} \hbox {if } |z| \notin (r_{1},r_{2}), \\ +\infty , &{} \hbox {if } |z| \in (r_{1},r_{2}), \end{array}\right. } \qquad b>0, \; \alpha > -1. \end{aligned}$$

(2.1)

In this work, we focus on the corresponding point process:

$$\begin{aligned} \frac{1}{n!\mathcal {Z}_{n}} \prod _{1 \le j < k \le n} |z_{k} -z_{j}|^{2} \prod _{j=1}^{n}e^{-n Q(z_{j})}\prod _{j=1}^n \frac{d^2 z_j}{\pi }. \end{aligned}$$

(2.2)

Fig. 1

Illustration of the point process (2.2) with $n=4096$, $r_{1}=\frac{3}{5}b^{-\frac{1}{2b}}$, $r_{2}=\frac{4}{5}b^{-\frac{1}{2b}}$, $\alpha =0$ and the indicated values of b

The method of balayage in [12] shows that the equilibrium measure associated with Eq. (2.1) is

$$\begin{aligned} \mu (d^{2}z) = 2b^{2}r^{2b-1}\chi _{[0,r_{1}]\cup [r_{2},b^{-\frac{1}{2b}}]}(r) dr\frac{d\theta }{2\pi } + \sigma _{1} \delta _{r_{1}}(r) dr \frac{d\theta }{2\pi } + \sigma _{2} \delta _{r_{2}}(r) dr \frac{d\theta }{2\pi }, \end{aligned}$$

(2.3)

where $z=re^{i\theta }$, $r>0$, $\theta \in (-\pi ,\pi ]$ and

$$\begin{aligned} \sigma _{1} = \sigma _{\star }-br_{1}^{2b}, \qquad \sigma _{2} = br_{2}^{2b}-\sigma _{\star }, \qquad \sigma _{\star } := \frac{r_{2}^{2b}-r_{1}^{2b}}{2\log (\frac{r_{2}}{r_{1}})}. \end{aligned}$$

(2.4)

The droplet is given by $S=\{z:|z|\in [0,r_{1}]\cup [r_{2},b^{-\frac{1}{2b}}]\}$. We write $U=\{z:|z|>b^{-\frac{1}{2b}}\}$ and $G=\{z:r_1<|z|<r_2\}$ for the unbounded and bounded components of $\mathbb {C}\setminus S$ respectively, and refer to G as the “spectral gap”, or “hard wall”. Note also that G is not simply connected.

Thus $\partial U = \{|z| = b^{-\frac{1}{2b}}\}$ is a soft edge while $\partial G = \partial S {\setminus } \partial U = \{|z|=r_{1}\}\cup \{|z|=r_{2}\}$ is the union of two hard edges (see also Fig. 1). The quantity $\sigma _{1}>0$ represents the proportion of the points swept out from G which accumulate near $\{|z|=r_{1}\}$, $\sigma _{2}>0$ is the proportion of the points accumulating near $\{|z|=r_{2}\}$, and $\sigma _{\star }=\mu (\{z:\, |z|\le r_1\})$.

Since Q is rotation-invariant, the canonical correlation kernel of Eq. (2.2) is given by

$$\begin{aligned}&K_{n}(z,w) = e^{-\frac{n}{2}Q(z)}e^{-\frac{n}{2}Q(w)} \sum _{j=1}^{n} \frac{z^{j-1}\overline{w}^{j-1}}{h_{j}}, \end{aligned}$$

(2.5)

where

$$\begin{aligned} h_{j}&:= \Vert z^{j-1}e^{-\frac{n}{2}Q(z)}\Vert _{L^{2}(\mathbb {C},d^{2}z/\pi )}^{2} = \int _{\mathbb {C}}|z|^{2j-2} e^{-nQ(z)} \frac{d^{2}z}{\pi } \nonumber \\&= \frac{1}{b n^{\frac{j+\alpha }{b}}} \bigg ( \gamma (\tfrac{j+\alpha }{b},nr_{1}^{2b}) - \gamma (\tfrac{j+\alpha }{b},nr_{2}^{2b}) + \Gamma (\tfrac{j+\alpha }{b}) \bigg ), \qquad j=1,2,\ldots ,n, \end{aligned}$$

(2.6)

and the gamma functions $\gamma$, $\Gamma$ are defined by

$$\begin{aligned} \Gamma (a):=\int _{0}^{\infty } t^{a-1}e^{-t}dt, \qquad \gamma (a,z) := \int _{0}^{z}t^{a-1}e^{-t}dt. \end{aligned}$$

(2.7)

We recall some asymptotic properties of $\gamma$ in Appendix 2.

Our main results are asymptotic formulas for $K_{n}(z,w)$ as $n\rightarrow +\infty$ for five regimes where $z,w\in S$ are “close" to $\partial G$. We will explore (i) the so-called “hard edge regime", i.e. when z, w are $\mathcal{O}(n^{-1})$ away from $\partial G$, and (ii) the “semi-hard edge regime", i.e. when z, w are at a distance of order $n^{-1/2}$ from $\partial G$. (If z, w are slightly further away from $\partial G$, for example if z, w are at a distance of order $\sqrt{\log n}/\sqrt{n}$ from $\partial G$, then the asymptotics of $K_{n}(z,w)$ are no longer affected by the hard wall and we recover the bulk regime [11]. This regime is not included here as it is standard by now.) The semi-hard edge regime was recently discovered in [11] in the study of a problem of counting statistics, but this regime has not yet been explored at the level of the correlation kernel. This regime is genuinely different from the bulk regime and from the hard edge regime.

For both the hard and semi-hard edge regimes, we study the asymptotics of $K_{n}(z,w)$ in the microscopic (i.e. when $z \approx w$) and the macroscopic (i.e. when $|z - w| \asymp 1$) cases. We find that the asymptotics of $K_{n}(z,w)$ oscillates in n in the hard edge regime, but not in the semi-hard edge regime. In the microscopic (or “diagonal”) case, the oscillations are described in terms of the Jacobi $\theta$ function, while in the macroscopic (or “off-diagonal”) case they are described in terms of weighted Szegő kernels. Our main results are stated in Sect. 3 and can be summarized as follows (see also Fig. 2):

• 1. Theorem 3.1 establishes the asymptotics of $K_{n}(z,w)$, up to and including the fourth term of order $\sqrt{n}$, when $|z-w|=\mathcal{O}(\frac{1}{n})$ with z, w being both at a distance $\mathcal{O}(\frac{1}{n})$ from $\partial G$,

• 2. Theorem 3.5 establishes the asymptotics of $K_{n}(z,w)$, up to and including the second term of order $\sqrt{n}$, when $|z-w| =\mathcal{O}(\frac{1}{\sqrt{n}})$ with z, w being both at a distance $\asymp \frac{1}{\sqrt{n}}$ from $\partial G$,

• 3. Theorem 3.11 establishes the leading order asymptotics of $K_{n}(z,w)$ when z, w are on different sides of G and are both at a distance $\mathcal{O}(\frac{1}{n})$ from $\partial G$,

• 4. Theorem 3.13 establishes the leading order asymptotics of $K_{n}(z,w)$ when $|z-w| \asymp 1$ with z, w being on the same side of G and being both at a distance $\mathcal{O}(\frac{1}{n})$ from $\partial G$,

• 5. Theorem 3.15 establishes an upper bound on $K_{n}(z,w)$ when $|z-w| \asymp 1$ with z, w being on the same side of G and being both at a distance $\asymp \frac{1}{\sqrt{n}}$ from $\partial G$.

Fig. 2

Summary of our main results on the asymptotics of $K_{n}(z,w)$

3 Main Results

Our first main result is an asymptotic formula for $K_{n}(z,w)$ in the microscopic regime $|z-w|=\mathcal{O}(n^{-1})$ when both z and w are in the hard edge regime. This asymptotic formula contains oscillations that are described in terms of the Jacobi $\theta$ function. We recall that this function is defined by

$$\begin{aligned} \theta (z;\tau ) := \sum _{\ell =-\infty }^{+\infty } e^{2\pi i \ell z}e^{\pi i \ell ^{2}\tau }, \qquad z \in \mathbb {C}, \qquad \tau \in i(0,+\infty ), \end{aligned}$$

(3.1)

and satisfies $\theta (-z;\tau )=\theta (z;\tau )$ and $\theta (z+1;\tau )=\theta (z;\tau )$. Other properties of $\theta$ can be found in e.g. [57, Chap. 20]. The statement of Theorem 3.1 also involves Euler’s gamma constant $\gamma _{\textrm{E}}\approx 0.5772$, as well as the exponential integral $E_{1}$ and the complementary error function ${\textrm{erfc}}$ (see e.g. [57, Eqs. 6.2.1 and 7.2.2]), which we recall are defined by

$$\begin{aligned} E_{1}(x) := \int _{x}^{+\infty } \frac{e^{-t}}{t}dt, \qquad {{\textrm{erfc}}} (z) = \frac{2}{\sqrt{\pi }}\int _{z}^{\infty } e^{-t^{2}}dt. \end{aligned}$$

(3.2)

Theorem 3.1

(“$r_{1}$ hard edge case”) Let $t_{1},t_{2}\ge 0$ and $\beta \in \mathbb {R}$ be fixed, and define

$$\begin{aligned} z = r_{1}\Big (1-\frac{t_{1}}{\sigma _{1}n}\Big )e^{i\beta }, \qquad w = r_{1}\Big (1-\frac{t_{2}}{\sigma _{1}n}\Big )e^{i\beta }. \end{aligned}$$

(3.3)

As $n \rightarrow + \infty$,

$$\begin{aligned}&K_{n}(z,w) = C_{1} n^{2} + C_{2} \, n \log n + \Big (C_{3}+\frac{\sigma _{1}}{r_{1}^{2}}e^{-t_{1}-t_{2}}\mathcal {F}_{n}\Big ) n + C_{4} \sqrt{n} + \mathcal{O}(n^{\frac{2}{5}}), \end{aligned}$$

(3.4)

where $\mathcal {F}_{n}$ is given by

$$\begin{aligned}&\mathcal {F}_{n} = \frac{(\log \theta )'\Big (n \sigma _{\star }+\frac{1}{2}-\alpha +\frac{\log (\sigma _{2}/\sigma _{1})}{2\log (r_{2}/r_{1})}; \frac{\pi i}{\log (r_{2}/r_{1})}\Big ) + \log (\sigma _{2}/\sigma _{1})}{2\log (r_{2}/r_{1})}, \end{aligned}$$

(3.5)

and $\sigma _{1},\sigma _{2}$ and $\sigma _{\star }$ are defined in (2.4). If $(t_{1},t_{2}) \ne (0,0)$, the constants $C_{1},\ldots ,C_{4}$ (constants with respect to n) are given by

$$\begin{aligned} C_{1}&= \sigma _{1}^{2}\frac{1-e^{-t_{1}-t_{2}}( 1+t_{1}+t_{2}) }{r_{1}^{2}(t_{1}+t_{2})^{2}}, \nonumber \\ C_{2}&= \frac{b^{2}r_{1}^{2b-2}}{2}, \nonumber \\ C_{3}&= - b^{2} r_{1}^{2b-2}\bigg ( E_{1}( t_{1}+t_{2} ) + \gamma _{\textrm{E}} + \log \bigg (\frac{br_{1}^{2b}(t_{1}+t_{2})}{\sigma _{1}\sqrt{2\pi }}\bigg ) \bigg ) \nonumber \\&\quad + \frac{\sigma _{1}}{ r_{1}^{2}(t_{1}+t_{2})^{3}} \bigg \{ \frac{(t_{1}+t_{2})^{2}(t_{1}^{2}+t_{2}^{2})}{2e^{t_{1}+t_{2}}} + \Big ( 2t_{1}t_{2} - \frac{b^{2}r_{1}^{2b}}{\sigma _{1}}(t_{1}+t_{2})(t_{1}^{2}+t_{2}^{2})\Big ) \bigg ( 1 - \frac{1+t_{1}+t_{2}}{e^{t_{1}+t_{2}}} \bigg ) \bigg \}, \nonumber \\ C_{4}&= \sqrt{2} b^{2}r_{1}^{b-2}\big ( 1-2br_{1}^{2b}\frac{t_{1}+t_{2}}{\sigma _{1}} \big )\mathcal {I}, \end{aligned}$$

(3.6)

and $\mathcal {I}\in \mathbb {R}$ is given by

$$\begin{aligned}&\mathcal {I} = \int _{-\infty }^{+\infty } \bigg \{ \frac{y\, e^{-y^{2}}}{\sqrt{\pi }\, {\textrm{erfc}}(y)} - \chi _{(0,+\infty )}(y) \bigg [ y^{2}+\frac{1}{2} \bigg ] \bigg \}dy \approx -0.81367. \end{aligned}$$

(3.7)

If $t_{1}=t_{2}=0$, then one simply needs to let $t_{1}+t_{2}\rightarrow 0$ in the above formulas for $C_{1},\ldots ,C_{4}$; more precisely, for $t_{1}=t_{2}=0$ the constants $C_{1},\ldots ,C_{4}$ are given by

$$\begin{aligned}&C_{1} = \frac{\sigma _{1}^{2}}{2 r_{1}^{2}}, \qquad C_{2} = \frac{b^{2}r_{1}^{2b-2}}{2}, \qquad C_{3} = \frac{b^{2}r_{1}^{2b-2}}{2}\log \Big ( \frac{2\pi \sigma _{1}^{2}}{b^{2}r_{1}^{2b}} \Big ), \qquad C_{4} = \sqrt{2} b^{2}r_{1}^{b-2} \mathcal {I}. \end{aligned}$$

Remark 3.2

If either $t_{1}<0$ or $t_{2}<0$, then either $z\in G$ or $w\in G$ and $K_{n}(z,w)$ is trivially 0 by Eqs. (2.1) and (2.5); this is why we restricted ourselves to $t_{1},t_{2}\ge 0$ in the statement of Theorem 3.1. Taking $t_{1}=t_{2}$ in Theorem 3.1 yields precise asymptotics for the one-point function $R_{n,1}\big (r_{1}e^{i\beta }(1-\frac{t_{1}}{\sigma _{1}n})\big )$ in the hard edge regime. For $t_{1}=t_{2}$, $C_{1}$ in Theorem 3.1 also appeared in [67, Eq. (21) with $L=1$] in the context of truncated unitary matrices, and for general $t_{1},t_{2}$ the quantity $C_{1}$ also appeared when considering perturbations of rank L to Hermitian matrices, see e.g. [23, Eq. (2.64)], [45, Sects. 2.2 and 3.1] and the references therein. (For truncated unitary matrices, the situation is somewhat simpler because there is no bulk and the droplet is connected. Therefore, the asymptotics for the correlation kernel are expected to be of a simpler form than Eq. (3.4), i.e. without oscillations, and without terms of order $n\log n$ – see also the discussion [13, end of Sect. 1].)

For comparison, in the soft edge setting, the asymptotics of the one-point function near a spectral gap take the following form: for $z=e^{i\beta }(r_{1} + \frac{t}{\sqrt{n\Delta Q(r_{1})}})$, we have

$$\begin{aligned} R_{n,1}(z)=\tilde{C}_{1}n + \sqrt{n}(\tilde{C}_{2} + e^{-2t^{2}}\tilde{\mathcal {F}}_{n}) + \mathcal{O}((\log n)^{4}), \qquad \hbox {as } n \rightarrow + \infty , \end{aligned}$$

where $\tilde{C}_{1},\tilde{C}_{2}$ are independent of n and $\tilde{\mathcal {F}}_{n}$ is independent of t and of order 1, see [10, Theorem 1.9].

Fig. 3

Numerical confirmations of Theorems 3.1 (left) and 3.5 (right). For both pictures, $b=1.3$, $\alpha =1.26$, $r_{1}=0.42\smash {b^{-\frac{1}{2b}}}$, $r_{2}=0.67\smash {b^{-\frac{1}{2b}}}$. Left: the function $n\mapsto \frac{1}{\log n}\big [K_{n}(z,w) - \big (C_{1} n^{2} + C_{2} \, n \log n + \big (C_{3}+\frac{\sigma _{1}}{r_{1}^{2}}e^{-t_{1}-t_{2}}\mathcal {F}_{n}\big ) n + C_{4} \sqrt{n}\big )\big ]$, with $z = r_{1}(1-\frac{t_{1}}{\sigma _{1}n})$, $w = r_{1}(1-\frac{t_{2}}{\sigma _{1}n})$, $t_{1}=0.21$, $t_{2}=0.45$, and $C_{1},C_{2},C_{3},C_{4}$ as in Eq. (3.6). This function seems to approach a constant as $n\rightarrow +\infty$, which is consistent with Theorem 3.1, and also suggests that the term $\mathcal{O}(n^{\frac{2}{5}})$ in Eq. (3.4) is actually $\mathcal{O}(\log n)$. Right: the function $n\mapsto K_{n}(z,w) - \big (C_{1} n + C_{2} \sqrt{n}\big )$, with $z = r_{1}\big (1-\frac{\mathfrak {s}_{1}}{br_{1}^{b}\sqrt{2n}}\big )$, $w = r_{1}\big (1-\frac{\mathfrak {s}_{2}}{br_{1}^{b}\sqrt{2n}}\big )$, $\mathfrak {s}_{1}=1.21$, $\mathfrak {s}_{2}=1.45$, and $C_{1},C_{2}$ as in Eq. (3.10). This function seems to approach a constant as $n\rightarrow +\infty$, which is consistent with Theorem 3.5

Remark 3.3

In this paper we focus on the case where the hard wall G lies entirely in the interior of $\{z:|z|\le b^{-\frac{1}{2b}}\}$. We do not cover the simpler case where the hard wall is of the form $\tilde{G}=\{z:|z| > r_{1}\}$; in this case we expect that a similar formula as Eq. (3.4) holds but with $\mathcal {F}_{n}=0$ (there should be no oscillations since the droplet is then given by $\{z:|z|\le r_{1}\}$, which is a connected set).

This expectation is supported by the works [61]. Consider a radially symmetric potential Q whose associated equilibrium measure is $\mu (d^{2}z)=2\Delta Q(z) \, \chi _{[\rho _{1},r_{1}]}(r) r\, dr \frac{d\theta }{2\pi } + \sigma _{1} \delta _{r_{1}}(r)dr \frac{d\theta }{2\pi }$ ($z=re^{i\theta }$), supported on a connected set of the form $\tilde{S}=\{z:|z|\in [\rho _{1},r_{1}] \}$ with $r_{1}>\rho _{1}\ge 0$. Let $p\in \partial \tilde{U} = \{z:|z|=r_{1}\}$. Under general conditions on Q, it is proved in [61, Theorem 2.1] that, as $n\rightarrow + \infty$,

$$\begin{aligned} K_{n}\Big ( r_{1}e^{i\beta } \Big [1 - \frac{t_{1}}{n \sigma _{1}}\Big ], r_{1}e^{i\beta }\Big [1 - \frac{t_{2}}{n \sigma _{1}} \Big ] \Big )&= \frac{n^{2}\sigma _{1}^{2}}{r_{1}^{2}} \int _{0}^{1} t e^{-x(t_{1}+t_{2})}dx + o(n^{2}) {= C_{1}n^{2}} + o(n^{2}). \end{aligned}$$

Comparing the above with Eq. (3.4), we conclude that the leading order behavior of $K_{n}\big ( r_{1}e^{i\beta } (1 - \frac{t_{1}}{\sigma _{1}n})$, $r_{1}e^{i\beta } (1 - \frac{t_{2}}{\sigma _{1}n} ) \big )$ in the two situations $G=\{z:|z|\in (r_{1},r_{2})\}$ (considered here) and $\tilde{G}=\{z:|z|\in (r_{1},+\infty )\}$ (covered in [61]) are identical.

Remark 3.4

For two dimensional point processes in the multi-component regime (i.e. when S consists of several disjoint components), the Jacobi $\theta$ function is known to also describe large gap fluctuations [30], smooth linear statistics and microscopic correlations for ensembles with soft edges [10], and disk counting statistics for ensembles with hard edges [12]. It is still an open problem to establish the emergence of the theta function for two-dimensional point processes that are not rotation-invariant; such ensembles include the so-called lemniscate ensemble (see e.g. [27]) and the elliptic Ginibre point process with a large point charge [21].

Our next theorem on the asymptotics of $K_{n}(z,w)$ concerns the semi-hard edge regime in the microscopic regime $|z-w|=\mathcal{O}(n^{-1/2})$. Let $\tilde{\Delta } Q(r_{1}):= \lim _{r\rightarrow r_{1}, r<r_{1}}\Delta Q(r) = b^{2}r_{1}^{2b-2}$.

Theorem 3.5

(“$r_{1}$ semi-hard edge case”) Let $\mathfrak {s}_{1}, \mathfrak {s}_{2} > 0$, $\beta \in \mathbb {R}$ be fixed parameters, and define

$$\begin{aligned}&z = r_{1}\Big (1-\frac{\mathfrak {s}_{1}}{br_{1}^{b}\sqrt{2n}}\Big )e^{i\beta },{} & {} w = r_{1}\Big (1-\frac{\mathfrak {s}_{2}}{br_{1}^{b}\sqrt{2n}}\Big )e^{i\beta } = \bigg (r_{1} - \frac{\mathfrak {s}_{2}}{(2 \tilde{\Delta } Q(r_{1}))^{\frac{1}{2}}\sqrt{ n}}\bigg )e^{i\beta }. \end{aligned}$$

(3.8)

As $n \rightarrow + \infty$,

$$\begin{aligned}&K_{n}(z,w) = C_{1} n + C_{2} \sqrt{n} + \mathcal{O}(1), \end{aligned}$$

(3.9)

where

$$\begin{aligned} C_{1}&= 2\tilde{\Delta } Q(r_{1}) \int _{-\infty }^{+\infty } \frac{e^{-\frac{(y+\mathfrak {s}_{1})^{2} + (y+\mathfrak {s}_{2})^{2}}{2}}}{\sqrt{\pi } {\textrm{erfc}}(y)}dy , \nonumber \\ C_{2}&= b\frac{(2 \tilde{\Delta } Q(r_{1}))^{\frac{1}{2}} }{r_{1}}\int _{-\infty }^{+\infty } \frac{e^{-\frac{(y+\mathfrak {s}_{1})^{2} + (y+\mathfrak {s}_{2})^{2}}{2}}}{\sqrt{\pi } {\textrm{erfc}}(y)} \bigg \{ \frac{(10y^{2}-2)e^{-y^{2}}}{3\sqrt{\pi }{\textrm{erfc}}(y)} + 5y - \frac{10y^{3}}{3} + \frac{\mathfrak {s}_{1}+\mathfrak {s}_{2}}{b} \nonumber \\&- y \frac{\mathfrak {s}_{1}^{2}+\mathfrak {s}_{2}^{2}}{2b}- 2y^{2}(\mathfrak {s}_{1}+\mathfrak {s}_{2}) + \frac{2b-3}{b} \frac{\mathfrak {s}_{1}^{3}+\mathfrak {s}_{2}^{3}}{6} \bigg \}dy. \end{aligned}$$

(3.10)

Fig. 4

The semi-hard edge density profile $\rho (x)$

Remark 3.6

In the statement of Theorem 3.5, note that $\mathfrak {s}_{1},\mathfrak {s}_{2}>0$. The case $\mathfrak {s}_{1}=\mathfrak {s}_{2}=0$ is very different and is covered as a special case of Theorem 3.1.

Remark 3.7

The constant $C_{1}$ in Eq. (3.9) is universal for general potentials in the semi-hard regime: if things are defined appropriately, the associated “microscopic semi-hard correlation kernel”

$$\tilde{K}_n(\mathfrak {s}_{1},\mathfrak {s}_{2}):=\frac{1}{2n\tilde{\Delta } Q(r_1)} K_n\Big (r_1-\tfrac{\mathfrak {s}_{1}}{\sqrt{2n\tilde{\Delta } Q(r_1)}},r_1-\tfrac{\mathfrak {s}_{2}}{\sqrt{2n\tilde{\Delta } Q(r_1)}}\Big )$$

satisfies

$$\begin{aligned} \tilde{K}_n(\mathfrak {s}_{1},\mathfrak {s}_{2})= \int _{-\infty }^{+\infty } \frac{e^{-\frac{(y+\mathfrak {s}_{1})^{2} + (y+\mathfrak {s}_{2})^{2}}{2}}}{\sqrt{\pi } {\textrm{erfc}}(y)}dy + o(1), \quad \hbox {as } n \rightarrow + \infty . \end{aligned}$$

(3.11)

For convenience, precise conditions implying the universality Eq. (3.11) are discussed in Appendix 1.

Passing to the limit as $n\rightarrow \infty$ in Eq. (3.11) with $\mathfrak {s}_{2} = \mathfrak {s}_{1}=x$, we obtain the microscopic semi-hard edge one-point density profile

$$\begin{aligned} \rho (x):=\int _{-\infty }^{+\infty } \frac{e^{-(y+x)^{2}}}{\sqrt{\pi } {{\textrm{erfc}}}(y)}\,dy,\qquad (x>0), \end{aligned}$$

(3.12)

which is believed to be universal at any kinds hard walls in the semi-hard regime, see Appendix 1. The density profile $\rho (x)$ is depicted in Fig. 4.

Remark 3.8

In the paper [58], the authors consider two independent plasmas whose droplets are annuli which are adjacent to each other, and where the common boundary between the plasmas is an impermeable membrane, i.e. it is a soft/hard edge for each of the plasmas.

Interestingly, a density profile for the $\mathcal{O}(1/\sqrt{n})$-interface about the edge is computed in [58, Eq. (3.4)]. The profile bears a formal resemblance to Eq. (3.12), and also to the soft/hard edge plasma function studied in [15, 16, 47].

Our next three theorems on the asymptotics of $K_{n}(z,w)$ treat some macroscopic regimes when $|z-w|\asymp 1$.

For comparison purposes, we first briefly recall the results from [10, 14] about asymptotics for $K_{n}(z,w)$ in some macroscopic regimes for ensembles with soft edges. When z, w are far from each other but both close to $\partial U$, the asymptotics of $K_{n}(z,w)$ involves the (unweighted) Szegő kernel $S^U_{\textrm{soft}}(z,w)$, which is the reproducing kernel for the Hardy space $H^2_0(U)$ of all holomorphic functions on U vanishing at infinity, supplied with the norm $\Vert f\Vert _{H^2_{0}(U)}^2:= \int _{\partial U}|f|^2\,|dz|$, see [14]. On the other hand, when z, w are far from each other and both close to a (soft) spectral gap of the form $G=\{z:|z|\in (r_{1},r_{2})\}$, there is no convergence towards a limiting kernel [10]: for example, if z and w are close to $\partial G$ but lie on different regions of $\mathbb {C}\setminus G$ (i.e. $|z|-r_{1} \asymp n^{-1/2}$ and $|w|-r_{2} \asymp n^{-1/2}$, or vice-versa), the asymptotics are described by the oscillatory Szegő kernel

$$\begin{aligned} \mathcal {S}^G_{\textrm{soft}}(z,w;n):=\frac{1}{2\pi }\sum _{\ell =-\infty }^{+\infty }\frac{(z\bar{w})^\ell }{\frac{{r_1}^{2\ell +1-2x}}{\sqrt{\Delta Q(r_1)}}+\frac{r_2^{2\ell +1-2x}}{\sqrt{\Delta Q(r_2)}}}, \end{aligned}$$

(3.13)

where $x=x(n):= n \int _{|z|\le r_{1}}\mu (d^{2}z)-\lfloor n \int _{|z|\le r_{1}}\mu (d^{2}z) \rfloor \in [0,1)$ and $\mu$ is the equilibrium measure associated with Q. $\mathcal {S}^G_{\textrm{soft}}(z,w;n)$ is an analytic function of $z,\bar{w}\in G$ which is also well-defined when $|z|-r_{1} \asymp n^{-1/2}$ and $|w|-r_{2} \asymp n^{-1/2}$ (or vice-versa). This function is also the reproducing kernel for the weighted Hardy space $H^2(G;n)$ consisting of all analytic functions on G such that

$$\begin{aligned} \Vert f\Vert _{H^2(G;n)}^2:=\int _{\partial G}|f(z)|^2|z|^{-2x(n)}(\Delta Q(z))^{-\frac{1}{2}}\,|dz|<\infty . \end{aligned}$$

(3.14)

The situation when z, w are far from each other but are on the same side of G gets more complicated because the series Eq. (3.13) is divergent if $z,w \in \partial G$ with $|z|=|w|$. For $z=r_{1}e^{i\theta _{1}}$ and $w=r_{1}e^{i\theta _{2}}$, the weighted Szegő kernel appearing in the asymptotics of $K_{n}(z,w)$ turns out to be

$$\begin{aligned} \mathcal {S}^G_{\textrm{soft}}(r_{1}e^{i\theta _{1}},r_{1}e^{i\theta _{2}};n):= \frac{1}{2\pi } \frac{\sqrt{\Delta Q(r_1)}}{r_1^{1-2x}} \bigg ( \frac{1}{e^{i( \theta _1-\theta _2)}-1} \\ + \sum \limits _{\ell =0}^{\infty } \frac{e^{i(\theta _1-\theta _2)\ell }}{1 + \sqrt{\frac{\Delta Q(r_{1})}{\Delta Q(r_{2})}} (\frac{r_{2}}{r_{1}})^{2\ell -2x+1}} -\sum \limits _{\ell =-\infty }^{-1} \frac{ e^{i(\theta _1-\theta _2)\ell } }{1+\sqrt{\frac{\Delta Q(r_{2})}{\Delta Q(r_{1})}} (\frac{r_{1}}{r_{2}})^{2\ell -2x+1}}\bigg ). \end{aligned}$$

(3.15)

As noticed in Ref. [10], the series in Eq. (3.15) can be recognized as the Abel limit of Eq. (3.13); more precisely, $\mathcal {S}^G_{\textrm{soft}}(r_{1}e^{i\theta _{1}},r_{1}e^{i\theta _{2}};n)$ in Eq. (3.15) is equal to $\lim _{r \searrow r_{1}} \mathcal {S}^G_{\textrm{soft}}(re^{\theta _{1}},re^{\theta _{2}};n)$, where $\mathcal {S}^G_{\textrm{soft}}(re^{\theta _{1}},re^{\theta _{2}};n)$ is as in Eq. (3.13) with $z=re^{i\theta _{1}}$ and $w=re^{i\theta _{2}}$.

It is apriori not clear at all whether analogues to the above results from [10, 14] exist near a hard edge. For example, $\Delta Q(r_{1})$ appears in Eqs. (3.13)–(3.15), but in our setting $\{|z|=r_{1}\}$ is a hard edge and $\Delta Q(r_{1})$ does not even make sense. It therefore came as a surprise to us that our findings completely mimic the results of [10]: the main difference is that the quantities $\sqrt{\Delta Q(r_{1})}$ and $\sqrt{\Delta Q(r_{2})}$ appearing in Eq. (3.13)–(3.15) should here be replaced by $\sigma _{1}/r_{1}$ and $\sigma _{2}/r_{2}$, respectively. More precisely, we have the following definition.

Definition 3.9

The oscillatory Szegő kernel $\mathcal {S}^{G}_{\textrm{hard}}(z,w;n)$ associated with $G = \{r_{1}< |z| < r_{2}\}$ and n is defined for $z, w \in G$ by

$$\begin{aligned} \mathcal {S}^{G}_{\textrm{hard}}(z,w;n) := \frac{1}{2\pi } \sum _{\ell =-\infty }^{+\infty } \frac{(z\bar{w})^{\ell }}{\frac{r_{1}^{2(\ell +1-x)}}{\sigma _{1}} + \frac{r_{2}^{2(\ell +1-x)}}{\sigma _{2}} }. \end{aligned}$$

(3.16)

This is an analytic function of $z, \bar{w} \in G$. This function is also well-defined as long as $|z\overline{w}|$ lies in a compact subset of $(r_{1}^{2},r_{2}^{2})$; for example, it is well-defined for $|z|=r_{1}$ and $|w|=r_{2}$ (and vice-versa, it is also well-defined for $|z|=r_{2}$ and $|w|=r_{1}$). For $|z|=|w|=r_{1}$, $z\ne w$, the above series does not converge absolutely. Nevertheless, for $\theta _{1} \ne \theta _{2}$, the Abel limit $\lim _{r \searrow 1} \mathcal {S}^{G}_{\textrm{hard}}(re^{\theta _{1}},re^{\theta _{2}};n)$ exists, we denote it by $\mathcal {S}^G_{\textrm{hard}}(r_{1}e^{i\theta _{1}},r_{1}e^{i\theta _{2}};n)$, and is equal to

$$\begin{aligned} \mathcal {S}^G_{\textrm{hard}}(r_{1}e^{i\theta _{1}},r_{1}e^{i\theta _{2}};n) = \\ \frac{1}{2\pi } \frac{\sigma _{1}}{r_{1}^{2(1-x)}} \bigg ( \frac{1}{e^{i(\theta _{1}-\theta _{2})}-1} + \sum _{\ell =0}^{+\infty } \frac{e^{i(\theta _{1}-\theta _{2})\ell }}{1+\frac{\sigma _{1}}{\sigma _{2}}(\frac{r_{2}}{r_{1}})^{2(\ell +1-x)}} - \sum _{\ell =-\infty }^{-1} \frac{e^{i(\theta _{1}-\theta _{2})\ell }}{1+\frac{\sigma _{2}}{\sigma _{1}}(\frac{r_{1}}{r_{2}})^{2(\ell +1-x)}} \bigg ). \end{aligned}$$

(3.17)

Because one had to take the Abel limit to make sense of $\mathcal {S}^G_{\textrm{hard}}(r_{1}e^{i\theta _{1}},r_{1}e^{i\theta _{2}};n)$, we call $\mathcal {S}^G_{\textrm{hard}}(r_{1}e^{i\theta _{1}}$, $r_{1}e^{i\theta _{2}};n)$ an oscillatory “regularized" Szegő kernel, see also Fig. 2.

Remark 3.10

The kernel Eq. (3.16) is the reproducing kernel for the weighted Hardy space $H^2(G;n)$ of all analytic functions g on G satisfying

$$\begin{aligned} \Vert f\Vert _{H^2(G;n)}^2:=\int _{\{|z|=r_{1}\}}|f(z)|^2 \frac{r_{1}^{1-2x(n)}}{\sigma _{1}}|dz| + \int _{\{|z|=r_{2}\}}|f(z)|^2 \frac{r_{2}^{1-2x(n)}}{\sigma _{2}}\,|dz|<\infty . \end{aligned}$$

(3.18)

We next state our results on the asymptotics of $K_{n}(z,w)$ in the macroscopic regime when z and w are in different regions of $\mathbb {C}\setminus G$.

Theorem 3.11

(“$r_{1}$-$r_{2}$ hard edge case”) Let $t_{1},t_{2}\ge 0$ and $\theta _{1},\theta _{2}\in \mathbb {R}$ be fixed, and define

$$\begin{aligned} z = r_{1}\Big (1-\frac{t_{1}}{\sigma _{1}n}\Big )e^{i\theta _{1}}, \qquad w = r_{2}\Big (1+\frac{t_{2}}{\sigma _{2}n}\Big )e^{i\theta _{2}}, \qquad t_{1}, t_{2} \ge 0, \;\; \theta _{1},\theta _{2}\in \mathbb {R}. \end{aligned}$$

(3.19)

As $n \rightarrow + \infty$, we have

$$\begin{aligned}&K_{n}(z,w) = 2 \pi n \cdot \mathcal {S}^{G}_{\textrm{hard}}(z,w;n) \cdot e^{i \lfloor j_{\star }\rfloor (\theta _{1}-\theta _{2})} (r_{1}r_{2})^{-x} e^{-t_{1}-t_{2}} + \mathcal{O}((\log n)^{2}), \end{aligned}$$

where $j_{\star }:= n \sigma _{\star }-\alpha$, and $x=x(n)$ is given by

$$\begin{aligned} x = j_{\star }-\lfloor j_{\star } \rfloor = n \sigma _{\star }-\alpha - \lfloor n \sigma _{\star }-\alpha \rfloor \in [0,1). \end{aligned}$$

(3.20)

Remark 3.12

The analogue of Theorem 3.11 near soft edges is given by [10, Corollary 1.14] and is as follows: for $z=\big (r_1 + \frac{t}{\sqrt{n\Delta Q(r_1)}}\big )e^{i\theta _1}$, $w=\big (r_2 + \frac{s}{\sqrt{n\Delta Q(r_2)}}\big )e^{i\theta _2}$ with $s,t\in \mathbb {R}$, we have

$$\begin{aligned} K_{n}(z,w) = \sqrt{2\pi n} \cdot \mathcal {S}_{\textrm{soft}}^G(z,w;n)\cdot e^{i\lfloor j_{\star }\rfloor (\theta _1-\theta _2)} (r_1r_2)^{-x} e^{-t^2-s^2} +\mathcal{O}((\log n)^{3}), \end{aligned}$$

where $\mathcal {S}_{\textrm{soft}}^G(z,w;n)$ is given by Eq. (3.13), $j_{\star }:=n \int _{|z|\le r_{1}}\mu (d^{2}z)$ and x and $\mu$ are as in Eq. (3.13).

Fig. 5

Numerical confirmations of Theorems 3.11 (left) and 3.13 (right). For both pictures, $b=1.3$, $\alpha =1.26$, $r_{1}=0.42\smash {b^{-\frac{1}{2b}}}$, $r_{2}=0.67\smash {b^{-\frac{1}{2b}}}$. Left: the function $n\mapsto \big |\big [K_{n}(z,w) - 2 \pi n \cdot \mathcal {S}^{G}_{\textrm{hard}}(z,w;n) \cdot e^{i \lfloor j_{\star }\rfloor (\theta _{1}-\theta _{2})} (r_{1}r_{2})^{-x} e^{-t_{1}-t_{2}}\big |$, with $z = r_{1}(1-\frac{t_{1}}{\sigma _{1}n})e^{i\theta _{1}}$, $w = r_{2}(1+\frac{t_{2}}{\sigma _{2}n})e^{i\theta _{2}}$, $t_{1}=0.21$, $t_{2}=0.45$, $\theta _{1}=0$, $\theta _{2}=0.312$. This function seems to grow very slowly as $n\rightarrow +\infty$, which is consistent with Theorem 3.11. Right: the function $n\mapsto \frac{1}{\sqrt{n}} \big |K_{n}(z,w) - 2\pi n \cdot \mathcal {S}^{G}_{\textrm{hard}}(r_{1}e^{\theta _{1}},r_{1}e^{\theta _{2}};n) \cdot e^{i \lfloor j_{\star } \rfloor (\theta _{1}-\theta _{2})} r_{1}^{-2x} e^{-t_{1}-t_{2}} \big |$, with $z = r_{1}(1-\frac{t_{1}}{\sigma _{1}n})e^{i\theta _{1}}$, $w = r_{1}(1-\frac{t_{2}}{\sigma _{1}n})e^{i\theta _{2}}$, $t_{1}=0.91$, $t_{2}=1.45$, $\theta _{1}=0$, $\theta _{2}=0.312$. This function seems to approach a constant as $n\rightarrow +\infty$, which is consistent with Theorem 3.13, and also suggests that the term $\mathcal{O}(\sqrt{n \log n} )$ in Eq. (3.22) is actually $\mathcal{O}(\sqrt{n} )$.

We next consider the macroscopic regime where z and w are on different sides of G.

Theorem 3.13

(“$r_{1}$-$r_{1}$ hard edge case”) Let $t_{1},t_{2}\ge 0$ and $\theta _{1}\ne \theta _{2} \in \mathbb {R}$ be fixed, and define

$$\begin{aligned} z = r_{1}\Big (1-\frac{t_{1}}{\sigma _{1}n}\Big )e^{i\theta _{1}}, \qquad w = r_{1}\Big (1-\frac{t_{2}}{\sigma _{1}n}\Big )e^{i\theta _{2}}. \end{aligned}$$

(3.21)

As $n \rightarrow \infty$, we have

$$\begin{aligned} K_{n}(z,w) = 2\pi n \cdot \mathcal {S}^{G}_{\textrm{hard}}(r_{1}e^{\theta _{1}},r_{1}e^{\theta _{2}};n) \cdot e^{i \lfloor j_{\star } \rfloor (\theta _{1}-\theta _{2})} r_{1}^{-2x} e^{-t_{1}-t_{2}} + \mathcal{O}(\sqrt{n \log n} ), \end{aligned}$$

(3.22)

where $j_{\star }:= n \sigma _{\star }-\alpha$, $x = j_{\star }-\lfloor j_{\star } \rfloor$.

Remark 3.14

The analogue of Theorem 3.13 near a soft edge is given by [10, Corollary 1.14] and is as follows: for $z=\big (r_1 + \frac{t}{\sqrt{n\Delta Q(r_1)}}\big )e^{i\theta _1}$, $w=\big (r_1 + \frac{s}{\sqrt{n\Delta Q(r_2)}}\big )e^{i\theta _2}$ with $s,t\in \mathbb {R}$, $\theta _{1}\ne \theta _{2}$, we have

$$\begin{aligned} K_{n}(z,w) = \sqrt{2\pi n} \cdot \mathcal {S}_{\textrm{soft}}^G(r_{1}e^{i\theta _{1}},r_{1}e^{i\theta _{2}};n)\cdot e^{i\lfloor j_{\star }\rfloor (\theta _1-\theta _2)} r_1^{-2x} e^{-t^2-s^2} +\mathcal{O}((\log n)^{5}), \end{aligned}$$

where $\mathcal {S}_{\textrm{soft}}^G(r_{1}e^{i\theta _{1}},r_{1}e^{i\theta _{2}};n)$ is given by Eq. (3.15), $j_{\star }=n \int _{|z|\le r_{1}}\mu (d^{2}z)$, and x and $\mu$ are as in Eq. (3.13).

Our last result shows that $K_{n}(z,w)$ is small whenever z and w are in the semi-hard edge regime and $|z-w|\asymp 1$.

Theorem 3.15

(“$r_{1}$-$r_{1}$ semi-hard edge case”) Let $\mathfrak {s}_{1}, \mathfrak {s}_{2} > 0$ and $\theta _{1}\ne \theta _{2} \in \mathbb {R}$ be fixed, and define

$$\begin{aligned}&z = r_{1}\Big (1-\frac{\mathfrak {s}_{1}}{br_{1}^{b}\sqrt{2n}}\Big )e^{i\theta _{1}}, \qquad w = r_{1}\Big (1-\frac{\mathfrak {s}_{2}}{br_{1}^{b}\sqrt{2n}}\Big )e^{i\theta _{2}}. \end{aligned}$$

As $n \rightarrow + \infty$,

$$\begin{aligned}&K_{n}(z,w) = \mathcal{O}(1). \end{aligned}$$

(3.23)

Remark 3.16

The estimate Eq. (3.23) can be probably be strengthened with more efforts. In fact, in the setting of Theorem 3.15, we believe that $K_{n}(z,w) = \mathcal{O}(n^{-K})$ holds as $n \rightarrow \infty$ for any fixed $K>0$.

Outline of the paper. In Sect. 4, we obtain large n estimates for $h_{j}$ valid uniformly in different ranges of $j\in \{1,\ldots ,n\}$. These estimates will be useful for the proofs of each of our five theorems. The proof of Theorem 3.11 is given in Sect. 5, the proofs of Theorems 3.1 and 3.13 are given in Sect. 6, and the proofs of Theorems 3.5 and 3.15 are given in Sect. 7.

4 Asymptotics of $h_{j}$

Recall that $h_{j}$ is defined in Eq. (2.6). Following [29, 30], we introduce the following quantities: for $j=1,\ldots ,n$ and $\ell =1,2$, we define

$$\begin{aligned} a_{j} := \frac{j+\alpha }{b}, \qquad \lambda _{j,\ell } := \frac{bnr_{\ell }^{2b}}{j+\alpha }, \qquad \eta _{j,\ell } := (\lambda _{j,\ell }-1)\sqrt{\frac{2 (\lambda _{j,\ell }-1-\log \lambda _{j,\ell })}{(\lambda _{j,\ell }-1)^{2}}}. \end{aligned}$$

(4.1)

Let $\epsilon > 0$ be a small constant independent of n. Define

$$\begin{aligned}&j_{\ell ,-}:=\lceil \tfrac{bnr_{\ell }^{2b}}{1+\epsilon } - \alpha \rceil ,{} & {} j_{\ell ,+} := \lfloor \tfrac{bnr_{\ell }^{2b}}{1-\epsilon } - \alpha \rfloor ,{} & {} \ell =1,2, \nonumber \\&j_{0,-}:=1,{} & {} j_{0,+}:=M',{} & {} j_{3,-}:=n+1, \end{aligned}$$

(4.2)

where $\lceil x \rceil$ denotes the smallest integer $\ge x$, $\lfloor x \rfloor$ denotes the largest integer $\le x$, and $M'$ is a large but fixed constant. We take $\epsilon$ sufficiently small such that

$$\begin{aligned} \frac{br_{1}^{2b}}{1-\epsilon }< \frac{br_{2}^{2b}}{1+\epsilon }, \qquad \hbox {and} \qquad \frac{br_{2}^{2b}}{1-\epsilon } < 1. \end{aligned}$$

(4.3)

The quantity $\sigma _{\star }$, which we recall is defined in Eq. (2.4), will appear naturally in our analysis. It is easy to check that $\sigma _{\star } \in (br_{1}^{2b},br_{2}^{2b})$. For technical reasons, we also assume $\epsilon >0$ is small enough so that

$$\begin{aligned} \frac{br_{1}^{2b}}{1-\epsilon }< \sigma _{\star } < \frac{br_{2}^{2b}}{1+\epsilon }. \end{aligned}$$

(4.4)

Recall also that $j_{\star }:= n \sigma _{\star } -\alpha$, and note that

$$\begin{aligned} \frac{a_{j}(\eta _{j,2}^{2}-\eta _{j,1}^{2})}{2} = 2(j_{\star }-j) \log \frac{r_{2}}{r_{1}}. \end{aligned}$$

(4.5)

Let M be such that $M'\sqrt{\log n}\le M \le n^{\frac{1}{10}}$, and define

$$\begin{aligned} g_{\ell ,-} := \bigg \lceil \frac{bnr_{\ell }^{2b}}{1+\frac{M}{\sqrt{n}}}-\alpha \bigg \rceil , \qquad g_{\ell ,+} := \bigg \lfloor \frac{bnr_{\ell }^{2b}}{1-\frac{M}{\sqrt{n}}}-\alpha \bigg \rfloor , \qquad \ell =1,2, \end{aligned}$$

and

$$\begin{aligned} M_{j,k} := \sqrt{n}(\lambda _{j,k}-1), \qquad \hbox {for all } k \in \{1,2\} \hbox { and } j \in \{g_{k,-},\ldots ,g_{k,+}\}. \end{aligned}$$

(4.6)

In this section, M can be arbitrary within the range $M \in [M'\sqrt{\log n}, n^{\frac{1}{10}}]$, but we already mention that in Sects. 5 and 6.2 we will take $M=M'\sqrt{\log n}$, in Sect. 6.1 we will take $M=n^{\frac{1}{10}}$, and in Sect. 7 we will take $M=M' \log n$. For $k=1,2$ and $g_{k,-}\le j \le g_{k,+}$, we also define

$$\begin{aligned} \chi _{j,k}^{-} := \chi _{(-\infty ,0)}(M_{j,k}), \qquad \chi _{j,k}^{+} := \chi _{(0,+\infty )}(M_{j,k}). \end{aligned}$$

We start with the following useful lemma.

Lemma 4.1

As $n \rightarrow + \infty$ with $M' \le j \le n$, we have

$$\begin{aligned} \frac{n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} = \frac{\sqrt{n} \, e^{\frac{n}{b}(j/n-j/n \log (\frac{j/n}{b}))}}{\sqrt{2\pi }}\bigg ( \frac{b}{j/n} \bigg )^{\frac{\alpha }{b}} \sqrt{\frac{j/n}{b}} \bigg ( 1 + \frac{-b^{2}+6b \alpha - 6 \alpha ^{2}}{12 b j/n} \frac{1}{n} + \mathcal{O}(j^{-2}) \bigg ). \end{aligned}$$

(4.7)

As $n \rightarrow + \infty$ with $g_{k,-} \le j \le g_{k,+}$, $k=1,2$, we have

$$\begin{aligned} \frac{n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})}&= e^{r_{k}^{2b}(1-\log (r_{k}^{2b}))n}e^{M_{j,k} r_{k}^{2b} \log (r_{k}^{2b})\sqrt{n}}e^{-r_{k}^{2b}(\frac{1}{2}+\log (r_{k}^{2b}))M_{j,k}^{2}} \frac{r_{k}^{b}\sqrt{n}}{\sqrt{2\pi }} \nonumber \\&\quad \times \bigg ( 1+\frac{M_{j,k}}{6}(-3+5M_{j,k}^{2}r_{k}^{2b}+6M_{j,k}^{2}r_{k}^{2b}\log (r_{k}^{2b})) \frac{1}{\sqrt{n}} \nonumber \\&\quad + \frac{1}{n}\bigg \{ r_{k}^{4b} M_{j,k}^{6} \bigg ( \frac{25}{72} + \frac{5\log (r_{k}^{2b})}{6} + \frac{(\log (r_{k}^{2b}))^{2}}{2} \bigg ) \nonumber \\&\quad - \frac{3}{2}r_{k}^{2b}M_{j,k}^{4} (1+\log (r_{k}^{2b})) \bigg \}\nonumber \\&\quad + \frac{125+450 \log (r_{k}^{2b}) + 540 \log ^{2}(r_{k}^{2b}) + 216\log ^{3}(r_{k}^{2b})}{1296}r_{k}^{6b}M_{j,k}^{9}n^{-3/2} \nonumber \\&\quad + \mathcal{O}\bigg (\frac{1+M_{j,k}^{2}}{n} + \frac{1+M_{j,k}^{7}}{n^{3/2}} + \frac{1+M_{j,k}^{12}}{n^{2}} \bigg ) \bigg ). \end{aligned}$$

(4.8)

Proof

Formula Eq. (4.7) directly follows from the well-known large j asymptotics of $\Gamma (\frac{j+\alpha }{b})$ (see e.g. [57, 5.11.1]), and Eq. (4.8) directly follows from the large j asymptotics of $\Gamma (\frac{j+\alpha }{b})$ and the fact that, by Eqs. (4.1) and (4.6), we have

$$\begin{aligned} \frac{j}{n} = -\frac{\alpha }{n} + \frac{br_{k}^{2b}}{1+\frac{M_{j,k}}{\sqrt{n}}}, \qquad g_{k,-} \le j \le g_{k,+}, \; k=1,2. \end{aligned}$$

$\square$

Lemma 4.2

$M'$ can be chosen sufficiently large such that the following asymptotics hold uniformly for $M'\sqrt{\log n}\le M \le n^{\frac{1}{10}}$.

• 1. Let j be fixed. As $n \rightarrow + \infty$, we have

  $$\begin{aligned} h_{j}^{-1} = \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \bigg ( 1 + \mathcal{O}(n^{\frac{j+\alpha }{b}-1}e^{-nr_{1}^{2b}}) \bigg ). \end{aligned}$$

  (4.9)

• 2. As $n \rightarrow + \infty$ with $M' \le j \le j_{1,-}$, we have

  $$\begin{aligned} h_{j}^{-1}&= \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \bigg ( 1 + \mathcal{O}(n^{-\frac{1}{2}}e^{-nr_{1}^{2b}\frac{\epsilon - \log (1+\epsilon )}{1+\epsilon }} \bigg ) \nonumber \\&= b \frac{\sqrt{n} \, e^{\frac{n}{b}(j/n-j/n \log (\frac{j/n}{b}))}}{\sqrt{2\pi }}\bigg ( \frac{b}{j/n} \bigg )^{\frac{\alpha }{b}} \sqrt{\frac{j/n}{b}} \bigg ( 1 + \frac{-b^{2}+6b \alpha - 6 \alpha ^{2}}{12 b j/n} \frac{1}{n} + \mathcal{O}(j^{-2}) \bigg ). \end{aligned}$$

  (4.10)

• 3. As $n \rightarrow + \infty$ with $j_{1,-} \le j \le g_{1,-}$, we have

  $$\begin{aligned} h_{j}^{-1}&= \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \bigg ( 1 + \mathcal{O}(n^{-100}) \bigg ) \nonumber \\&= b \frac{\sqrt{n} \, e^{\frac{n}{b}(j/n-j/n \log (\frac{j/n}{b}))}}{\sqrt{2\pi }}\bigg ( \frac{b}{j/n} \bigg )^{\frac{\alpha }{b}} \sqrt{\frac{j/n}{b}} \bigg ( 1 + \frac{-b^{2}+6b \alpha - 6 \alpha ^{2}}{12 b j/n} \frac{1}{n} + \mathcal{O}(n^{-2}) \bigg ). \end{aligned}$$

  (4.11)

• 4. As $n \rightarrow + \infty$ with $g_{1,-} \le j \le g_{1,+}$, we have

  $$\begin{aligned} h_{j}^{-1}&= \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \frac{1}{\frac{1}{2}{\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}}) } \bigg ( 1 + \frac{(5M_{j,1}^{2}r_{1}^{2b}-2)e^{-\frac{r_{1}^{2b}M_{j,1}^{2}}{2}}}{3\sqrt{2\pi } r_{1}^{b} {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}}) \sqrt{n}} + \frac{\chi _{j,1}^{-}}{n} \bigg \{ \frac{25r_{1}^{4b}M_{j,1}^{6}}{72} \nonumber \\&\quad + \frac{3r_{1}^{2b}M_{j,1}^{4}}{2} \bigg \} - \chi _{j,1}^{-}\frac{125r_{1}^{6b}M_{j,1}^{9}}{1296n^{3/2}} + \mathcal{O}\bigg (\frac{1+M_{j,1}^{2}}{n}+\frac{1+|M_{j,1}^{7}|}{n^{3/2}} + \frac{1+M_{j,1}^{12}}{n^{2}}\bigg ) \bigg ) \nonumber \\&= b \frac{e^{r_{1}^{2b}(1-\log (r_{1}^{2b}))n}e^{M_{j,1} r_{1}^{2b} \log (r_{1}^{2b})\sqrt{n}}e^{-r_{1}^{2b}(\frac{1}{2}+\log (r_{1}^{2b}))M_{j,1}^{2}} r_{1}^{b}\sqrt{n}}{\frac{1}{2}{\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}})\sqrt{2\pi }} \nonumber \\&\quad \times \bigg ( 1 + \frac{1}{\sqrt{n}}\bigg \{\frac{(5M_{j,1}^{2}r_{1}^{2b}-2)e^{-\frac{r_{1}^{2b}M_{j,1}^{2}}{2}}}{3\sqrt{2\pi } r_{1}^{b} {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}}) } + M_{j,1}^{3}r_{1}^{2b}\log (r_{1}^{2b}) - \frac{M_{j,1}}{2} + \frac{5M_{j,1}^{3}r_{1}^{2b}}{6} \bigg \} \nonumber \\&\quad + \frac{\frac{r_{1}^{4b}}{2} (\log (r_{1}^{2b}))^{2} M_{j,1}^{6} - 2 r_{1}^{2b} \log (r_{1}^{2b})M_{j,1}^{4}}{n} + \frac{r_{1}^{6b} (\log (r_{1}^{2b}))^{3}M_{j,1}^{9}}{6n^{3/2}} \nonumber \\&\quad + \mathcal{O}\Big (\frac{1+M_{j,1}^{2}+M_{j,1}^{6}\chi _{j,1}^{+}}{n}+\frac{1+|M_{j,1}^{7}|+M_{j,1}^{9}\chi _{j,1}^{+}}{n^{3/2}} + \frac{1+M_{j,1}^{12}}{n^{2}}\Big ) \bigg ). \end{aligned}$$

  (4.12)

• 5. As $n \rightarrow + \infty$ with $g_{1,+} \le j \le j_{1,+}$, we have

  $$\begin{aligned} h_{j}^{-1}&= \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \sqrt{n} \, e^{\frac{n}{b}(br_{1}^{2b}-j/n+j/n \log (\frac{j/n}{br_{1}^{2b}}))} \frac{\sqrt{2\pi }}{\sqrt{b \, j/n}} \bigg ( \frac{j/n}{br_{1}^{2b}} \bigg )^{\frac{\alpha }{b}} ( j/n - br_{1}^{2b} ) \nonumber \\&\quad \times \bigg ( 1 + \frac{12b^{3} r_{1}^{2b} j/n + 6b((j/n)^{2}-(b r_{1}^{2b})^{2})\alpha + (b^{2}+6\alpha ^{2}) (j/n-br_{1}^{2b})^{2}}{12 b j/n (j/n-br_{1}^{2b})^{2} n} \nonumber \\&\quad - \frac{2 b^{4} r_{1}^{4b}}{n^{2}(j/n-br_{1}^{2b})^{4}} + \frac{10b^{6}r_{1}^{6b}}{n^{3}(j/n-br_{1}^{2b})^{6}}\nonumber \\&\quad + \mathcal{O}\Big ( \frac{1}{n^{2}(j/n-br_{1}^{2b})^{3}} + \frac{1}{n^{4}(j/n-br_{1}^{2b})^{8}} \Big ) \bigg ) \nonumber \\&= \frac{n(j/n-br_{1}^{2b})}{r_{1}^{2\alpha }}e^{n(r_{1}^{2b}-2 j/n \log (r_{1}))} \bigg ( 1 + \frac{b^{2}r_{1}^{2b}+(j/n-br_{1}^{2b})\alpha }{n(j/n-br_{1}^{2b})^{2}} - \frac{2 b^{4} r_{1}^{4b}}{n^{2}(j/n-br_{1}^{2b})^{4}} \nonumber \\&+ \frac{10b^{6}r_{1}^{6b}}{n^{3}(j/n-br_{1}^{2b})^{6}} + \mathcal{O}\Big ( \frac{1}{n^{2}(j/n-br_{1}^{2b})^{3}} + \frac{1}{n^{4}(j/n-br_{1}^{2b})^{8}} \Big ) \bigg ). \end{aligned}$$

  (4.13)

• 6. As $n \rightarrow + \infty$ with $j_{1,+} \le j \le \lfloor j_{\star }\rfloor$, we have

  $$\begin{aligned}&h_{j}^{-1} = \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \sqrt{n} \, e^{\frac{n}{b}(br_{1}^{2b}-j/n+j/n \log (\frac{j/n}{br_{1}^{2b}}))} \frac{\sqrt{2\pi }}{\sqrt{b \, j/n}} \frac{\big ( \frac{j/n}{br_{1}^{2b}} \big )^{\frac{\alpha }{b}}}{\frac{1}{j/n-br_{1}^{2b}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{br_{2}^{2b}-j/n}} \nonumber \\&\hspace{1cm} \times \bigg ( 1 + \frac{12b^{3} r_{1}^{2b} j/n + 6b((j/n)^{2}-(b r_{1}^{2b})^{2})\alpha + (b^{2}+6\alpha ^{2}) (j/n-br_{1}^{2b})^{2}}{12 b j/n (j/n-br_{1}^{2b})^{2} n} \nonumber \\&\hspace{1cm} + \mathcal{O}\Big ( \frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{n} + \frac{1}{n^{2}} \Big ) \bigg ) \nonumber \\&= \frac{n}{r_{1}^{2\alpha }}\frac{e^{n(r_{1}^{2b}-2 j/n \log (r_{1}))}}{\frac{1}{j/n-br_{1}^{2b}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{br_{2}^{2b}-j/n}} \bigg ( 1 + \frac{b^{2}r_{1}^{2b}+(j/n-br_{1}^{2b})\alpha }{n(j/n-br_{1}^{2b})^{2}} + \mathcal{O}\Big ( \frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{n} + \frac{1}{n^{2}} \Big ) \bigg ). \end{aligned}$$

  (4.14)

• 7. As $n \rightarrow + \infty$ with $\lfloor j_{\star }\rfloor +1 \le j \le j_{2,-}$, we have

  $$\begin{aligned} h_{j}^{-1}&= \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \sqrt{n} \, e^{\frac{n}{b}(br_{2}^{2b}-j/n+j/n \log (\frac{j/n}{br_{2}^{2b}}))} \frac{\sqrt{2\pi }}{\sqrt{b \, j/n}} \frac{\big ( \frac{j/n}{br_{2}^{2b}} \big )^{\frac{\alpha }{b}}}{\frac{1}{br_{2}^{2b} - j/n}+\frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{j/n-br_{1}^{2b}}} \nonumber \\&\quad \times \bigg ( 1 + \frac{12b^{3} r_{2}^{2b} j/n + 6b((j/n)^{2}-(b r_{2}^{2b})^{2})\alpha + (b^{2}+6\alpha ^{2}) (j/n-br_{2}^{2b})^{2}}{12 b j/n (j/n-br_{2}^{2b})^{2} n} \nonumber \\&\quad + \mathcal{O}\Big ( \frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{n} + \frac{1}{n^{2}} \Big ) \bigg ) \nonumber \\&= \frac{n}{r_{2}^{2\alpha }} \frac{e^{n(r_{2}^{2b}-2j/n \log (r_{2}))}}{\frac{1}{br_{2}^{2b} - j/n}+\frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{j/n-br_{1}^{2b}}} \bigg ( 1 + \frac{b^{2}r_{2}^{2b}+(j/n-br_{2}^{2b})\alpha }{n(j/n-br_{2}^{2b})^{2}} + \mathcal{O}\Big ( \frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{n} + \frac{1}{n^{2}} \Big ) \bigg ). \end{aligned}$$

  (4.15)

• 8. As $n \rightarrow + \infty$ with $j_{2,-} \le j \le g_{2,-}$, we have

  $$\begin{aligned}&h_{j}^{-1} = \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \sqrt{n} \, e^{\frac{n}{b}(br_{2}^{2b}-j/n+j/n \log (\frac{j/n}{br_{2}^{2b}}))} \frac{\sqrt{2\pi }}{\sqrt{b \, j/n}} \bigg ( \frac{j/n}{br_{2}^{2b}} \bigg )^{\frac{\alpha }{b}} ( br_{2}^{2b} - j/n ) \nonumber \\&\hspace{1cm} \times \bigg ( 1 + \frac{12b^{3} r_{2}^{2b} j/n + 6b((j/n)^{2}-(b r_{2}^{2b})^{2})\alpha + (b^{2}+6\alpha ^{2}) (j/n-br_{2}^{2b})^{2}}{12 b j/n (j/n-br_{2}^{2b})^{2} n} \nonumber \\&\hspace{1cm} + \mathcal{O}\Big ( \frac{1}{n^{2}(j/n-br_{2}^{2b})^{4}} \Big ) \bigg ) \nonumber \\&= \frac{n(br_{2}^{2b}-j/n)}{r_{2}^{2\alpha }} e^{n(r_{2}^{2b}-2j/n \log (r_{2}))} \bigg ( 1 + \frac{b^{2}r_{2}^{2b}+(j/n-br_{2}^{2b})\alpha }{n(j/n-br_{2}^{2b})^{2}} + \mathcal{O}\Big ( \frac{1}{n^{2}(j/n-br_{2}^{2b})^{4}} \Big ) \bigg ). \end{aligned}$$

  (4.16)

• 9. As $n \rightarrow + \infty$ with $g_{2,-} \le j \le g_{2,+}$, we have

  $$\begin{aligned} h_{j}^{-1}&= \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \frac{1}{1-\frac{1}{2}{\textrm{erfc}}(-\frac{M_{j,2}r_{2}^{b}}{\sqrt{2}}) } \bigg ( 1 + \frac{(2-5M_{j,2}^{2}r_{2}^{2b})e^{-\frac{r_{2}^{2b}M_{j,2}^{2}}{2}}}{3\sqrt{2\pi } r_{2}^{b} (2-{\textrm{erfc}}(-\frac{M_{j,2}r_{2}^{b}}{\sqrt{2}})) \sqrt{n}} \nonumber \\&\quad + \frac{\chi _{j,2}^{+}}{n} \bigg \{ \frac{25r_{2}^{4b}M_{j,2}^{6}}{72}+ \frac{3r_{2}^{2b}M_{j,2}^{4}}{2} \bigg \} - \chi _{j,2}^{+}\frac{125r_{2}^{6b}M_{j,2}^{9}}{1296n^{3/2}}\nonumber \\&\quad + \mathcal{O}\bigg (\frac{1+M_{j,2}^{2}}{n} + \frac{1+|M_{j,2}^{7}|}{n^{3/2}} + \frac{1+M_{j,2}^{12}}{n^{2}} \bigg ) \bigg ) \nonumber \\&= b \frac{e^{r_{2}^{2b}(1-\log (r_{2}^{2b}))n}e^{M_{j,2} r_{2}^{2b} \log (r_{2}^{2b})\sqrt{n}}e^{-r_{2}^{2b}(\frac{1}{2}+\log (r_{2}^{2b}))M_{j,2}^{2}} r_{2}^{b}\sqrt{n}}{(1-\frac{1}{2}{\textrm{erfc}}(-\frac{M_{j,2}r_{2}^{b}}{\sqrt{2}}))\sqrt{2\pi }} \nonumber \\&\quad + \frac{\frac{r_{2}^{4b}}{2} (\log (r_{2}^{2b}))^{2} M_{j,2}^{6} - 2 r_{2}^{2b} \log (r_{2}^{2b})M_{j,2}^{4}}{n} + \frac{r_{2}^{6b} (\log (r_{2}^{2b}))^{3}M_{j,2}^{9}}{6n^{3/2}} \nonumber \\&\quad + \mathcal{O}\Big (\frac{1+M_{j,2}^{2}+M_{j,2}^{6}\chi _{j,2}^{-}}{n}+\frac{1+|M_{j,2}^{7}|+M_{j,2}^{9}\chi _{j,2}^{-}}{n^{3/2}} + \frac{1+M_{j,2}^{12}}{n^{2}}\Big ) \bigg ). \end{aligned}$$

  (4.17)

• 10. As $n \rightarrow + \infty$ with $g_{2,+} \le j \le j_{2,+}$, we have

  $$\begin{aligned} h_{j}^{-1}&= \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \bigg ( 1 + \mathcal{O}(n^{-100}) \bigg ) \nonumber \\&= b \frac{\sqrt{n} \, e^{\frac{n}{b}(j/n-j/n \log (\frac{j/n}{b}))}}{\sqrt{2\pi }}\bigg ( \frac{b}{j/n} \bigg )^{\frac{\alpha }{b}} \sqrt{\frac{j/n}{b}} \bigg ( 1 + \frac{-b^{2}+6b \alpha - 6 \alpha ^{2}}{12 b j/n} \frac{1}{n} + \mathcal{O}(n^{-2}) \bigg ). \end{aligned}$$

  (4.18)

• 11. As $n \rightarrow + \infty$ with $j_{2,+} \le j \le n$, we have

  $$\begin{aligned} h_{j}^{-1}&= \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \bigg ( 1 + \mathcal{O}(n^{-\frac{1}{2}}e^{-nr_{2}^{2b}\frac{-\epsilon - \log (1-\epsilon )}{1-\epsilon }} \bigg ) \nonumber \\&= b \frac{\sqrt{n} \, e^{\frac{n}{b}(j/n-j/n \log (\frac{j/n}{b}))}}{\sqrt{2\pi }}\bigg ( \frac{b}{j/n} \bigg )^{\frac{\alpha }{b}} \sqrt{\frac{j/n}{b}} \bigg ( 1 + \frac{-b^{2}+6b \alpha - 6 \alpha ^{2}}{12 b j/n} \frac{1}{n} + \mathcal{O}(n^{-2}) \bigg ). \end{aligned}$$

  (4.19)

Proof

Note from Eqs. (2.6) and (4.1) that

$$\begin{aligned} h_{j}^{-1}&= \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\tfrac{j+\alpha }{b})} \bigg ( \frac{\gamma (\tfrac{j+\alpha }{b},nr_{1}^{2b})}{\Gamma (\tfrac{j+\alpha }{b})} - \frac{\gamma (\tfrac{j+\alpha }{b},nr_{2}^{2b})}{\Gamma (\tfrac{j+\alpha }{b})} + 1 \bigg )^{-1} \nonumber \\&= \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (a_{j})} \bigg ( \frac{\gamma (a_{j},\lambda _{j,1}a_{j})}{\Gamma (a_{j})} - \frac{\gamma (a_{j},\lambda _{j,2}a_{j})}{\Gamma (a_{j})} + 1 \bigg )^{-1}, \qquad j=1,2,\ldots ,n. \end{aligned}$$

(4.20)

For fixed j, Lemma B.1 implies that $\gamma (\tfrac{j+\alpha }{b},nr_{k}^{2b})=\Gamma (\tfrac{j+\alpha }{b})+\mathcal{O}(n^{\tfrac{j+\alpha }{b}-1}e^{-nr_{k}^{2b}})$ as $n \rightarrow \infty$, and Eq. (4.9) follows. Now we turn to Eq. (4.10). By Lemma B.3 (i),

$$\begin{aligned} h_{j}^{-1}&= \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \bigg ( 1 + \mathcal{O}(n^{-\frac{1}{2}}e^{-\frac{a_{j}\eta _{j,1}^{2}}{2}}) + \mathcal{O}(n^{-\frac{1}{2}}e^{-\frac{a_{j}\eta _{j,2}^{2}}{2}}) \bigg ) \\&= \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \bigg ( 1 + \mathcal{O}(n^{-\frac{1}{2}}e^{-\frac{a_{j}\eta _{j,1}^{2}}{2}}) \bigg ) \end{aligned}$$

as $n \rightarrow \infty$, $M' \le j \le j_{1,-}$. Moreover,

$$\begin{aligned} \frac{a_{j}\eta _{j,1}^{2}}{2} = n \frac{j/n}{b}\bigg (\frac{br_{1}^{2b}}{j/n}-1-\log \Big (\frac{br_{1}^{2b}}{j/n}\Big )\bigg ) + \mathcal{O}(1) \ge nr_{1}^{2b}\frac{\epsilon - \log (1+\epsilon )}{1+\epsilon } + \mathcal{O}(1) \end{aligned}$$

as $n \rightarrow \infty , \; M' \le j \le j_{1,-}$, which implies the first line in Eq. (4.10). The second line in Eq. (4.10) then directly follows from Lemma 4.1. Now we prove Eq. (4.11). By Eq. (4.20) and Lemmas B.2 and B.3 (i),

$$\begin{aligned} h_{j}^{-1} = \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \bigg ( \frac{1}{\frac{1}{2}{{\textrm{erfc}}}(-\eta _{j,1}\sqrt{\frac{a_{j}}{2}}) } + \mathcal{O}(n^{-\frac{1}{2}}e^{-\frac{1}{2}a_{j}\eta _{j,1}^{2}}) \bigg ) \end{aligned}$$

as $n \rightarrow + \infty$ with $j_{1,-} \le j \le g_{1,-}$. On the other hand,

$$\begin{aligned} -\eta _{j,1}\sqrt{\frac{a_{j}}{2}} \le -\frac{Mr_{1}^{2}}{\sqrt{2}} + \mathcal{O}\Big (\frac{M^{2}}{\sqrt{n}}\Big ), \qquad \hbox {as } n \rightarrow + \infty , \; j_{1,-} \le j \le g_{1,-}, \end{aligned}$$

which implies the first line in Eq. (4.11) provided $M'$ is chosen large enough (recall that $M'\sqrt{\log n}\le M$). The second line in Eq. (4.11) then directly follows from Lemma 4.1. Now we prove Eq. (4.12). By Lemmas B.2 and B.3 (i), as $n \rightarrow + \infty$ with $g_{1,-} \le j \le g_{1,+}$, we have

$$\begin{aligned} h_{j}^{-1} = \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \frac{1 + \mathcal{O}(e^{-\frac{1}{2}\frac{a_{j}}{2}(\eta _{j,2}^{2}-\eta _{j,1}^{2})})}{\frac{1}{2}{\textrm{erfc}}(-\eta _{j,1}\sqrt{\frac{a_{j}}{2}}) - \frac{e^{-\frac{1}{2}a_{j}\eta _{j,1}^{2}}}{\sqrt{2\pi a_{j}}}\big ( c_{0}(\eta _{j,1}) + \frac{c_{1}(\eta _{j,1})}{a_{j}} + \mathcal{O}(\frac{1}{n^{2}}) \big )} . \end{aligned}$$

The above can be expanded in terms of $M_{j,1}$ using Eqs. (4.1) and (4.6), namely using

$$\begin{aligned} \eta _{j,1} = (\lambda _{j,1}-1)\sqrt{\frac{2(\lambda _{j,1}-1-\log \lambda _{j,1})}{(\lambda _{j,1}-1)^{2}}}, \qquad \lambda _{j,1} = 1 + \frac{M_{j,1}}{\sqrt{n}}, \qquad j = \frac{br_{1}^{2b} n}{\lambda _{j,1}}-\alpha . \end{aligned}$$

Using also

$$\begin{aligned} \frac{e^{-\frac{r_{1}^{2b}M_{j,1}^{2}}{2}}}{\sqrt{\pi } {{\textrm{erfc}}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}}) } = {\left\{ \begin{array}{ll} - \frac{r_{1}^{b}M_{j,1}}{\sqrt{2}} - \frac{1}{\sqrt{2}r_{1}^{b}M_{j,1}} + \mathcal{O}(M_{j,1}^{-3}), &{} \hbox {as } M_{j,1}\rightarrow -\infty , \\ \mathcal{O}(e^{-\frac{r_{1}^{2b}M_{j,1}^{2}}{2}}), &{} \hbox {as } M_{j,1}\rightarrow +\infty , \end{array}\right. } \end{aligned}$$

we obtain

$$\begin{aligned} \frac{1}{\frac{1}{2}{\textrm{erfc}}(-\eta _{j,1}\sqrt{\frac{a_{j}}{2}}) - \frac{e^{-\frac{1}{2}a_{j}\eta _{j,1}^{2}}}{\sqrt{2\pi a_{j}}}\big ( c_{0}(\eta _{j,1}) + \frac{c_{1}(\eta _{j,1})}{a_{j}} + \mathcal{O}(\frac{1}{n^{2}}) \big )} \\ = \frac{1}{\frac{1}{2}{\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}})} \bigg \{ 1 + \frac{(5M_{j,1}^{2}r_{1}^{2b}-2)e^{-\frac{r_{1}^{2b}M_{j,1}^{2}}{2}}}{3\sqrt{2\pi } r_{1}^{b} {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}}) \sqrt{n}} + \frac{\chi _{j,1}^{-}}{n} \bigg ( \frac{25r_{1}^{4b}M_{j,1}^{6}}{72} + \frac{3r_{1}^{2b}M_{j,1}^{4}}{2} \bigg ) \nonumber \\ - \chi _{j,1}^{-} \frac{125r_{1}^{6b}M_{j,1}^{9}}{1296n^{3/2}} + \mathcal{O}\bigg (\frac{1+M_{j,1}^{2}}{n} + \frac{1+|M_{j,1}^{7}|}{n^{3/2}} + \frac{1+M_{j,1}^{12}}{n^{2}} \bigg ) \bigg ) \bigg \} \end{aligned}$$

as $n \rightarrow + \infty$ uniformly for $g_{1,-} \le j \le g_{1,+}$, and the first expansion in Eq. (4.12) follows. The second expansion in Eq. (4.12) then follows from Lemma 4.1. Now we prove Eq. (4.13). By Eqs. (4.4), (4.20), Lemmas B.2 and B.3 (i), we have

$$\begin{aligned} h_{j}^{-1} = \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \frac{1}{\frac{1}{2}{\textrm{erfc}}(-\eta _{j,1}\sqrt{a_{j}/2})-\frac{e^{-\frac{a_{j}\eta _{j,1}^{2}}{2}}}{\sqrt{2\pi a_{j}}}(c_{0}(\eta _{j,1})+\frac{c_{1}(\eta _{j,1})}{a_{j}})} \bigg ( 1 + \mathcal{O}\big (n^{-2} + e^{-\frac{a_{j}}{2}(\eta _{j,2}^{2}-\eta _{j,1}^{2})}\big ) \bigg ), \end{aligned}$$

and the first expansion in Eq. (4.13) follows from a long but direct computation using Eqs. (4.1), (B.4) and (B.5). The second expansion in Eq. (4.13) then follows from Lemma 4.1. Now we prove Eqs. (4.14) and (4.15). Using Lemma B.3 (i)–(ii), we find

$$\begin{aligned}&h_{j}^{-1} = \frac{b n^{\frac{j+\alpha }{b}}}{\Gamma (\frac{j+\alpha }{b})} \Big ( 1+\mathcal{O}(n^{-2}) \Big ) \\&\times \frac{1}{\frac{e^{-\frac{a_{j}\eta _{j,1}^{2}}{2}}}{\sqrt{2\pi }}(\frac{1}{1-\lambda _{j,1}}\frac{1}{\sqrt{a_{j}}}+\frac{1+10 \lambda _{j,1}+\lambda _{j,1}^{2}}{12(\lambda _{j,1}-1)^{3}}\frac{1}{a_{j}^{3/2}}) - \frac{e^{-\frac{a_{j}\eta _{j,2}^{2}}{2}}}{\sqrt{2\pi }}(\frac{-1}{\lambda _{j,2}-1}\frac{1}{\sqrt{a_{j}}}+\frac{1+10 \lambda _{j,2}+\lambda _{j,2}^{2}}{12(\lambda _{j,2}-1)^{3}}\frac{1}{a_{j}^{3/2}})} \end{aligned}$$

as $n \rightarrow + \infty$ with $j_{1,+} \le j \le j_{2,-}$. Then Eqs. (4.14) and (4.15) follow from a long but direct computation using Eqs. (4.1), (4.5) and Lemma 4.1.

The proofs of Eqs. (4.16)–(4.19) are similar to that of Eqs. (4.10)–(4.13), respectively, so we omit them. $\square$

5 Proof of Theorem 3.11

In this section z and w are given by

$$\begin{aligned} z = r_{1}\Big (1-\frac{t_{1}}{\sigma _{1}n}\Big )e^{i\theta _{1}}, \qquad w = r_{2}\Big (1+\frac{t_{2}}{\sigma _{2}n}\Big )e^{i\theta _{2}}, \qquad t_{1}, t_{2} \ge 0, \theta _{1},\theta _{2}\in \mathbb {R}. \end{aligned}$$

(5.1)

Let us rewrite Eq. (2.5) as $K_{n}(z,w) = \sum _{j=1}^{n}F_{j}$, where

$$\begin{aligned}&F_{j} := e^{-\frac{n}{2}Q(z)}e^{-\frac{n}{2}Q(w)} \frac{z^{j-1}\overline{w}^{j-1}}{h_{j}} = e^{-\frac{n}{2}|z|^{2b}}e^{-\frac{n}{2}|w|^{2b}}|z|^{\alpha }|w|^{\alpha }\frac{z^{j-1}\overline{w}^{j-1}}{h_{j}}. \end{aligned}$$

(5.2)

We will obtain the asymptotics of the summand $F_{j}$ as $n\rightarrow +\infty$ for several regimes of the indice j; such splitting can also be found in e.g. [5, 25, 40] in the context of large gap probabilities. In this section we take $M=M' \sqrt{\log n}$. The following lemma is proved using Lemma 4.2.

Lemma 5.1

Let $\delta \in (0,\frac{1}{100})$ be fixed. $M'$ can be chosen sufficiently large and independently of $\delta$ such that the following hold.

• 1. Let j be fixed. As $n \rightarrow + \infty$, we have

  $$\begin{aligned} F_{j} = \mathcal{O}(e^{-\frac{n}{2}(1-\delta )(r_{1}^{2b}+r_{2}^{2b})}). \end{aligned}$$

  (5.3)

• 2. As $n \rightarrow + \infty$ with $M' \le j \le j_{1,-}$, we have

  $$\begin{aligned} F_{j}&= \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})\sigma _{1}}). \end{aligned}$$

  (5.4)

• 3. As $n \rightarrow + \infty$ with $j_{1,-} \le j \le g_{1,-}$, we have

  $$\begin{aligned} F_{j}&= \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})\sigma _{1}}). \end{aligned}$$

  (5.5)

• 4. As $n \rightarrow + \infty$ with $g_{1,-} \le j \le g_{1,+}$, we have

  $$\begin{aligned}&F_{j} = \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})\sigma _{1}}). \end{aligned}$$

  (5.6)

• 5. As $n \rightarrow + \infty$ with $g_{1,+} \le j \le j_{1,+}$, we have

  $$\begin{aligned}&F_{j} = \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})(\sigma _{\star }-\frac{br_{1}^{2b}}{1-\epsilon })}). \end{aligned}$$

  (5.7)

• 6. As $n \rightarrow + \infty$ with $j_{1,+} \le j \le \lfloor j_{\star }\rfloor$, we have

  $$\begin{aligned}&F_{j} = \frac{n}{r_{1}r_{2}}\frac{(\frac{r_{1}}{r_{2}})^{j_{\star }-j}e^{(j-1)i(\theta _{1}-\theta _{2})}}{\frac{1}{j/n-br_{1}^{2b}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{br_{2}^{2b}-j/n}}e^{-(j/n-br_{1}^{2b})\frac{t_{1}}{\sigma _{1}}-(br_{2}^{2b}-j/n)\frac{t_{2}}{\sigma _{2}}} \Big ( 1 + \mathcal{O}\big ( n^{-1} \big ) \Big ). \end{aligned}$$

  (5.8)

• 7. As $n \rightarrow + \infty$ with $\lfloor j_{\star }\rfloor +1 \le j \le j_{2,-}$, we have

  $$\begin{aligned}&F_{j} = \frac{n}{r_{1}r_{2}}\frac{(\frac{r_{1}}{r_{2}})^{j-j_{\star }}e^{(j-1)i(\theta _{1}-\theta _{2})}}{\frac{1}{br_{2}^{2b}-j/n}+\frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{j/n-br_{1}^{2b}}}e^{-(j/n-br_{1}^{2b})\frac{t_{1}}{\sigma _{1}}-(br_{2}^{2b}-j/n)\frac{t_{2}}{\sigma _{2}}} \Big ( 1 + \mathcal{O}\big ( n^{-1} \big ) \Big ). \end{aligned}$$

  (5.9)

• 8. As $n \rightarrow + \infty$ with $j_{2,-} \le j \le g_{2,-}$, we have

  $$\begin{aligned}&F_{j} = \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})(\frac{br_{2}^{2b}}{1+\epsilon }-\sigma _{\star })}). \end{aligned}$$

  (5.10)

• 9. As $n \rightarrow + \infty$ with $g_{2,-} \le j \le g_{2,+}$, we have

  $$\begin{aligned}&F_{j} = \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})\sigma _{2}}). \end{aligned}$$

  (5.11)

• 10. As $n \rightarrow + \infty$ with $g_{2,+} \le j \le j_{2,+}$, we have

  $$\begin{aligned} F_{j}&= \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})\sigma _{2}}). \end{aligned}$$

  (5.12)

• 11. As $n \rightarrow + \infty$ with $j_{2,+} \le j \le n$, we have

  $$\begin{aligned} F_{j}&= \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})\sigma _{2}}). \end{aligned}$$

  (5.13)

Proof

Eq. (5.3) directly follows from Eqs. (4.9) and (5.2). Let us now prove Eq. (5.4). By Eqs. (4.10) and (5.2), as $n \rightarrow + \infty$ with $M' \le j \le j_{1,-}$, we have

$$\begin{aligned} F_{j} = \mathcal{O}\bigg (\exp \bigg \{ n(1-\delta ) \bigg [ \frac{1}{b}(j/n-j/n \log (\frac{j/n}{b})) - \frac{1}{2}r_{1}^{2b} - \frac{1}{2}r_{2}^{2b} + j/n \log (r_{1}r_{2}) \bigg ] \bigg \} \bigg ). \end{aligned}$$

It is easy to check that the function $[0,br_{1}^{b}r_{2}^{b}] \in y \mapsto y-y \log (\frac{y}{b}) + y \log (r_{1}^{b}r_{2}^{b})$ is increasing. Since $j/n \le br_{1}^{2b}$, we thus have

$$\begin{aligned} F_{j}&= \mathcal{O}\bigg (\exp \bigg \{ n(1-\delta ) \bigg [ r_{1}^{2b}-r_{1}^{2b} \log (r_{1}^{2b}) - \frac{1}{2}r_{1}^{2b} - \frac{1}{2}r_{2}^{2b} + br_{1}^{2b} \log (r_{1}r_{2}) \bigg ] \bigg \} \bigg ), \end{aligned}$$

which is Eq. (5.4) (recall the definitions of $\sigma _{1}$ and $\sigma _{\star }$ in Eq. (2.4)). The proof of Eq. (5.5) is identical (but uses Eq. (4.11) instead of Eq. (4.10)). We now turn to the proof of Eq. (5.6). By Eqs. (4.12) and (5.2), as $n \rightarrow +\infty$ with $g_{1,-} \le j \le g_{1,+}$ we have

$$\begin{aligned} F_{j} = \mathcal{O}\bigg ( \exp \bigg \{ n(1-\delta /2) \bigg [ r_{1}^{2b} - r_{1}^{2b}\log (r_{1}^{2b}) - \frac{1}{2}r_{1}^{2b} - \frac{1}{2}r_{2}^{2b} + j/n \log (r_{1}r_{2}) \bigg ] \bigg \} \bigg ). \end{aligned}$$

Since $j/n = br_{1}^{2b} + \mathcal{O}(n^{-\frac{1}{2}})$,

$$\begin{aligned} F_{j} = \mathcal{O}\bigg ( \exp \bigg \{ n(1-\delta ) \bigg [ r_{1}^{2b} - r_{1}^{2b}\log (r_{1}^{2b}) - \frac{1}{2}r_{1}^{2b} - \frac{1}{2}r_{2}^{2b} + br_{1}^{2b} \log (r_{1}r_{2}) \bigg ] \bigg \} \bigg ), \end{aligned}$$

and Eq. (5.6) follows. Now we prove Eq. (5.7). By Eqs. (4.13) and (5.2), as $n \rightarrow + \infty$ with $g_{1,+} \le j \le j_{1,+}$, we have

$$\begin{aligned} F_{j} = \mathcal{O}\bigg ( \exp \bigg \{ n(1-\delta ) \bigg [ r_{1}^{2b} - 2j/n \log (r_{1}) - \frac{1}{2}r_{1}^{2b} - \frac{1}{2}r_{2}^{2b} + j/n \log (r_{1}r_{2}) \bigg ] \bigg \} \bigg ). \end{aligned}$$

Recall from Eq. (4.4) that $\sigma _{\star } > \frac{br_{1}^{2b}}{1-\epsilon }$. Since $j/n \le \frac{br_{1}^{2b}}{1-\epsilon } + \mathcal{O}(n^{-1})$ by Eq. (4.2), we have

$$\begin{aligned} F_{j} = \mathcal{O}\bigg ( \exp \bigg \{ n(1-\delta ) \bigg [ - \frac{r_{2}^{2b}-r_{1}^{2b}}{2} + \frac{br_{1}^{2b}}{1-\epsilon } \log (\frac{r_{2}}{r_{1}}) \bigg ] \bigg \} \bigg ), \end{aligned}$$

which is Eq. (5.7). Now we prove Eq. (5.8). As $n \rightarrow + \infty$ with $j_{1,+} \le j \le \lfloor j_{\star }\rfloor$, we have

$$\begin{aligned}&e^{n(r_{1}^{2b}-2 j/n \log (r_{1}))} e^{-\frac{n}{2}|z|^{2b}}e^{-\frac{n}{2}|w|^{2b}}|z|^{\alpha }|w|^{\alpha } z^{j-1}\overline{w}^{j-1} \\&\quad = e^{-\frac{n}{2}(r_{2}^{2b}-r_{1}^{2b}-2j/n \log (\frac{r_{2}}{r_{1}}))} e^{(j-1)i(\theta _{1}-\theta _{2})} e^{- (j/n-br_{1}^{2b})\frac{t_{1}}{\sigma _{1}}}e^{-(br_{2}^{2b}-j/n)\frac{t_{2}}{\sigma _{2}}}r_{1}^{\alpha -1}r_{2}^{\alpha -1} \big ( 1+\mathcal{O}(n^{-1}) \big ) \end{aligned}$$

Now Eq. (5.8) follows directly from Eqs. (4.14), (5.2) and the identity (recall $j_{\star } = n \sigma _{\star } -\alpha$)

$$\begin{aligned} \frac{r_{2}^{\alpha }}{r_{1}^{\alpha }} e^{-\frac{n}{2}(r_{2}^{2b}-r_{1}^{2b}-2j/n \log (\frac{r_{2}}{r_{1}}))} = \Big (\frac{r_{2}}{r_{1}}\Big )^{-(j_{\star }-j)}. \end{aligned}$$

The proofs of Eqs. (5.9)–(5.13) are similar to that of Eqs. (5.4)–(5.8), respectively, so we omit them. $\square$

Theorem 5.2

Let $z,w,t_{1},t_{2},\theta _{1},\theta _{2}$ be as in Eq. (5.1). As $n \rightarrow + \infty$, we have

$$\begin{aligned}&K_{n}(z,w) = \frac{n}{r_{1}r_{2}} e^{-t_{1}-t_{2}} e^{i(\theta _{1}-\theta _{2})\lfloor j_{\star }\rfloor } \nonumber \\&\times \Bigg \{ \sum _{j=0}^{+\infty } \frac{(\frac{r_{1}}{r_{2}})^{j+x}e^{(j+1)i(\theta _{2}-\theta _{1})}}{\frac{1}{\sigma _{1}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j+x)}}{\sigma _{2}}} + \sum _{j=0}^{+\infty } \frac{(\frac{r_{1}}{r_{2}})^{j+1-x}e^{j i(\theta _{1}-\theta _{2})}}{\frac{1}{\sigma _{2}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j+1-x)}}{\sigma _{1}}} \Bigg \} + \mathcal{O}((\log n)^{2}), \end{aligned}$$

where $\sigma _{\star }, \sigma _{1}, \sigma _{2}$ are given by Eq. (2.4), $j_{\star } = n \sigma _{\star }-\alpha$, and $x=x(n)$ is given by Eq. (3.20).

Proof

By Lemma 5.1,

$$\begin{aligned} K_{n} = \sum _{j=j_{1,+}}^{j_{2,-}}F_{j} + \mathcal{O}(e^{-cn}), \qquad \hbox {as } n \rightarrow + \infty \end{aligned}$$

for some $c>0$. Furthermore, by Eqs. (5.8) and (5.9),

$$\begin{aligned}&\sum _{j=j_{1,+}}^{\lfloor j_{\star } \rfloor }F_{j} = \sum _{j=j_{1,+}}^{\lfloor j_{\star } \rfloor } \frac{n}{ r_{1}r_{2}}\frac{(\frac{r_{1}}{r_{2}})^{j_{\star }-j}e^{(j-1)i(\theta _{1}-\theta _{2})}}{\frac{1}{j/n-br_{1}^{2b}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{br_{2}^{2b}-j/n}}e^{-(j/n-br_{1}^{2b})\frac{t_{1}}{\sigma _{1}}-(br_{2}^{2b}-j/n)\frac{t_{2}}{\sigma _{2}}} \Big ( 1 + \mathcal{O}\big ( n^{-1} \big ) \Big ), \\&\sum _{j=\lfloor j_{\star } \rfloor +1}^{j_{2,-}}F_{j} = \sum _{j=\lfloor j_{\star } \rfloor +1}^{j_{2,-}} \frac{n}{ r_{1}r_{2}}\frac{(\frac{r_{1}}{r_{2}})^{j-j_{\star }}e^{(j-1)i(\theta _{1}-\theta _{2})}}{\frac{1}{br_{2}^{2b}-j/n}+\frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{j/n-br_{1}^{2b}}}e^{-(j/n-br_{1}^{2b})\frac{t_{1}}{\sigma _{1}}-(br_{2}^{2b}-j/n)\frac{t_{2}}{\sigma _{2}}} \Big ( 1 + \mathcal{O}\big ( n^{-1} \big ) \Big ). \end{aligned}$$

For $j-\lfloor j_{\star } \rfloor \ge M' \log n$, we have $(\frac{r_{1}}{r_{2}})^{j-\lfloor j_{\star }}=\mathcal{O}(n^{-200})$ provided that $M'$ is chosen large enough. We infer that

$$\begin{aligned} \sum _{j=j_{1,+}}^{\lfloor j_{\star } \rfloor }F_{j}&= \mathcal{O}(n^{-100})\\&\quad + \hspace{-0.1cm} \sum _{j=\lfloor j_{\star } \rfloor - M' \log n}^{\lfloor j_{\star } \rfloor } \hspace{-0.2cm} \frac{n}{r_{1}r_{2}}\frac{(\frac{r_{1}}{r_{2}})^{j_{\star }-j}e^{(j-1)i(\theta _{1}-\theta _{2})}}{\frac{1}{j/n-br_{1}^{2b}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{br_{2}^{2b}-j/n}}e^{-(j/n-br_{1}^{2b})\frac{t_{1}}{\sigma _{1}}-(br_{2}^{2b}-j/n)\frac{t_{2}}{\sigma _{2}}} \Big ( 1 + \mathcal{O}\big ( n^{-1} \big ) \Big ), \\ \sum _{j=\lfloor j_{\star } \rfloor +1}^{j_{2,-}}F_{j}&= \mathcal{O}(n^{-100})\\&\quad + \hspace{-0.1cm} \sum _{j=\lfloor j_{\star } \rfloor +1}^{\lfloor j_{\star } \rfloor +M' \log n} \hspace{-0.2cm} \frac{n}{r_{1}r_{2}}\frac{(\frac{r_{1}}{r_{2}})^{j-j_{\star }}e^{(j-1)i(\theta _{1}-\theta _{2})}}{\frac{1}{br_{2}^{2b}-j/n}+\frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{j/n-br_{1}^{2b}}}e^{-(j/n-br_{1}^{2b})\frac{t_{1}}{\sigma _{1}}-(br_{2}^{2b}-j/n)\frac{t_{2}}{\sigma _{2}}} \Big ( 1 + \mathcal{O}\big ( n^{-1} \big ) \Big ). \end{aligned}$$

provided that $M'$ is chosen sufficiently large. For $|j-\lfloor j_{\star } \rfloor | \le M' \log n$, we have $j/n = \sigma _{\star } + \mathcal{O}(\frac{\log n}{n})$, and thus

$$\begin{aligned}&\sum _{j=j_{1,+}}^{\lfloor j_{\star } \rfloor }F_{j} = \mathcal{O}(n^{-100}) + \frac{n}{r_{1}r_{2}} e^{- t_{1}- t_{2}} \hspace{-0.1cm} \sum _{j=\lfloor j_{\star } \rfloor - M' \log n}^{\lfloor j_{\star } \rfloor } \hspace{-0cm} \frac{(\frac{r_{1}}{r_{2}})^{j_{\star }-j}e^{(j-1)i(\theta _{1}-\theta _{2})}}{\frac{1}{\sigma _{1}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{\sigma _{2}}} \Big ( 1 + \mathcal{O}\big ( \frac{\log n}{n} \big ) \Big ), \\&\sum _{j=\lfloor j_{\star } \rfloor +1}^{j_{2,-}}F_{j} = \mathcal{O}(n^{-100}) + \frac{n}{r_{1}r_{2}} e^{- t_{1}- t_{2}} \hspace{-0.1cm} \sum _{j=\lfloor j_{\star } \rfloor +1}^{\lfloor j_{\star } \rfloor +M' \log n} \hspace{-0cm} \frac{(\frac{r_{1}}{r_{2}})^{j-j_{\star }}e^{(j-1)i(\theta _{1}-\theta _{2})}}{\frac{1}{\sigma _{2}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{\sigma _{1}}} \Big ( 1 + \mathcal{O}\big ( \frac{\log n}{n} \big ) \Big ). \end{aligned}$$

We can rewrite this as

$$\begin{aligned}&\sum _{j=j_{1,+}}^{\lfloor j_{\star } \rfloor }F_{j} = \mathcal{O}\big ((\log n)^{2}\big ) + \frac{n}{r_{1}r_{2}} e^{- t_{1}- t_{2}} \sum _{j=\lfloor j_{\star } \rfloor - M' \log n}^{\lfloor j_{\star } \rfloor } \frac{(\frac{r_{1}}{r_{2}})^{j_{\star }-j}e^{(j-1)i(\theta _{1}-\theta _{2})}}{\frac{1}{\sigma _{1}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{\sigma _{2}}}, \\&\sum _{j=\lfloor j_{\star } \rfloor +1}^{j_{2,-}}F_{j} = \mathcal{O}\big ((\log n)^{2}\big ) + \frac{n}{r_{1}r_{2}} e^{- t_{1}- t_{2}} \sum _{j=\lfloor j_{\star } \rfloor +1}^{\lfloor j_{\star } \rfloor +M' \log n} \frac{(\frac{r_{1}}{r_{2}})^{j-j_{\star }}e^{(j-1)i(\theta _{1}-\theta _{2})}}{\frac{1}{\sigma _{2}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{\sigma _{1}}}. \end{aligned}$$

In the above sums appearing on the right-hand sides, we can replace $M' \log n$ by $+\infty$ at the cost of an error $\mathcal{O}(n^{-100})$; this error can in turn be absorbed in $\mathcal{O}\big ((\log n)^{2}\big )$. We then find the claim after changing indices. $\square$

A computation using Eq. (3.16) gives

$$\begin{aligned} \frac{1}{2\pi r_{1}r_{2}} \Bigg \{ \sum _{j=0}^{+\infty } \frac{(\frac{r_{1}}{r_{2}})^{j+x}e^{(j+1)i(\theta _{2}-\theta _{1})}}{\frac{1}{\sigma _{1}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j+x)}}{\sigma _{2}}} + \sum _{j=0}^{+\infty } \frac{(\frac{r_{1}}{r_{2}})^{j+1-x}e^{j i(\theta _{1}-\theta _{2})}}{\frac{1}{\sigma _{2}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j+1-x)}}{\sigma _{1}}} \Bigg \} \\ = (r_{1}r_{2})^{-x} \mathcal {S}^{G}_{\textrm{hard}}(r_{1}e^{i\theta _{1}},r_{2}e^{i\theta _{2}};n) = (r_{1}r_{2})^{-x} \mathcal {S}^{G}_{\textrm{hard}}(z,w;n) + \mathcal{O}(n^{-1}), \end{aligned}$$

and Theorem 3.11 follows.

6 Proofs of Theorems 3.1 and 3.13

In this section z and w are given by

$$\begin{aligned} z = r_{1}\Big (1-\frac{t_{1}}{\sigma _{1}n}\Big )e^{i\theta _{1}}, \qquad w = r_{1}\Big (1-\frac{t_{2}}{\sigma _{1}n}\Big )e^{i\theta _{2}}, \qquad t_{1}, t_{2} \ge 0, \theta _{1},\theta _{2}\in \mathbb {R}, \end{aligned}$$

(6.1)

and the parameters $t_{1},t_{2},\theta _{1},\theta _{2}$ are independent of n. At this point, $\theta _{1},\theta _{2}\in \mathbb {R}$ can be arbitrary, but we already mention that in Sect. 6.1 below, we will focus on the case $\theta _{1}=\theta _{2}$, and in Sect. 6.2 we will consider the case $\theta _{1}\ne \theta _{2} \mod 2\pi$. As in Eq. (5.2) we write $K_{n}(z,w) = \sum _{j=1}^{n}F_{j}$, where

$$\begin{aligned}&F_{j} := e^{-\frac{n}{2}Q(z)}e^{-\frac{n}{2}Q(w)} \frac{z^{j-1}\overline{w}^{j-1}}{h_{j}} = e^{-\frac{n}{2}|z|^{2b}}e^{-\frac{n}{2}|w|^{2b}}|z|^{\alpha }|w|^{\alpha }\frac{z^{j-1}\overline{w}^{j-1}}{h_{j}}. \end{aligned}$$

(6.2)

Here M can be arbitrary within the range $M \in [M'\sqrt{\log n}, n^{\frac{1}{10}}]$ (the values of M will be more restricted in Sects. 6.1 and 6.2 below). The following lemma is proved using Lemma 4.2.

Lemma 6.1

Let $\delta \in (0,\frac{1}{100})$ be fixed. $M'$ can be chosen sufficiently large and independently of $\delta$ such that the following hold.

• 1. Let j be fixed. As $n \rightarrow + \infty$, we have

  $$\begin{aligned} F_{j} = \mathcal{O}(e^{-n (1-\delta )r_{1}^{2b}}). \end{aligned}$$

  (6.3)

• 2. As $n \rightarrow + \infty$ with $M' \le j \le j_{1,-}$, we have

  $$\begin{aligned} F_{j}&= \mathcal{O}(e^{-nr_{1}^{2b}(1-\delta ) \frac{\epsilon - \log (1+\epsilon )}{1+\epsilon }}). \end{aligned}$$

  (6.4)

• 3. As $n \rightarrow + \infty$ with $j_{1,-} \le j \le g_{1,-}$, we have

  $$\begin{aligned} F_{j}&= \mathcal{O}(n^{-100}). \end{aligned}$$

  (6.5)

• 4. As $n \rightarrow + \infty$ with $g_{1,-} \le j \le g_{1,+}$, we have

  $$\begin{aligned}&F_{j} = e^{i(j-1)(\theta _{1}-\theta _{2})} \sqrt{n}\frac{\sqrt{2} \, b r_{1}^{b-2} e^{-\frac{M_{j,1}^{2}r_{1}^{2b}}{2}}}{\sqrt{\pi } {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}})} \bigg ( 1 + \frac{1}{\sqrt{n}}\bigg \{\frac{(5M_{j,1}^{2}r_{1}^{2b}-2)e^{-\frac{r_{1}^{2b}M_{j,1}^{2}}{2}}}{3\sqrt{2\pi } r_{1}^{b} {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}}) } \nonumber \\&\qquad + \frac{M_{j,1}}{6} \Big ( 5M_{j,1}^{2}r_{1}^{2b}-3+6br_{1}^{2b}\frac{t_{1}+t_{2}}{\sigma _{1}} \Big ) \bigg \} \nonumber \\&+ \mathcal{O}\bigg (\frac{1+M_{j,1}^{2}+M_{j,1}^{6}\chi _{j,1}^{+}}{n}+\frac{1+M_{j,1}^{7}+M_{j,1}^{9}\chi _{j,1}^{+}}{n^{3/2}} + \frac{1+M_{j,1}^{12}}{n^{2}}\bigg ) \bigg ). \end{aligned}$$

  (6.6)

• 5. As $n \rightarrow + \infty$ with $g_{1,+} \le j \le j_{1,+}$, we have

  $$\begin{aligned} F_{j}&= e^{i(j-1)(\theta _{1}-\theta _{2})} n \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}(j/n-br_{1}^{2b})}{r_{1}^{2}}\bigg ( 1 + \frac{1}{n} \bigg \{ \frac{t_{1}+t_{2}}{\sigma _{1}}(1-\alpha ) \nonumber \\&\quad - \frac{j/n}{2 \sigma _{1}^{2}}(t_{1}^{2}+t_{2}^{2})+ \frac{b}{2\sigma _{1}^{2}}(1-2b)r_{1}^{2b}(t_{1}^{2}+t_{2}^{2}) + \frac{b^{2}r_{1}^{2b}+(j/n-br_{1}^{2b})\alpha }{(j/n-br_{1}^{2b})^{2}} \bigg \} \nonumber \\&\quad - \frac{2 b^{4} r_{1}^{4b}}{n^{2}(j/n-br_{1}^{2b})^{4}}+ \frac{10b^{6}r_{1}^{6b}}{n^{3}(j/n-br_{1}^{2b})^{6}} + \mathcal{O}\Big ( \frac{1}{n^{2}(j/n-br_{1}^{2b})^{3}} \nonumber \\&\quad + \frac{1}{n^{4}(j/n-br_{1}^{2b})^{8}} \Big ) \bigg ). \end{aligned}$$

  (6.7)

• 6. As $n \rightarrow + \infty$ with $j_{1,+} \le j \le \lfloor j_{\star }\rfloor$, we have

  $$\begin{aligned}&F_{j} = e^{i(j-1)(\theta _{1}-\theta _{2})} \frac{n}{r_{1}^{2}} \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}}{\frac{1}{j/n-br_{1}^{2b}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{br_{2}^{2b}-j/n}} \bigg ( 1 + \frac{1}{n} \bigg \{ \frac{t_{1}+t_{2}}{\sigma _{1}}(1-\alpha ) - \frac{j/n}{2\sigma _{1}^{2}}(t_{1}^{2}+t_{2}^{2}) \nonumber \\&+ \frac{b}{2\sigma _{1}^{2}}(1-2b)r_{1}^{2b}(t_{1}^{2}+t_{2}^{2}) + \frac{b^{2}r_{1}^{2b}+(j/n-br_{1}^{2b})\alpha }{(j/n-br_{1}^{2b})^{2}} \bigg \} + \mathcal{O}\Big ( \frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{n} + \frac{1}{n^{2}} \Big ) \bigg ). \end{aligned}$$

  (6.8)

• 7. As $n \rightarrow + \infty$ with $\lfloor j_{\star }\rfloor +1 \le j \le j_{2,-}$, we have

  $$\begin{aligned}&F_{j} = e^{i(j-1)(\theta _{1}-\theta _{2})}\frac{n}{r_{1}^{2}} \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{\frac{1}{br_{2}^{2b} - j/n}+\frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{j/n-br_{1}^{2b}}} \bigg ( 1 + \frac{1}{n} \bigg \{ \frac{t_{1}+t_{2}}{\sigma _{1}}(1-\alpha ) - \frac{j/n}{2\sigma _{1}^{2}}(t_{1}^{2}+t_{2}^{2}) \nonumber \\&+ \frac{b}{2\sigma _{1}^{2}}(1-2b)r_{1}^{2b}(t_{1}^{2}+t_{2}^{2}) + \frac{b^{2}r_{2}^{2b}+(j/n-br_{2}^{2b})\alpha }{(j/n-br_{2}^{2b})^{2}} \bigg \} + \mathcal{O}\Big ( \frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{n} + \frac{1}{n^{2}} \Big ) \bigg ). \end{aligned}$$

  (6.9)

• 8. As $n \rightarrow + \infty$ with $j_{2,-} \le j \le g_{2,-}$, we have

  $$\begin{aligned}&F_{j} = \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})(\frac{br_{2}^{2b}}{1+\epsilon }-\sigma _{\star })}). \end{aligned}$$

  (6.10)

• 9. As $n \rightarrow + \infty$ with $g_{2,-} \le j \le g_{2,+}$, we have

  $$\begin{aligned}&F_{j} = \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})\sigma _{2}}). \end{aligned}$$

  (6.11)

• 10. As $n \rightarrow + \infty$ with $g_{2,+} \le j \le j_{2,+}$, we have

  $$\begin{aligned} F_{j}&= \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})\sigma _{2}}). \end{aligned}$$

  (6.12)

• 11. As $n \rightarrow + \infty$ with $j_{2,+} \le j \le n$, we have

  $$\begin{aligned} F_{j}&= \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})\sigma _{2}}). \end{aligned}$$

  (6.13)

Proof

Eq. (6.3) follows from Eqs. (4.9) and (6.2). Let us now prove Eq. (6.4). By Eqs. (4.10) and (6.2), as $n \rightarrow + \infty$ with $M' \le j \le j_{1,-}$, we have

$$\begin{aligned} F_{j} = \mathcal{O}\bigg (\exp \bigg \{ n(1-\delta ) \bigg [ \frac{1}{b}(j/n-j/n \log (\frac{j/n}{b})) - r_{1}^{2b} + 2j/n \log (r_{1}) \bigg ] \bigg \} \bigg ). \end{aligned}$$

It is easy to check that the function $y \mapsto y-y \log (\frac{y}{b}) + y \log (r_{1}^{2b})$ is increasing for $y\in [0,br_{1}^{2b}]$. Since $j/n \le \frac{br_{1}^{2b}}{1+\epsilon }+\mathcal{O}(n^{-1})$, we thus have

$$\begin{aligned} F_{j}&= \mathcal{O}\bigg (\exp \bigg \{ -nr_{1}^{2b}(1-\delta ) \frac{\epsilon - \log (1+\epsilon )}{1+\epsilon } \bigg \} \bigg ), \end{aligned}$$

which is Eq. (6.4). Now we prove Eq. (6.5). By Eqs. (4.11) and (6.2), as $n \rightarrow + \infty$ with $j_{1,-} \le j \le g_{1,-}$, we have

$$\begin{aligned} F_{j} = \mathcal{O}\bigg ( \exp \bigg \{ n(1-\delta ) \bigg [ \frac{j/n - j/n \log (\frac{j/n}{b})}{b}-r_{1}^{2b} + 2j/n \log (r_{1}) \bigg ] \bigg \} \bigg ). \end{aligned}$$

Since $[0,br_{1}^{2b}] \in y \mapsto y-y \log (\frac{y}{b}) + y \log (r_{1}^{2b})$ is increasing and $j/n \le \frac{b r_{1}^{2b}}{1+\frac{M}{\sqrt{n}}} + \mathcal{O}(n^{-1})$, we have

$$\begin{aligned} F_{j} = \mathcal{O}\bigg ( \exp \bigg \{ -(1-\delta ) \frac{M^{2}r_{1}^{2b}}{2} \bigg \} \bigg ). \end{aligned}$$

Since $M\ge M' \sqrt{\log n}$, Eq. (6.5) holds provided that $M'$ is chosen sufficiently large. Now we prove Eq. (6.6). As $n \rightarrow + \infty$ with $g_{1,-} \le j \le g_{1,+}$, we have

$$\begin{aligned}&e^{r_{1}^{2b}(1-\log (r_{1}^{2b}))n}e^{M_{j,1} r_{1}^{2b} \log (r_{1}^{2b})\sqrt{n}}e^{-r_{1}^{2b}(\frac{1}{2}+\log (r_{1}^{2b}))M_{j,1}^{2}} e^{-\frac{n}{2}|z|^{2b}}e^{-\frac{n}{2}|w|^{2b}}|z|^{\alpha }|w|^{\alpha }z^{j-1}\overline{w}^{j-1} \\&= e^{i(j-1)(\theta _{1}-\theta _{2})}e^{-\frac{1}{2}M_{j,1}^{2}r_{1}^{2b}}r_{1}^{-2} \bigg \{ 1 - \frac{b M_{j,1}r_{1}^{2b}}{\sqrt{n}}\Big (2M_{j,1}^{2}\log r_{1}-\frac{t_{1}+t_{2}}{\sigma _{1}}\Big ) \\&+ \frac{1}{n}\bigg ( 2b^{2}r_{1}^{4b}(\log r_{1})^{2} M_{j,1}^{6} + 2br_{1}^{2b}\big ( 1-br_{1}^{2b} \frac{t_{1}+t_{2}}{\sigma _{1}} \big ) \log (r_{1})M_{j,1}^{4} \bigg ) - \frac{r_{1}^{6b}(\log (r_{1}^{2b}))^{3}M_{j,1}^{9}}{6n^{3/2}} \\&\quad + \mathcal{O}\bigg (\frac{1+M_{j,1}^{2}}{n}+\frac{1+M_{j,1}^{7}}{n^{3/2}} + \frac{1+M_{j,1}^{12}}{n^{2}}\bigg ) \bigg \}. \end{aligned}$$

Now Eq. (6.6) follows directly from Eqs. (4.12) and (6.2). The proofs of Eqs. (6.7)–(6.9) follow in a similar way, using Eqs. (4.13)–(4.15), respectively.

Let us now prove Eq. (6.10). By Eqs. (4.16) and (6.2), as $n \rightarrow + \infty$ with $j_{2,-} \le j \le g_{2,-}$, we have

$$\begin{aligned} F_{j} = \mathcal{O}\bigg (\exp \bigg \{ n(1-\delta ) \bigg [ r_{2}^{2b} - 2 j/n \log r_{2} - r_{1}^{2b} + 2j/n \log (r_{1}) \bigg ] \bigg \} \bigg ). \end{aligned}$$

Since $j/n = \frac{br_{2}^{2b}}{1+\epsilon }$, Eq. (6.10) follows. The proof of Eq. (6.11) is similar, and uses the fact that for $g_{2,-} \le j \le g_{2,+}$, we have $j/n = br_{2}^{2b} + \mathcal{O}(n^{-1/2})$. The proofs of Eqs. (6.12) and (6.13) are also similar, and uses the fact that for $g_{2,+} \le j \le n$, we have $\frac{1}{b}(j/n - j/n \log (\frac{j/n}{b})) - r_{1}^{2b} + 2j/n \log (r_{1}) \le -2 \log (\frac{r_{2}}{r_{1}})\sigma _{2}$. $\square$

Lemma 6.2

As $n \rightarrow + \infty$, we have

$$\begin{aligned}&K_{n}(z,w) = S_{4} + S_{5} + S_{6} + S_{7} + \mathcal{O}(n^{-100}), \end{aligned}$$

where

$$\begin{aligned} S_{4} := \sum _{j=g_{1,-}}^{g_{1,+}}F_{j}, \qquad S_{5} := \sum _{j=g_{1,+}+1}^{j_{1,+}}F_{j}, \qquad S_{6} := \sum _{j=j_{1,+}+1}^{\lfloor j_{\star }\rfloor }F_{j}, \qquad S_{7} := \sum _{j=\lfloor j_{\star }\rfloor +1}^{j_{2,-}} F_{j}. \end{aligned}$$

Furthermore, as $n \rightarrow + \infty$ we have

$$\begin{aligned} S_{4}&= \sum _{j=g_{1,-}}^{g_{1,+}} e^{i(j-1)(\theta _{1}-\theta _{2})} \sqrt{n}\frac{\sqrt{2} \, b r_{1}^{b-2} e^{-\frac{M_{j,1}^{2}r_{1}^{2b}}{2}}}{\sqrt{\pi } {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}})} \bigg ( 1 + \frac{1}{\sqrt{n}}\bigg \{\frac{(5M_{j,1}^{2}r_{1}^{2b}-2)e^{-\frac{r_{1}^{2b}M_{j,1}^{2}}{2}}}{3\sqrt{2\pi } r_{1}^{b} {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}}) } \nonumber \\&\quad + \frac{M_{j,1}}{6} \Big ( 5M_{j,1}^{2}r_{1}^{2b}-3+6br_{1}^{2b}\frac{t_{1}+t_{2}}{\sigma _{1}} \Big ) \bigg \} \nonumber \\&\quad + \mathcal{O}\bigg (\frac{1+M_{j,1}^{2}+M_{j,1}^{6}\chi _{j,1}^{+}}{n}+\frac{1+M_{j,1}^{7}+M_{j,1}^{9}\chi _{j,1}^{+}}{n^{3/2}} + \frac{1+M_{j,1}^{12}}{n^{2}}\bigg ) \bigg ), \end{aligned}$$

(6.14)

$$\begin{aligned} S_{5}&= \sum _{j=g_{1,+}+1}^{j_{1,+}} e^{i(j-1)(\theta _{1}-\theta _{2})} n \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}(j/n-br_{1}^{2b})}{r_{1}^{2}}\bigg ( 1 + \frac{1}{n} \bigg \{ \frac{t_{1}+t_{2}}{\sigma _{1}}(1-\alpha ) \nonumber \\&\quad - \frac{j/n}{2\sigma _{1}^{2}}(t_{1}^{2}+t_{2}^{2}) + \frac{b}{2\sigma _{1}^{2}}(1-2b)r_{1}^{2b}(t_{1}^{2}+t_{2}^{2}) + \frac{b^{2}r_{1}^{2b}+(j/n-br_{1}^{2b})\alpha }{(j/n-br_{1}^{2b})^{2}} \bigg \} \nonumber \\&\quad - \frac{2 b^{4} r_{1}^{4b}}{n^{2}(j/n-br_{1}^{2b})^{4}}+ \frac{10b^{6}r_{1}^{6b}}{n^{3}(j/n-br_{1}^{2b})^{6}} + \mathcal{O}\Big ( \frac{1}{n^{2}(j/n-br_{1}^{2b})^{3}}\nonumber \\&\quad + \frac{1}{n^{4}(j/n-br_{1}^{2b})^{8}} \Big ) \bigg ), \end{aligned}$$

(6.15)

$$\begin{aligned} S_{6}&= \sum _{j=j_{1,+}+1}^{\lfloor j_{\star }\rfloor }e^{i(j-1)(\theta _{1}-\theta _{2})} \frac{n}{r_{1}^{2}} \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}}{\frac{1}{j/n-br_{1}^{2b}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{br_{2}^{2b}-j/n}} \bigg ( 1 + \frac{1}{n} \bigg \{ \frac{t_{1}+t_{2}}{\sigma _{1}}(1-\alpha ) \nonumber \\&\quad - \frac{j/n}{2\sigma _{1}^{2}}(t_{1}^{2}+t_{2}^{2}) + \frac{b}{2 \sigma _{1}^{2}}(1-2b)r_{1}^{2b}(t_{1}^{2}+t_{2}^{2}) + \frac{b^{2}r_{1}^{2b}+(j/n-br_{1}^{2b})\alpha }{(j/n-br_{1}^{2b})^{2}} \bigg \}\nonumber \\&\quad + \mathcal{O}\Big ( \frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{n} + \frac{1}{n^{2}} \Big ) \bigg ), \end{aligned}$$

(6.16)

$$\begin{aligned} S_{7}&= \sum _{j=\lfloor j_{\star }\rfloor +1}^{j_{2,-}} e^{i(j-1)(\theta _{1}-\theta _{2})}\frac{n}{r_{1}^{2}} \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{\frac{1}{br_{2}^{2b} - j/n}+\frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{j/n-br_{1}^{2b}}} \bigg ( 1 + \frac{1}{n} \bigg \{ \frac{t_{1}+t_{2}}{\sigma _{1}}(1-\alpha ) \nonumber \\&\quad - \frac{j/n}{2\sigma _{1}^{2}}(t_{1}^{2}+t_{2}^{2})+ \frac{b}{2\sigma _{1}^{2}}(1-2b)r_{1}^{2b}(t_{1}^{2}+t_{2}^{2}) + \frac{b^{2}r_{2}^{2b}+(j/n-br_{2}^{2b})\alpha }{(j/n-br_{2}^{2b})^{2}} \bigg \}\nonumber \\&\quad + \mathcal{O}\Big ( \frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{n} + \frac{1}{n^{2}} \Big ) \bigg ). \end{aligned}$$

(6.17)

Proof

This is a straightforward consequence of Lemma 6.1. $\square$

Lemma 6.3

As $n \rightarrow + \infty$,

$$\begin{aligned} S_{7} = \frac{n e^{i \lfloor j_{\star } \rfloor (\theta _{1}-\theta _{2})} \sigma _{1}}{r_{1}^{2} e^{t_{1}+t_{2}}}\sum _{j=0}^{+\infty } \frac{e^{i j(\theta _{1}-\theta _{2})}}{1 + \frac{\sigma _{1}}{\sigma _{2}}(\frac{r_{2}}{r_{1}})^{2(j+1-x)}} + \mathcal{O}\big ( (\log n)^{2} \big ), \end{aligned}$$

where $x=x(n)$ is given by Eq. (3.20).

Proof

By Eq. (6.17), as $n \rightarrow + \infty$,

$$\begin{aligned} S_{7} = \sum _{j=\lfloor j_{\star }\rfloor +1}^{j_{2,-}} e^{i(j-1)(\theta _{1}-\theta _{2})}\frac{n}{r_{1}^{2}} \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{\frac{1}{br_{2}^{2b} - j/n}+\frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{j/n-br_{1}^{2b}}} + \sum _{j=\lfloor j_{\star }\rfloor +1}^{j_{2,-}} \mathcal{O}\bigg ( \Big (\frac{r_{1}}{r_{2}}\Big )^{2(j-j_{\star })} \bigg ). \end{aligned}$$

The last sum is $\mathcal{O}(\log n)$. For the first sum, thanks to $(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}$ appearing in the numerator, the upper bound of summation $j_{2,-}$ can be replaced by $\lfloor j_{\star } \rfloor + M' \log n$ at the cost of an error $\mathcal{O}(\log n)$. We thus have

$$\begin{aligned} S_{7} = \sum _{\lfloor j_{\star }\rfloor +1}^{\lfloor j_{\star } \rfloor + M' \log n} e^{i(j-1)(\theta _{1}-\theta _{2})}\frac{n}{r_{1}^{2}} \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{\frac{1}{br_{2}^{2b} - j/n}+\frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{j/n-br_{1}^{2b}}} + \mathcal{O}\big ( \log n \big ). \end{aligned}$$

For $|j-\lfloor j_{\star } \rfloor | \le M' \log n$, we have $j/n = \sigma _{\star } + \mathcal{O}(\frac{\log n}{n})$, and thus

$$\begin{aligned} S_{7} = \frac{n e^{-t_{1}-t_{2}}}{r_{1}^{2}}\sum _{\lfloor j_{\star }\rfloor +1}^{\lfloor j_{\star } \rfloor + M' \log n} e^{i(j-1)(\theta _{1}-\theta _{2})} \frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{\frac{1}{\sigma _{2}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}}{\sigma _{1}}} + \mathcal{O}\big ( (\log n)^{2} \big ). \end{aligned}$$

In the above sum, we can replace $\lfloor j_{\star } \rfloor + M' \log n$ by $+\infty$ at the cost of an error $\mathcal{O}(\log n)$ (again thanks to $(\frac{r_{1}}{r_{2}})^{2(j-j_{\star })}$ appearing in the numerator). Then the claim follows from a simple shift of the indice of summation. $\square$

Lemma 6.4

As $n \rightarrow + \infty$,

$$\begin{aligned} S_{6}&= -\frac{n e^{i \lfloor j_{\star } \rfloor (\theta _{1}-\theta _{2})}\sigma _{1}}{r_{1}^{2}e^{t_{1}+t_{2}} } \sum _{j=0}^{+\infty } \frac{e^{-i(j+1)(\theta _{1}-\theta _{2})}}{1+\frac{\sigma _{2}}{\sigma _{1}}(\frac{r_{2}}{r_{1}})^{2(j+x)}} \\&\quad + \sum _{j=j_{1,+}+1}^{\lfloor j_{\star }\rfloor }e^{i(j-1)(\theta _{1}-\theta _{2})} \frac{n (j/n-br_{1}^{2b})}{r_{1}^{2}e^{(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}} \bigg ( 1 + \frac{1}{n} \bigg \{ \frac{t_{1}+t_{2}}{\sigma _{1}}(1-\alpha ) - \frac{j/n}{2\sigma _{1}^{2}}(t_{1}^{2}+t_{2}^{2}) \\&\quad + \frac{b}{2\sigma _{1}^{2}}(1-2b)r_{1}^{2b}(t_{1}^{2}+t_{2}^{2}) + \frac{b^{2}r_{1}^{2b}+(j/n-br_{1}^{2b})\alpha }{(j/n-br_{1}^{2b})^{2}} \bigg \} + \mathcal{O}\Big ( \frac{1}{n^{2}} \Big ) \bigg ) + \mathcal{O}\big ( (\log n)^{2} \big ). \end{aligned}$$

Proof

Let us split $S_{6} = S_{6}^{(1)}+S_{6}^{(2)}$, where

$$\begin{aligned} S_{6}^{(1)} = \sum _{j=j_{1,+}+1}^{\lfloor j_{\star } \rfloor - (M'+1) \log n} F_{j}, \qquad S_{6}^{(2)} = \sum _{j=\lfloor j_{\star } \rfloor - M' \log n}^{\lfloor j_{\star } \rfloor } F_{j}. \end{aligned}$$

Using Eq. (6.16) and a similar analysis as in the proof of Lemma 6.3, we obtain

$$\begin{aligned} S_{6}^{(2)}&= \mathcal{O}\big ( (\log n)^{2} \big ) + \frac{n}{ r_{1}^{2}} e^{-t_{1}-t_{2}}\sum _{j=\lfloor j_{\star } \rfloor - M' \log n}^{\lfloor j_{\star } \rfloor } \frac{e^{i(j-1)(\theta _{1}-\theta _{2})}}{\frac{1}{\sigma _{1}}} \nonumber \\&\quad + \frac{n}{r_{1}^{2}} e^{-t_{1}-t_{2}}\sum _{j=\lfloor j_{\star } \rfloor - M' \log n}^{\lfloor j_{\star } \rfloor } \bigg \{ \frac{e^{i(j-1)(\theta _{1}-\theta _{2})}}{\frac{1}{\sigma _{1}}+\frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{\sigma _{2}}} - \frac{e^{i(j-1)(\theta _{1}-\theta _{2})}}{\frac{1}{\sigma _{1}}} \bigg \} \end{aligned}$$

(6.18)

as $n \rightarrow + \infty$. The last term in Eq. (6.18) is easily shown to be

$$\begin{aligned} -\frac{n e^{i \lfloor j_{\star } \rfloor (\theta _{1}-\theta _{2})}\sigma _{1}}{r_{1}^{2}e^{t_{1}+t_{2}} } \sum _{j=0}^{+\infty } \frac{e^{-i(j+1)(\theta _{1}-\theta _{2})}}{1+\frac{\sigma _{2}}{\sigma _{1}}(\frac{r_{2}}{r_{1}})^{2(j+x)}} + \mathcal{O}(\log n) \end{aligned}$$

as $n \rightarrow + \infty$. For the first sum in Eq. (6.18), since $|j-\lfloor j_{\star } \rfloor | \le M' \log n$, we can replace $\sigma _{1}$ by $j/n-br_{1}^{2b}$ at the cost of an error $\mathcal{O}((\log n)^{2})$ in the asymptotics of $S_{6}^{(2)}$; we thus find

$$\begin{aligned} S_{6}^{(2)}&= \mathcal{O}\big ( (\log n)^{2} \big ) + \frac{n}{ r_{1}^{2}} e^{-t_{1}-t_{2}} \hspace{-0.25cm} \sum _{j=\lfloor j_{\star } \rfloor - M' \log n}^{\lfloor j_{\star } \rfloor } \frac{e^{i(j-1)(\theta _{1}-\theta _{2})}}{\frac{1}{j/n-br_{1}^{2b}}}\\&\quad -\frac{n e^{i \lfloor j_{\star } \rfloor (\theta _{1}-\theta _{2})}\sigma _{1}}{r_{1}^{2}e^{t_{1}+t_{2}} } \sum _{j=0}^{+\infty } \frac{e^{-i(j+1)(\theta _{1}-\theta _{2})}}{1+\frac{\sigma _{2}}{\sigma _{1}}(\frac{r_{2}}{r_{1}})^{2(j+x)}} \\&= \mathcal{O}\big ( (\log n)^{2} \big ) -\frac{n e^{i \lfloor j_{\star } \rfloor (\theta _{1}-\theta _{2})}\sigma _{1}}{r_{1}^{2}e^{t_{1}+t_{2}} } \sum _{j=0}^{+\infty } \frac{e^{-i(j+1)(\theta _{1}-\theta _{2})}}{1+\frac{\sigma _{2}}{\sigma _{1}}(\frac{r_{2}}{r_{1}})^{2(j+x)}} \\&\quad + \hspace{-0.25cm} \sum _{j=\lfloor j_{\star } \rfloor - M' \log n}^{\lfloor j_{\star } \rfloor } e^{i(j-1)(\theta _{1}-\theta _{2})} \frac{n}{r_{1}^{2}} \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}}{\frac{1}{j/n-br_{1}^{2b}}} \bigg ( 1 + \frac{1}{n} \bigg \{ \frac{t_{1}+t_{2}}{\sigma _{1}}(1-\alpha ) - \frac{j/n}{2\sigma _{1}^{2}}(t_{1}^{2}+t_{2}^{2}) \\&\quad + \frac{b}{2 \sigma _{1}^{2}}(1-2b)r_{1}^{2b}(t_{1}^{2}+t_{2}^{2}) + \frac{b^{2}r_{1}^{2b}+(j/n-br_{1}^{2b})\alpha }{(j/n-br_{1}^{2b})^{2}} \bigg \} + \mathcal{O}\Big ( \frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{n} + \frac{1}{n^{2}} \Big ) \bigg ) \end{aligned}$$

For $j_{1,+}+1 \le j \le \lfloor j_{\star } \rfloor - (M'+1) \log n$, we can replace $(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}$ by 0 in the asymptotics Eq. (6.8) of $F_{j}$ at the cost of an error $\mathcal{O}(n^{-100})$ (provided $M'$ is chosen large enough). We thus have

$$\begin{aligned}&S_{6}^{(1)} = \sum _{j=j_{1,+}+1}^{\lfloor j_{\star } \rfloor - (M'+1) \log n} e^{i(j-1)(\theta _{1}-\theta _{2})} \frac{n}{r_{1}^{2}} \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}}{\frac{1}{j/n-br_{1}^{2b}}} \bigg ( 1 + \frac{1}{n} \bigg \{ \frac{t_{1}+t_{2}}{\sigma _{1}}(1-\alpha ) - \frac{j/n}{2\sigma _{1}^{2}}(t_{1}^{2}+t_{2}^{2}) \nonumber \\&+ \frac{b}{2 \sigma _{1}^{2}}(1-2b)r_{1}^{2b}(t_{1}^{2}+t_{2}^{2}) + \frac{b^{2}r_{1}^{2b}+(j/n-br_{1}^{2b})\alpha }{(j/n-br_{1}^{2b})^{2}} \bigg \} + \mathcal{O}\Big ( \frac{(\frac{r_{1}}{r_{2}})^{2(j_{\star }-j)}}{n} + \frac{1}{n^{2}} \Big ) \bigg ), \end{aligned}$$

The claim follows after combining the above two expansions. $\square$

As a direct consequence of Lemmas 6.2, 6.3 and 6.4, we obtain the following.

Lemma 6.5

As $n \rightarrow + \infty$,

$$\begin{aligned}&S_{5} + S_{6} + S_{7} = S_{567} + \frac{n e^{i \lfloor j_{\star } \rfloor (\theta _{1}-\theta _{2})} \sigma _{1}}{ r_{1}^{2} e^{t_{1}+t_{2}}} \mathcal {Q}_{n} + \mathcal{O}\big ( (\log n)^{2} \big ), \end{aligned}$$

where

$$\begin{aligned}&S_{567} = \sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor }e^{i(j-1)(\theta _{1}-\theta _{2})} \frac{n (j/n-br_{1}^{2b})}{r_{1}^{2}e^{(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}} \bigg ( 1 + \frac{1}{n} \bigg \{ \frac{t_{1}+t_{2}}{\sigma _{1}}(1-\alpha ) - \frac{j/n}{2\sigma _{1}^{2}}(t_{1}^{2}+t_{2}^{2}) \nonumber \\&+ \frac{b}{2\sigma _{1}^{2}}(1-2b)r_{1}^{2b}(t_{1}^{2}+t_{2}^{2}) + \frac{b^{2}r_{1}^{2b}+(j/n-br_{1}^{2b})\alpha }{(j/n-br_{1}^{2b})^{2}} \bigg \} - \frac{2 b^{4} r_{1}^{4b}}{n^{2}(j/n-br_{1}^{2b})^{4}} \nonumber \\&+ \frac{10b^{6}r_{1}^{6b}}{n^{3}(j/n-br_{1}^{2b})^{6}} + \mathcal{O}\Big ( \frac{1}{n^{2}(j/n-br_{1}^{2b})^{3}} + \frac{1}{n^{4}(j/n-br_{1}^{2b})^{8}} \Big ) \bigg ), \end{aligned}$$

(6.19)

$$\begin{aligned}&\mathcal {Q}_{n} := \sum _{j=0}^{+\infty } \frac{e^{i j(\theta _{1}-\theta _{2})}}{1 + \frac{\sigma _{1}}{\sigma _{2}}(\frac{r_{2}}{r_{1}})^{2(j+1-x)}} - \sum _{j=0}^{+\infty } \frac{e^{-i(j+1)(\theta _{1}-\theta _{2})}}{1+\frac{\sigma _{2}}{\sigma _{1}}(\frac{r_{2}}{r_{1}})^{2(j+x)}}, \end{aligned}$$

(6.20)

and $x=x(n)$ is given by Eq. (3.20).

6.1 Proof of Theorem 3.1

In this subsection, we take $\theta _{1}=\theta _{2}$ and $M=n^{\frac{1}{10}}$.

Lemma 6.6

For $\theta _{1}=\theta _{2}$, we have

$$\begin{aligned} \mathcal {Q}_{n} = \frac{(\log \theta )'(n \sigma _{\star }-\alpha +\frac{\log (\frac{\sigma _{2}}{\sigma _{1}}\frac{r_{2}}{r_{1}})}{2\log \frac{r_{2}}{r_{1}}}; \frac{\pi i}{\log \frac{r_{2}}{r_{1}}}) + (2x-1)\log \frac{r_{2}}{r_{1}} + \log \frac{\sigma _{2}}{\sigma _{1}}}{2\log \frac{r_{2}}{r_{1}}}, \end{aligned}$$

(6.21)

where $x=x(n)$ is given by Eq. (3.20).

Proof

For $y \in \mathbb {R}$, $\rho \in (0,1)$ and $a>0$, define

$$\begin{aligned} \Theta (y;\rho ,a)&= y(y-1)\log (\rho ) + y \log (a) \nonumber \\&+ \sum _{j=0}^{+\infty } \log \bigg ( 1+a\, \rho ^{2(j+y)} \bigg ) + \sum _{j=0}^{+\infty } \log \bigg ( 1+a^{-1} \rho ^{2(j+1-y)} \bigg ). \end{aligned}$$

(6.22)

It is proved in [30, Lemma 3.25] that

$$\begin{aligned} \Theta (y;\rho ,a)&= \frac{1}{2}\log \bigg ( \frac{\pi a \rho ^{-\frac{1}{2}}}{\log (\rho ^{-1})} \bigg ) + \frac{(\log a)^{2}}{4\log (\rho ^{-1})} - \sum _{j=1}^{+\infty } \log (1-\rho ^{2j}) \\&\quad + \log \theta \bigg ( y + \frac{\log (a \rho )}{2\log (\rho )} \bigg | \frac{\pi i}{\log (\rho ^{-1})} \bigg ), \end{aligned}$$

where $\theta$ is the Jacobi theta function defined in Eq. (3.1). By taking the derivative with respect to y in the above identity, by replacing $(y,\rho ,a)$ by $(x,\rho ^{-1},a^{-1})$, we get

$$\begin{aligned}&\sum _{j=0}^{+\infty } \frac{1}{1+a^{-1}\rho ^{2(j+1-x)}} - \sum _{j=0}^{+\infty } \frac{1}{1+a\rho ^{2(j+x)}}\nonumber \\&\quad = \frac{(\log \theta )'(x+\frac{\log (a\rho )}{2\log \rho }; \frac{\pi i}{\log \rho }) + (2x-1)\log \rho + \log a}{2\log \rho }. \end{aligned}$$

(6.23)

The claim follows from Eq. (6.23) with $\rho = \frac{r_{2}}{r_{1}}>1$ and $a=\frac{\sigma _{2}}{\sigma _{1}}$, and from the facts that $x = n \sigma _{\star }-\alpha - \lfloor n \sigma _{\star }-\alpha \rfloor$ and $\theta (z+1|\tau )=\theta (z|\tau )$. $\square$

To expand $S_{567}$, we will need the following Riemann sum lemma.

Lemma 6.7

(taken from [30, Lemma 3.4]) Let $A,a_{0}$, $B,b_{0}$ be bounded function of $n \in \{1,2,\ldots \}$, such that

$$\begin{aligned}&a_{n} := An + a_{0} \qquad \hbox { and } \qquad b_{n} := Bn + b_{0} \end{aligned}$$

are integers. Assume also that $B-A$ is positive and remains bounded away from 0. Let f be a function independent of n, and which is $C^{4}([\min \{\frac{a_{n}}{n},A\},\max \{\frac{b_{n}}{n},B\}])$ for all $n\in \{1,2,\ldots \}$. Then as $n \rightarrow + \infty$, we have

$$\begin{aligned}&\sum _{j=a_{n}}^{b_{n}}f(\tfrac{j}{n}) = n \int _{A}^{B}f(y)dy + \frac{(1-2a_{0})f(A)+(1+2b_{0})f(B)}{2} \nonumber \\&\quad + \frac{(-1+6a_{0}-6a_{0}^{2})f'(A)+(1+6b_{0}+6b_{0}^{2})f'(B)}{12n}\nonumber \\&\quad + \frac{(-a_{0}+3a_{0}^{2}-2a_{0}^{3})f''(A)+(b_{0}+3b_{0}^{2}+2b_{0}^{3})f''(B)}{12n^{2}} \nonumber \\&\quad + \mathcal{O}\bigg ( \frac{\mathfrak {m}_{A,n}(f''')+\mathfrak {m}_{B,n}(f''')}{n^{3}} + \sum _{j=a_{n}}^{b_{n}-1} \frac{\mathfrak {m}_{j,n}(f'''')}{n^{4}} \bigg ), \end{aligned}$$

(6.24)

where, for a given function g continuous on $[\min \{\frac{a_{n}}{n},A\},\max \{\frac{b_{n}}{n},B\}]$,

$$\begin{aligned} \mathfrak {m}_{A,n}(g) := \max _{y \in [\min \{\frac{a_{n}}{n},A\},\max \{\frac{a_{n}}{n},A\}]}|g(y)|, \quad \mathfrak {m}_{B,n}(g) := \max _{y \in [\min \{\frac{b_{n}}{n},B\},\max \{\frac{b_{n}}{n},B\}]}|g(y)|, \end{aligned}$$

and for $j \in \{a_{n},\ldots ,b_{n}-1\}$, $\mathfrak {m}_{j,n}(g):= \max _{y \in [\frac{j}{n},\frac{j+1}{n}]}|g(y)|$.

It turns out that $S_{567}$ has oscillatory asymptotics as $n \rightarrow + \infty$. To handle these oscillations, we follow [29] and introduce the following quantities:

$$\begin{aligned}&\theta _{-}^{(n,M)} := g_{-} - \bigg ( \frac{bn r^{2b}}{1+\frac{M}{\sqrt{n}}} - \alpha \bigg ) = \bigg \lceil \frac{bn r^{2b}}{1+\frac{M}{\sqrt{n}}} - \alpha \bigg \rceil - \bigg ( \frac{bn r^{2b}}{1+\frac{M}{\sqrt{n}}} - \alpha \bigg ), \\&\theta _{+}^{(n,M)} := \bigg ( \frac{bn r^{2b}}{1-\frac{M}{\sqrt{n}}} - \alpha \bigg ) - g_{+} = \bigg ( \frac{bn r^{2b}}{1-\frac{M}{\sqrt{n}}} - \alpha \bigg ) - \bigg \lfloor \frac{bn r^{2b}}{1-\frac{M}{\sqrt{n}}} - \alpha \bigg \rfloor . \end{aligned}$$

Note that $\theta _{-}^{(n,M)},\theta _{+}^{(n,M)} \in [0,1)$ are oscillatory but remain bounded as $n \rightarrow + \infty$. Recall that in this section, we take $\theta _{2} = \theta _{1}$.

Lemma 6.8

If $(t_{1},t_{2})\ne (0,0)$, then as $n \rightarrow +\infty$,

$$\begin{aligned}&S_{567} = D_{1} n^{2} + D_{2} \, n \log n + D_{3}^{(n,M)} n + D_{4}^{(n,M)}\sqrt{n} + \mathcal{O}\Big (M^{4} + \frac{n}{M^{6}} + \frac{\sqrt{n}}{M} \Big ), \nonumber \\&D_{1} := \sigma _{1}^{2}\frac{1-e^{-t_{1}-t_{2}}\big ( 1+t_{1}+t_{2} \big )}{r_{1}^{2}(t_{1}+t_{2})^{2}}, \qquad D_{2} := \frac{b^{2}r_{1}^{2b-2}}{2}, \nonumber \\&D_{3}^{(n,M)} := -b^{2} r_{1}^{2b-2}\Big ( E_{1}\big ( t_{1}+t_{2} \big ) + \gamma _{\textrm{E}} + \log (br_{1}^{2b}\frac{t_{1}+t_{2}}{\sigma _{1}}M) \Big ) - \frac{b^{2}M^{2}r_{1}^{4b-2}}{2} \nonumber \\&\quad + \frac{(1-2x)\sigma _{1}}{2 r_{1}^{2}e^{t_{1}+t_{2}}} + \frac{\sigma _{1}}{r_{1}^{2}(t_{1}+t_{2})^{3}} \bigg \{ \frac{(t_{1}+t_{2})^{2}(t_{1}^{2}+t_{2}^{2})}{2e^{t_{1}+t_{2}}} \nonumber \\&\quad + \Big ( 2t_{1}t_{2} - b^{2}r_{1}^{2b}\frac{t_{1}+t_{2}}{\sigma _{1}}(t_{1}^{2}+t_{2}^{2}) \Big ) \bigg ( 1 - \frac{1+t_{1}+t_{2}}{e^{t_{1}+t_{2}}} \bigg ) \bigg \} - \frac{b^{2}}{M^{2} r_{1}^{2}} + \frac{5b^{2}}{2M^{4} r_{1}^{2+2b}}, \nonumber \\&D_{4}^{(n,M)} := \frac{b^{2} r_{1}^{4b} \big ( br_{1}^{2b}\frac{t_{1}+t_{2}}{\sigma _{1}}-3 \big )}{3 r_{1}^{2}}M^{3} + \frac{br_{1}^{2b} \big ( 2\theta _{+}^{(n,M)} - 1 - 2b + 2b^{2}r_{1}^{2b}\frac{t_{1}+t_{2}}{\sigma _{1}} \big )}{2 r_{1}^{2}}M\nonumber \\&\quad \quad \quad + \frac{(2\alpha +2\theta _{+}^{(n,M)}-1)b}{2 r_{1}^{2} M}. \end{aligned}$$

(6.25)

If $t_{1}=t_{2}=0$, then Eq. (6.25) holds with

$$\begin{aligned}&D_{1} = \frac{\sigma _{1}^{2}}{2 r_{1}^{2}}, \quad D_{2} = \frac{b^{2}r_{1}^{2b-2}}{2}, \\&D_{3}^{(n,M)} = - \frac{b^{2}M^{2}r_{1}^{4b-2}}{2} + \frac{(\frac{1}{2}-x)\sigma _{1}}{ r_{1}^{2}} - b^{2}r_{1}^{2b-2}\log \frac{Mbr_{1}^{2b}}{\sigma _{1}} - \frac{b^{2}}{M^{2} r_{1}^{2}} + \frac{5b^{2}}{2M^{4} r_{1}^{2+2b}}, \\&D_{4}^{(n,M)} = -b^{2} r_{1}^{4b-2} M^{3} + \frac{br_{1}^{2b} \big ( 2\theta _{+}^{(n,M)} - 1 - 2b \big )}{2 r_{1}^{2}}M + \frac{(2\alpha +2\theta _{+}^{(n,M)}-1)b}{2 r_{1}^{2} M}. \end{aligned}$$

Proof

By Eq. (6.19) with $\theta _{1}=\theta _{2}$, as $n \rightarrow + \infty$ we have

$$\begin{aligned} S_{567}&= n \sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor } f_{1}(j/n) + \sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor } f_{2}(j/n) \nonumber \\&\quad + \frac{1}{n} \sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor } \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}}{r_{1}^{2}}\frac{-2 b^{4} r_{1}^{4b}}{(j/n-br_{1}^{2b})^{3}} \nonumber \\&\quad + \frac{1}{n^{2}} \sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor } \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}10b^{6}}{ r_{1}^{2-6b}(j/n-br_{1}^{2b})^{5}}\nonumber \\&\quad + \mathcal{O}\Big ( \frac{1}{n}\sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor }\frac{1}{(j/n-br_{1}^{2b})^{2}} + \frac{1}{n^{3}}\sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor } \frac{1}{(j/n-br_{1}^{2b})^{7}} \Big ), \end{aligned}$$

(6.26)

where $f_{1}(x):= \frac{x-br_{1}^{2b}}{r_{1}^{2}e^{(x-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}}$ and

$$\begin{aligned} f_{2}(x)&:= \frac{x-br_{1}^{2b}}{r_{1}^{2}e^{(x-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}} \bigg ( \frac{t_{1}+t_{2}}{\sigma _{1}}(1-\alpha ) + \frac{b(1-2b)r_{1}^{2b}-x}{2\sigma _{1}^{2}}(t_{1}^{2}+t_{2}^{2})\\&\quad + \frac{b^{2}r_{1}^{2b}+(x-br_{1}^{2b})\alpha }{(x-br_{1}^{2b})^{2}} \bigg ). \end{aligned}$$

Since $g_{1,+} = \big \lfloor \frac{bnr_{1}^{2b}}{1-\frac{M}{\sqrt{n}}}-\alpha \big \rfloor$, for the error term in Eq. (6.26) we have the following estimate:

$$\begin{aligned} \frac{1}{n}\sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor }\frac{1}{(j/n-br_{1}^{2b})^{2}} + \frac{1}{n^{3}}\sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor } \frac{1}{(j/n-br_{1}^{2b})^{7}} \\ = \mathcal{O}\bigg ( \int _{M/\sqrt{n}}^{1} \frac{dx}{x^{2}} + \frac{1}{n^{2}}\int _{M/\sqrt{n}}^{1} \frac{dx}{x^{7}} \bigg ) = \mathcal{O}\bigg ( \frac{\sqrt{n}}{M} + \frac{n}{M^{6}} \bigg ). \end{aligned}$$

Since $f_{1}$ has no pole at $br_{1}^{2b}$ and $f_{2}$ has a pole of order 1 at $br_{1}^{2b}$, long but direct calculations using Lemma 6.7 with $A=\frac{br_{1}^{2b}}{1-\frac{M}{\sqrt{n}}}$, $a_{0}=1-\alpha -\theta _{+}^{(n,M)}$, $B=\sigma _{\star }$ and $b_{0}=-\alpha -x$ show that

$$\begin{aligned}&n \sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor } f_{1}(j/n) = d_{1} n^{2} + (d_{2,1}M^{2}+d_{2,2})n + (d_{3,1}M^{3} + d_{3,2}M)\sqrt{n} + \mathcal{O}(M^{4}), \\&\sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor } f_{2}(j/n) = e_{1} n \log n + (e_{2,1} \log M + e_{2,2}) n + (e_{3,1}M + e_{3,2}M^{-1})\sqrt{n} + \mathcal{O}(M^{2}), \\&\frac{1}{n} \sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor } \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}}{r_{1}^{2}}\frac{-2 b^{4} r_{1}^{4b}}{(j/n-br_{1}^{2b})^{3}} = - \frac{b^{2}n}{ r_{1}^{2} M^{2}} + \mathcal{O}\Big ( \frac{\sqrt{n}}{M} \Big ), \\&\frac{1}{n^{2}} \sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor } \frac{e^{-(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}10b^{6}}{r_{1}^{2-6b}(j/n-br_{1}^{2b})^{5}} = \frac{5b^{2}n}{2 r_{1}^{2+2b}M^{4}} + \mathcal{O}\Big ( \frac{\sqrt{n}}{M^{3}} \Big ), \end{aligned}$$

as $n \rightarrow + \infty$, for some explicit coefficients $d_{1},d_{2,1},d_{2,2},d_{3,1},d_{3,2},e_{1}, e_{2,1},e_{2,2},e_{3,1},e_{3,2}$ independent of M that we do not write down. We then find the claim by combining the above two asymptotic formulas and simplifying (using primitives). $\square$

To expand $S_{4}$, we will need the following lemma.

Lemma 6.9

(Taken from [30, Lemma 3.10]) Let $h \in C^{3}(\mathbb {R})$ and $k =1$. As $n \rightarrow + \infty$, we have

$$\begin{aligned}&\sum _{j=g_{k,-}}^{g_{k,+}} \hspace{-0.12cm} h(M_{j,k}) = br_{k}^{2b} \hspace{-0.07cm} \int _{-M}^{M} \hspace{-0.07cm} h(t) dt \; \sqrt{n} - 2 b r_{k}^{2b} \hspace{-0.07cm}\int _{-M}^{M} \hspace{-0.07cm} th(t) dt + \hspace{-0.07cm} \bigg ( \frac{1}{2}-\theta _{k,-}^{(n,M)} \bigg )h(M)\nonumber \\&\quad + \hspace{-0.07cm} \bigg ( \frac{1}{2}-\theta _{k,+}^{(n,M)} \bigg )h(-M) \nonumber \\&\quad + \frac{1}{\sqrt{n}}\bigg [ 3br_{k}^{2b} \hspace{-0.07cm} \int _{-M}^{M} \hspace{-0.1cm} t^{2}h(t)dt + \hspace{-0.07cm} \bigg ( \hspace{-0.04cm}\frac{1}{12}+\frac{\theta _{k,-}^{(n,M)}(\theta _{k,-}^{(n,M)}\hspace{-0.04cm}-\hspace{-0.04cm}1)}{2} \bigg )\frac{h'(M)}{br_{k}^{2b}} \nonumber \\&\quad - \bigg ( \hspace{-0.04cm}\frac{1}{12}+\frac{\theta _{k,+}^{(n,M)}(\theta _{k,+}^{(n,M)}-1)}{2} \bigg )\frac{h'(-M)}{br_{k}^{2b}} \bigg ] \nonumber \\&\quad + \mathcal{O}\Bigg ( \frac{1}{n^{3/2}} \hspace{-0.07cm} \sum _{j=g_{k,-}+1}^{g_{k,+}} \bigg ( \hspace{-0.07cm} (1+|M_{j,1}|^{3}) \tilde{\mathfrak {m}}_{j,n}(h) + (1+M_{j,1}^{2})\tilde{\mathfrak {m}}_{j,n}(h') \nonumber \\&\quad + (1+|M_{j,1}|) \tilde{\mathfrak {m}}_{j,n}(h'') + \tilde{\mathfrak {m}}_{j,n}(h''') \bigg ) \hspace{-0.07cm} \Bigg ), \end{aligned}$$

(6.27)

where, for $\tilde{h} \in C(\mathbb {R})$ and $j \in \{g_{k,-}+1,\ldots ,g_{k,+}\}$, we define $\tilde{\mathfrak {m}}_{j,n}(\tilde{h}):= \max _{x \in [M_{j,k},M_{j-1,k}]}|\tilde{h}(x)|$.

Lemma 6.10

As $n \rightarrow + \infty$,

$$\begin{aligned}&S_{4} = E_{3}^{(M)} n + E_{4}^{(M)} \sqrt{n} + \mathcal{O}\Big (M^{4} + \frac{M^{9}}{\sqrt{n}} + \frac{M^{14}}{n^{3/2}} \Big ), \\&E_{3}^{(M)} := br_{1}^{2b} \int _{-M}^{M} h_{1}(t)dt, \\&E_{4}^{(M)} := br_{1}^{2b} \int _{-M}^{M} (h_{2}(t)-2th_{1}(t))dt + \bigg ( \frac{1}{2}-\theta _{-}^{(n,M)} \bigg )h_{1}(M) + \bigg ( \frac{1}{2} - \theta _{+}^{(n,M)} \bigg )h_{1}(-M), \end{aligned}$$

where

$$\begin{aligned}&h_{1}(y) := \frac{\sqrt{2}\, b r_{1}^{b-2}e^{-\frac{r_{1}^{2b}}{2}y^{2}}}{\sqrt{\pi } {\textrm{erfc}}(-\frac{r_{1}^{b}y}{\sqrt{2}})}, \\&h_{2}(y) := \frac{\sqrt{2}\, b r_{1}^{b-2}e^{-\frac{r_{1}^{2b}}{2}y^{2}}}{\sqrt{\pi } {\textrm{erfc}}(-\frac{r_{1}^{b}y}{\sqrt{2}})} \Bigg ( \frac{y}{6}\Big ( 5y^{2}r_{1}^{2b}-3+6br_{1}^{2b}(t_{1}+t_{2}) \Big ) + \frac{e^{-\frac{r_{1}^{2b}}{2}y^{2}}(5y^{2}r_{1}^{2b}-2)}{3\sqrt{2\pi } r_{1}^{b} {\textrm{erfc}} (-\frac{r_{1}^{b}y}{\sqrt{2}})} \Bigg ). \end{aligned}$$

Proof

By Eq. (6.14), as $n \rightarrow + \infty$ we have

$$\begin{aligned} S_{4}&= \sqrt{n} \sum _{j=g_{1,-}}^{g_{1,+}}h_{1}(M_{j,1}) + \sum _{j=g_{1,-}}^{g_{1,+}}h_{2}(M_{j,1})\\&\quad + \mathcal{O}\bigg ( \sum _{j=g_{1,-}}^{g_{1,+}} \Big ( \frac{1+M_{j,1}^{3}}{\sqrt{n}} + \frac{1+M_{j,1}^{8}}{n} + \frac{1+M_{j,1}^{13}}{n^{3/2}} \Big ) \bigg ). \end{aligned}$$

Since $\sum _{j=g_{1,-}}^{g_{1,+}} 1 = \mathcal{O}(M\sqrt{n})$, the above error term can be rewritten as $\mathcal{O}(M^{4} + \frac{M^{9}}{\sqrt{n}} + \frac{M^{14}}{n^{3/2}})$. We then use Lemma 6.9 to expand the two sums involving $h_{1}$ and $h_{2}$, and the claim follows. $\square$

Recall that $\mathcal {I}$ is define by Eq. (3.7). Define also

$$\begin{aligned}&\mathcal {I}_1 = \int _{-\infty }^{+\infty } \bigg \{ \frac{e^{-y^{2}}}{\sqrt{\pi }\, {\textrm{erfc}}(y)} - \chi _{(0,+\infty )}(y) \bigg [ y + \frac{y}{2(1+y^{2})} \bigg ] \bigg \}dy, \end{aligned}$$

(6.28)

$$\begin{aligned}&\mathcal {I}_{2} = \int _{-\infty }^{+\infty } \bigg \{ \frac{y^{3}e^{-y^{2}}}{\sqrt{\pi } \, {\textrm{erfc}}(y)} - \chi _{(0,+\infty )}(y) \bigg [ y^{4}+\frac{y^{2}}{2}-\frac{1}{2} \bigg ] \bigg \}dy, \end{aligned}$$

(6.29)

$$\begin{aligned}&\mathcal {I}_{3} = \int _{-\infty }^{+\infty } \bigg \{ \bigg ( \frac{e^{-y^{2}}}{\sqrt{\pi } \, {\textrm{erfc}}(y)} \bigg )^{2} - \chi _{(0,+\infty )}(y) \bigg [ y^{2}+1 \bigg ] \bigg \} dy, \end{aligned}$$

(6.30)

$$\begin{aligned}&\mathcal {I}_{4} = \int _{-\infty }^{+\infty } \bigg \{ \bigg ( \frac{y \, e^{-y^{2}}}{\sqrt{\pi } \, {\textrm{erfc}}(y)} \bigg )^{2} - \chi _{(0,+\infty )}(y)\bigg [ y^{4}+y^{2}-\frac{3}{4} \bigg ] \bigg \} dy. \end{aligned}$$

(6.31)

Lemma 6.11

If $(t_{1},t_{2}) \ne (0,0)$, then as $n \rightarrow + \infty$,

$$\begin{aligned} S_{4} + S_{567}&= C_{1} n^{2} + C_{2} \, n \log n + \bigg ( \tilde{C}_{3}+\frac{(1-2x)\sigma _{1}}{2\pi r_{1}^{2}e^{t_{1}+t_{2}}} \bigg ) n \nonumber \\&+ \tilde{C}_{4} \sqrt{n} + \mathcal{O}\Big (M^{4} + \frac{M^{9}}{\sqrt{n}} + \frac{M^{14}}{n^{3/2}}+\frac{n}{M^{6}} + \frac{\sqrt{n}}{M}\Big ), \end{aligned}$$

(6.32)

where $C_{1},C_{2}$ are as in the statement of Theorem 3.1 and $\tilde{C}_{3},\tilde{C}_{4}$ are given by

$$\begin{aligned}&\tilde{C}_{3} = 2b^{2}r_{1}^{2b-2} \mathcal {I}_{1} - \frac{b^{2}r_{1}^{2b-2}}{2}\log \Big ( \frac{2}{r_{1}^{2b}} \Big ) - b^{2} r_{1}^{2b-2}\Big ( E_{1}( t_{1}+t_{2}) + \gamma _{\textrm{E}} + \log (br_{1}^{2b}\frac{t_{1}+t_{2}}{\sigma _{1}}) \Big ) \\&+ \frac{\sigma _{1}}{r_{1}^{2}(t_{1}+t_{2})^{3}} \bigg \{ \frac{(t_{1}+t_{2})^{2}(t_{1}^{2}+t_{2}^{2})}{2e^{t_{1}+t_{2}}} \\&\hspace{2.7cm} + \Big ( 2t_{1}t_{2} - \frac{b^{2}r_{1}^{2b}}{\sigma _{1}}(t_{1}+t_{2})(t_{1}^{2}+t_{2}^{2})\Big ) \bigg ( 1 - \frac{1+t_{1}+t_{2}}{e^{t_{1}+t_{2}}} \bigg ) \bigg \}, \\&\tilde{C}_{4} = \frac{\sqrt{2} b^{2} r_{1}^{b-2}}{3} \bigg \{ \big [15-6br_{1}^{2b}\frac{t_{1}+t_{2}}{\sigma _{1}}\big ]\mathcal {I} - 10 \, \mathcal {I}_{2} - 2 \, \mathcal {I}_{3} + 10 \, \mathcal {I}_{4} \bigg \}. \end{aligned}$$

If $t_{1}=t_{2}=0$, then Eq. (6.32) holds with

$$\begin{aligned}&C_{1} = \frac{\sigma _{1}^{2}}{2 r_{1}^{2}}, \qquad C_{2} = \frac{b^{2}r_{1}^{2b-2}}{2}, \qquad \tilde{C}_{3} = 2b^{2}r_{1}^{2b-2} \mathcal {I}_{1} + \frac{b^{2}r_{1}^{2b-2}}{2} \log \Big ( \frac{\sigma _{1}^{2}}{2b^{2}r_{1}^{2b}} \Big ), \\&\tilde{C}_{4} = \frac{\sqrt{2} b^{2} r_{1}^{b-2}}{3} \bigg \{ 15\, \mathcal {I} - 10 \, \mathcal {I}_{2} - 2 \, \mathcal {I}_{3} + 10 \, \mathcal {I}_{4} \bigg \}. \end{aligned}$$

Proof

Combining Lemmas 6.8 and 6.10, we infer that as $n \rightarrow + \infty$,

$$\begin{aligned} S_{4} + S_{567} = D_{1} n^{2} + D_{2} \, n \log n + (D_{3}^{(n,M)}+E_{3}^{(M)}) n + (D_{4}^{(M)}+E_{4}^{(M)})\sqrt{n} + \mathcal{O}\Big (M^{4}\Big ). \end{aligned}$$

On the other hand, as $n \rightarrow + \infty$,

$$\begin{aligned}&(D_{3}^{(n,M)}+E_{3}^{(M)})n = \bigg ( \tilde{C}_{3}+\frac{(1-2x)\sigma _{1}}{2 r_{1}^{2}e^{t_{1}+t_{2}}} \bigg )n + \mathcal{O}\Big (\frac{n}{M^{2}}\Big ), \\&(D_{4}^{(M)}+E_{4}^{(M)})\sqrt{n} = \tilde{C}_{4}\sqrt{n} + \mathcal{O}\Big (\frac{\sqrt{n}}{M}\Big ), \end{aligned}$$

and the claim follows. $\square$

Since $M=n^{\frac{1}{10}}$, we have $M^{4} = \frac{M^{9}}{\sqrt{n}} = \frac{M^{14}}{n^{3/2}} = \frac{n}{M^{6}} = \frac{\sqrt{n}}{M} = n^{\frac{2}{5}}$, and thus the $\mathcal{O}$-term in Eq. (6.32) can be written as $\mathcal{O}(n^{\frac{2}{5}})$. Hence, combining Lemmas 6.2, 6.5 and 6.11 yields

$$\begin{aligned} K_{n}(z,w) = C_{1} n^{2} + C_{2} \, n \log n + \bigg ( \tilde{C}_{3}+\frac{(\frac{1}{2}-x+\mathcal {Q}_{n})\sigma _{1}}{r_{1}^{2}e^{t_{1}+t_{2}}} \bigg ) n + \tilde{C}_{4} \sqrt{n} + \mathcal{O}(n^{\frac{2}{5}}). \end{aligned}$$

Furthermore, by Eq. (3.5) and Lemma 6.6, we have $\frac{1}{2}-x+\mathcal {Q}_{n}=\mathcal {F}_{n}$, where $\mathcal {F}_{n}$ is given by Eq. (3.5). Finally, using the following identities (proved in [11, Lemma 2.10])

$$\begin{aligned}&\mathcal {I}_{1} = \frac{\log (2\sqrt{\pi })}{2},{} & {} \mathcal {I}_{3} = \mathcal {I},{} & {} \mathcal {I}_{4} = \mathcal {I}_{2}-\mathcal {I}, \end{aligned}$$

(6.33)

we obtain that $\tilde{C}_{3}=C_{3}$ and $\tilde{C}_{4}=C_{4}$, where $C_{3}$ and $C_{4}$ are as in the statement of Theorem 3.1. This finishes the proof of Theorem 3.1.

6.2 Proof of Theorem 3.13

In this subsection we assume that $\theta _{1}\ne \theta _{2} \mod 2\pi$, and we take $M=M' \sqrt{\log n}$. By Eq. (6.19) we have

$$\begin{aligned} S_{567}&= \frac{n}{r_{1}^{2}} \sum _{j=g_{1,+}+1}^{\lfloor j_{\star }\rfloor } \hspace{-0.15cm} e^{i(j-1)(\theta _{1}-\theta _{2})} \frac{j/n-br_{1}^{2b}}{e^{(j/n-br_{1}^{2b})\frac{t_{1}+t_{2}}{\sigma _{1}}}} \bigg ( \hspace{-0.07cm} 1+ \mathcal{O}\bigg ( \frac{1}{n(j/n-br_{1}^{2b})^{2}} \bigg ) \hspace{-0.07cm} \bigg ) \hspace{-0.07cm}\\&= \frac{n e^{i \lfloor j_{\star }\rfloor (\theta _{1}-\theta _{2})}}{r_{1}^{2}} \sum _{\ell = 1}^{N} c_{\ell }d_{\ell } \end{aligned}$$

where $N:= \lfloor j_{\star }\rfloor -g_{1,+}$, $d_{\ell }:= p^{\ell }$, $p:=e^{-i (\theta _{1}-\theta _{2})}$, and $c_{\ell }$ satisfies

$$\begin{aligned} c_{\ell }&= \frac{-\ell /n-br_{1}^{2b} + \frac{\lfloor j_{\star }\rfloor +1}{n}}{e^{(-\ell /n-br_{1}^{2b} + \frac{\lfloor j_{\star }\rfloor +1}{n})\frac{t_{1}+t_{2}}{\sigma _{1}}}}\bigg ( 1 + \frac{1}{n} \bigg \{ \frac{t_{1}+t_{2}}{\sigma _{1}}(1-\alpha ) - \frac{-\ell /n + \frac{\lfloor j_{\star }\rfloor +1}{n}}{2}\frac{t_{1}^{2}+t_{2}^{2}}{\sigma _{1}^{2}} \nonumber \\&\quad + \frac{b}{2\sigma _{1}^{2}}(1-2b)r_{1}^{2b}(t_{1}^{2}+t_{2}^{2}) + \frac{b^{2}r_{1}^{2b}+(-\ell /n-br_{1}^{2b} + \frac{\lfloor j_{\star }\rfloor +1}{n})\alpha }{(-\ell /n-br_{1}^{2b} + \frac{\lfloor j_{\star }\rfloor +1}{n})^{2}} \bigg \}\nonumber \\&\quad - \frac{2 b^{4} r_{1}^{4b}}{n^{2}(-\ell /n-br_{1}^{2b} + \frac{\lfloor j_{\star }\rfloor +1}{n})^{4}} \nonumber \\&+ \frac{10b^{6}r_{1}^{6b}}{n^{3}(-\ell /n-br_{1}^{2b} + \frac{\lfloor j_{\star }\rfloor +1}{n})^{6}} + \mathcal{O}\Big ( \frac{1}{n^{2}(\ell /n-\sigma _{1})^{3}} + \frac{1}{n^{4}(\ell /n-\sigma _{1})^{8}} \Big ) \bigg ) \end{aligned}$$

(6.34)

as $n\rightarrow +\infty$. Using summation by parts, we get

$$\begin{aligned} S_{567} = \frac{n e^{i \lfloor j_{\star }\rfloor (\theta _{1}-\theta _{2})}}{r_{1}^{2}} \bigg ( c_{N}D_{N}-c_{1}D_{0}- \frac{c_{N}-c_{1}}{1-p} + \sum _{\ell =1}^{N-1} (c_{\ell +1}-c_{\ell }) \frac{p^{\ell +1}}{1-p} \bigg ), \end{aligned}$$

(6.35)

where $D_{\ell }:= \sum _{j=0}^{\ell }d_{j}$. Note that

$$\begin{aligned}&\frac{n e^{i \lfloor j_{\star }\rfloor (\theta _{1}-\theta _{2})}}{r_{1}^{2}} \bigg ( -c_{1}D_{0}- \frac{-c_{1}}{1-p} \bigg ) = \frac{n e^{i \lfloor j_{\star }\rfloor (\theta _{1}-\theta _{2})}}{r_{1}^{2}} \frac{p}{1-p} \frac{\sigma _{1}}{e^{t_{1}+t_{2}}}\big ( 1+ \mathcal{O}(n^{-1}) \big ), \end{aligned}$$

(6.36)

$$\begin{aligned}&\frac{n e^{i \lfloor j_{\star }\rfloor (\theta _{1}-\theta _{2})}}{r_{1}^{2}} \bigg ( c_{N}D_{N} - \frac{c_{N}}{1-p} \bigg ) = \mathcal{O}(M\sqrt{n}), \end{aligned}$$

(6.37)

where for Eq. (6.37) we have used that $-N/n-br_{1}^{2b} + \frac{\lfloor j_{\star }\rfloor +1}{n} = \mathcal{O}(M/\sqrt{n})$. Let us now prove that

$$\begin{aligned} \bigg | \frac{n e^{i \lfloor j_{\star }\rfloor (\theta _{1}-\theta _{2})}}{r_{1}^{2}} \sum _{\ell =1}^{N-1} (c_{\ell +1}-c_{\ell }) \frac{p^{\ell +1}}{1-p} \bigg | = \mathcal{O}\bigg ( \frac{\sqrt{n}}{M} \bigg ). \end{aligned}$$

(6.38)

From Eq. (6.34) we deduce that

$$\begin{aligned} c_{\ell +1}-c_{\ell } = c_{\ell } \bigg ( \frac{1+(\ell /n-\sigma _{1})\frac{t_{1}+t_{2}}{\sigma _{1}}}{(\ell /n-\sigma _{1})n} + \mathcal{O}\bigg ( \frac{1}{n^{2}(\ell /n-\sigma _{1})^{3}} \bigg ) \bigg ), \qquad \hbox {as } n \rightarrow + \infty . \end{aligned}$$

(6.39)

Using Eq. (6.39) and then Eq. (6.34), we get

$$\begin{aligned}&\sum _{\ell =1}^{N-1} (c_{\ell +1}-c_{\ell }) \frac{p^{\ell +1}}{1-p} \nonumber \\&\quad = \frac{1}{n}\frac{p}{1-p} \sum _{\ell =1}^{N-1} c_{\ell } \frac{1+(\ell /n-\sigma _{1})\frac{t_{1}+t_{2}}{\sigma _{1}}}{\ell /n-\sigma _{1}} p^{\ell } + \sum _{\ell =1}^{N-1} \mathcal{O}\bigg ( \frac{1}{n^{2}(\ell /n-\sigma _{1})^{2}} \bigg ) \nonumber \\&\quad = \frac{1}{n}\frac{p}{1-p} \sum _{\ell =1}^{N-1} \tilde{c}_{\ell } p^{\ell } + \mathcal{O}\bigg ( \frac{1}{M\sqrt{n}} \bigg ), \qquad \tilde{c}_{\ell } := \frac{1+(\ell /n-\sigma _{1})\frac{t_{1}+t_{2}}{\sigma _{1}}}{e^{(\ell /n-\sigma _{1})\frac{t_{1}+t_{2}}{\sigma _{1}}}}. \end{aligned}$$

(6.40)

Summing again by parts, we get

$$\begin{aligned} \sum _{\ell =1}^{N-1} \tilde{c}_{\ell } p^{\ell } = \tilde{c}_{N-1}D_{N-1}-\tilde{c}_{1}D_{0}- \frac{\tilde{c}_{N-1}-c_{1}}{1-p} + \sum _{\ell =1}^{N-2} (\tilde{c}_{\ell +1}-\tilde{c}_{\ell }) \frac{p^{\ell +1}}{1-p} = \mathcal{O}(1), \end{aligned}$$

where we have used that $\tilde{c}_{\ell +1}-\tilde{c}_{\ell }=\mathcal{O}(n^{-1})$ uniformly for $\ell =1,\ldots ,N-2$. Substituting the above in Eq. (6.40), this proves (6.38). Substituting now Eqs. (6.36)–(6.38) in Eq. (6.35), we get

$$\begin{aligned} S_{567} = \frac{n e^{i \lfloor j_{\star }\rfloor (\theta _{1}-\theta _{2})}}{r_{1}^{2}} \frac{p}{1-p} \frac{\sigma _{1}}{e^{t_{1}+t_{2}}} + \mathcal{O}\big ( M \sqrt{n} \big ). \end{aligned}$$

(6.41)

Now we analyze $S_{4}$. By Eq. (6.14), we have

$$\begin{aligned}&S_{4} = \sqrt{n} \sum _{j=g_{1,-}}^{g_{1,+}} p^{-(j-1)} \hat{c}_{j} = p^{2-g_{1,-}}\sqrt{n} \sum _{\ell =1}^{\hat{N}} p^{-\ell } \hat{c}_{\ell }', \qquad \hbox {as } n \rightarrow + \infty , \end{aligned}$$

where $\hat{N}:=g_{1,+}-g_{1,-}+1$ and $\hat{c}_{j}, \hat{c}_{\ell }'$ satisfy

$$\begin{aligned} \hat{c}_{j}&= \frac{\sqrt{2} \, b r_{1}^{b-2} e^{-\frac{M_{j,1}^{2}r_{1}^{2b}}{2}}}{\sqrt{\pi } {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}})} \bigg ( 1 + \frac{1}{\sqrt{n}}\bigg \{\frac{(5M_{j,1}^{2}r_{1}^{2b}-2)e^{-\frac{r_{1}^{2b}M_{j,1}^{2}}{2}}}{3\sqrt{2\pi } r_{1}^{b} {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}}) } \nonumber \\&\quad + \frac{M_{j,1}}{6} \Big ( 5M_{j,1}^{2}r_{1}^{2b}-3+6br_{1}^{2b}(t_{1}+t_{2}) \Big ) \bigg \} \nonumber \\&\quad + \mathcal{O}\bigg (\frac{1+M_{j,1}^{2}+M_{j,1}^{6}\chi _{j,1}^{+}}{n}+\frac{1+M_{j,1}^{7}+M_{j,1}^{9}\chi _{j,1}^{+}}{n^{3/2}} + \frac{1+M_{j,1}^{12}}{n^{2}}\bigg ) \bigg ), \\&\hat{c}_{\ell }' := \hat{c}_{j}, \quad \hbox {where } j = \ell +g_{1,-}-1. \end{aligned}$$

Summing by parts, we get

$$\begin{aligned} \sum _{\ell =1}^{\hat{N}} p^{-\ell } \hat{c}_{\ell }' = \hat{c}_{\hat{N}}'\hat{D}_{\hat{N}}-\hat{c}_{1}'\hat{D}_{0}- \frac{\hat{c}'_{\hat{N}}-\hat{c}'_{1}}{1-p^{-1}} + \sum _{\ell =1}^{\hat{N}-1} (\hat{c}'_{\ell +1}-\hat{c}'_{\ell }) \frac{p^{-\ell -1}}{1-p^{-1}} \end{aligned}$$

where $\hat{D}_{\ell }:= \sum _{j=0}^{\ell }p^{-j}$. Since $\hat{c}_{1}' = \mathcal{O}(e^{-M^{2}})$, $\hat{c}'_{\hat{N}} = \mathcal{O}(1)$, $\hat{c}'_{\ell +1}-\hat{c}'_{\ell } = \mathcal{O}(n^{-\frac{1}{2}})$ and $\hat{N} = \mathcal{O}(M\sqrt{n})$, we get $\sum _{\ell =1}^{\hat{N}} p^{-\ell } \hat{c}_{\ell }' = \mathcal{O}(M)$, and thus

$$\begin{aligned} S_{4} = \mathcal{O}(M\sqrt{n}), \qquad \hbox {as } n \rightarrow + \infty . \end{aligned}$$

(6.42)

Combining Eqs. (6.41), (6.42) and Lemmas 6.2 and 6.5 yields

$$\begin{aligned} K_{n}(z,w) = \frac{n e^{i \lfloor j_{\star } \rfloor (\theta _{1}-\theta _{2})} \sigma _{1}}{r_{1}^{2} e^{t_{1}+t_{2}}} \bigg ( \frac{1}{p^{-1}-1} + \mathcal {Q}_{n} \bigg ) + \mathcal{O}(M\sqrt{n}). \end{aligned}$$

(6.43)

Using Eqs. (3.17) and (6.20), we infer that

$$\begin{aligned} \frac{\sigma _{1}}{r_{1}^{2}} \bigg ( \frac{1}{p^{-1}-1} + \mathcal {Q}_{n} \bigg ) = 2\pi \cdot r_{1}^{-2x} \cdot \mathcal {S}^{G}_{\textrm{hard}}(r_{1}e^{\theta _{1}},r_{1}e^{\theta _{2}};n). \end{aligned}$$

(6.44)

Substituting Eq. (6.44) in Eq. (6.43) yields

$$\begin{aligned} K_{n}(z,w) = 2\pi n \cdot \mathcal {S}^{G}_{\textrm{hard}}(r_{1}e^{\theta _{1}},r_{1}e^{\theta _{2}};n) \cdot \frac{ e^{i \lfloor j_{\star } \rfloor (\theta _{1}-\theta _{2})}}{e^{t_{1}+t_{2}}} r_{1}^{-2x} + \mathcal{O}(M\sqrt{n}). \end{aligned}$$

Since $M=M' \sqrt{\log n}$, this finishes the proof of Theorem 3.13.

7 Proofs of Theorems 3.5 and 3.15

In this section z and w are given by

$$\begin{aligned} z = r_{1}\Big (1-\frac{s_{1}}{\sqrt{n}}\Big )e^{i\theta _{1}}, \qquad w = r_{1}\Big (1-\frac{s_{2}}{\sqrt{n}}\Big )e^{i\theta _{2}}, \qquad s_{1}, s_{2} > 0, \theta _{1},\theta _{2}\in \mathbb {R}, \end{aligned}$$

(7.1)

and $s_{1}, s_{2}, \theta _{1},\theta _{2}$ are independent of n. At this point $\theta _{1},\theta _{2}\in \mathbb {R}$ can be arbitrary (in Sect. 7.1 we will take $\theta _{2}=\theta _{1}$, and in Sect. 7.2 we will take $\theta _{2}\ne \theta _{1} \mod 2\pi$). We emphasize that $s_{1}, s_{2}$ are strictly positive (the case $s_{1}=s_{2}=0$ is very different and covered in Sect. 6). The parameters $s_{1},s_{2}$ in Eq. (7.1) are related to $\mathfrak {s}_{1}, \mathfrak {s}_{2}$ in Eq. (3.8) via $s_{j} = \mathfrak {s}_{j}/(\sqrt{2}br_{1}^{b})$, $j=1,2$. As in Eq. (5.2) we write $K_{n}(z,w) = \sum _{j=1}^{n}F_{j}$, where

$$\begin{aligned}&F_{j} := e^{-\frac{n}{2}Q(z)}e^{-\frac{n}{2}Q(w)} \frac{z^{j-1}\overline{w}^{j-1}}{h_{j}} = e^{-\frac{n}{2}|z|^{2b}}e^{-\frac{n}{2}|w|^{2b}}|z|^{\alpha }|w|^{\alpha }\frac{z^{j-1}\overline{w}^{j-1}}{h_{j}}. \end{aligned}$$

(7.2)

In this section we take $M=M' \log n$. The following lemma is proved using Lemma 4.2.

Lemma 7.1

Let $\delta \in (0,\frac{1}{100})$ be fixed. $M'$ can be chosen sufficiently large and independently of $\delta$ such that the following hold.

• 1. Let j be fixed. As $n \rightarrow + \infty$, we have

  $$\begin{aligned} F_{j} = \mathcal{O}(e^{-n (1-\delta )r_{1}^{2b}}). \end{aligned}$$

  (7.3)

• 2. As $n \rightarrow + \infty$ with $M' \le j \le j_{1,-}$, we have

  $$\begin{aligned} F_{j}&= \mathcal{O}(e^{-nr_{1}^{2b}(1-\delta ) \frac{\epsilon - \log (1+\epsilon )}{1+\epsilon } }). \end{aligned}$$

  (7.4)

• 3. As $n \rightarrow + \infty$ with $j_{1,-} \le j \le g_{1,-}$, we have

  $$\begin{aligned} F_{j}&= \mathcal{O}(n^{-100}). \end{aligned}$$

  (7.5)

• 4. As $n \rightarrow + \infty$ with $g_{1,-} \le j \le g_{1,+}$, we have

  $$\begin{aligned}&F_{j} = e^{i(j-1)(\theta _{1}-\theta _{2})} \sqrt{n}\frac{\sqrt{2} \, b r_{1}^{b-2} e^{-\frac{1}{2}r_{1}^{2b}(M_{j,1}^{2}-2bM_{j,1}(s_{1}+s_{2})+2b^{2}(s_{1}^{2}+s_{2}^{2}))}}{\sqrt{\pi } {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}})} \nonumber \\&\qquad \times \bigg ( 1 + \frac{1}{\sqrt{n}}\bigg \{\frac{(5M_{j,1}^{2}r_{1}^{2b}-2)e^{-\frac{r_{1}^{2b}M_{j,1}^{2}}{2}}}{3\sqrt{2\pi } r_{1}^{b} {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}}) } - \frac{M_{j,1}}{2}+s_{1}+s_{2} + r_{1}^{2b} \bigg [ \frac{5}{6}M_{j,1}^{3} \nonumber \\&\qquad - bM_{j,1}^{2}(s_{1}+s_{2}) + bM_{j,1} \frac{s_{1}^{2}+s_{2}^{2}}{2} + \frac{b^{2}(2b-3)}{3}(s_{1}^{3}+s_{2}^{3}) \bigg ] \bigg \} + \mathcal{O}\Big (\frac{1+M_{j,1}^{6}}{n} \Big ) \bigg ). \end{aligned}$$

  (7.6)

• 5. As $n \rightarrow + \infty$ with $g_{1,+} \le j \le j_{1,+}$, we have

  $$\begin{aligned}&F_{j} = \mathcal{O}(n^{-100}). \end{aligned}$$

  (7.7)

• 6. As $n \rightarrow + \infty$ with $j_{1,+} \le j \le \lfloor j_{\star }\rfloor$, we have

  $$\begin{aligned}&F_{j} = \mathcal{O}\Big ( e^{-(1-\delta )\frac{\epsilon br_{1}^{2b}}{1-\epsilon }(s_{1}+s_{2})\sqrt{n}} \Big ). \end{aligned}$$

  (7.8)

• 7. As $n \rightarrow + \infty$ with $\lfloor j_{\star }\rfloor +1 \le j \le j_{2,-}$, we have

  $$\begin{aligned}&F_{j} = \mathcal{O}\Big ( e^{-(1-\delta )\sigma _{1}(s_{1}+s_{2})\sqrt{n}} \Big ). \end{aligned}$$

  (7.9)

• 8. As $n \rightarrow + \infty$ with $j_{2,-} \le j \le g_{2,-}$, we have

  $$\begin{aligned}&F_{j} = \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})(\frac{br_{2}^{2b}}{1+\epsilon }-\sigma _{\star })}). \end{aligned}$$

  (7.10)

• 9. As $n \rightarrow + \infty$ with $g_{2,-} \le j \le g_{2,+}$, we have

  $$\begin{aligned}&F_{j} = \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})\sigma _{2}}). \end{aligned}$$

  (7.11)

• 10. As $n \rightarrow + \infty$ with $g_{2,+} \le j \le j_{2,+}$, we have

  $$\begin{aligned} F_{j}&= \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})\sigma _{2}}). \end{aligned}$$

  (7.12)

• 11. As $n \rightarrow + \infty$ with $j_{2,+} \le j \le n$, we have

  $$\begin{aligned} F_{j}&= \mathcal{O}(e^{-n(1-\delta ) \log (\frac{r_{2}}{r_{1}})\sigma _{2}}). \end{aligned}$$

  (7.13)

Proof

Eq. (7.3) directly follows from Eqs. (4.9) and (7.2). Let us now prove Eq. (7.4). By Eqs. (4.10) and (7.2), as $n \rightarrow + \infty$ with $M' \le j \le j_{1,-}$, we have

$$\begin{aligned} F_{j} = \mathcal{O}\bigg (\exp \bigg \{ n(1-\delta ) \bigg [ \frac{1}{b}(j/n-j/n \log (\frac{j/n}{b})) - r_{1}^{2b} + 2j/n \log (r_{1}) \bigg ] \bigg \} \bigg ). \end{aligned}$$

Since the function $[0,br_{1}^{2b}] \in y \mapsto y-y \log (\frac{y}{b}) + y \log (r_{1}^{2b})$ is increasing, and since $j/n \le \frac{br_{1}^{2b}}{1+\epsilon }+\mathcal{O}(n^{-1})$, we have

$$\begin{aligned} F_{j}&= \mathcal{O}\bigg (\exp \bigg \{ -nr_{1}^{2b}(1-\delta ) \frac{\epsilon - \log (1+\epsilon )}{1+\epsilon } \bigg \} \bigg ), \end{aligned}$$

which is Eq. (7.4). Now we prove Eq. (7.5). By Eqs. (4.11) and (7.2), as $n \rightarrow + \infty$ with $j_{1,-} \le j \le g_{1,-}$, we have

$$\begin{aligned} F_{j} = \mathcal{O}\bigg ( \exp \bigg \{ n(1-\delta ) \bigg [ \frac{j/n - j/n \log (\frac{j/n}{b})}{b}-r_{1}^{2b} + 2j/n \log (r_{1}) + \frac{s_{1}+s_{2}}{\sqrt{n}}(br_{1}^{2b}-j/n) \bigg ] \bigg \} \bigg ). \end{aligned}$$

For $j_{1,-} \le j \le g_{1,-}$, the above exponential attains its maximum at $j=g_{1,-} = n(\frac{br_{1}^{2b}}{1+\frac{M}{\sqrt{n}}}+\mathcal{O}(n^{-1}))$, and therefore

$$\begin{aligned} F_{j} = \mathcal{O}\bigg ( \exp \bigg \{ -(1-\delta ) \bigg [\frac{M^{2}r_{1}^{2b}}{2}-M br_{1}^{2b}(s_{1}+s_{2})\bigg ] \bigg \} \bigg ). \end{aligned}$$

Since $M=M' \log n$ and $s_{1},s_{2}$ are fixed, Eq. (7.5) holds provided $M'$ is chosen large enough. Now we prove Eq. (7.6). As $n \rightarrow + \infty$ with $g_{1,-} \le j \le g_{1,+}$, we have

$$\begin{aligned}&e^{r_{1}^{2b}(1-\log (r_{1}^{2b}))n}e^{M_{j,1} r_{1}^{2b} \log (r_{1}^{2b})\sqrt{n}}e^{-r_{1}^{2b}(\frac{1}{2}+\log (r_{1}^{2b}))M_{j,1}^{2}} e^{-\frac{n}{2}|z|^{2b}}e^{-\frac{n}{2}|w|^{2b}}|z|^{\alpha }|w|^{\alpha }z^{j-1}\overline{w}^{j-1} \\&= e^{i(j-1)(\theta _{1}-\theta _{2})}e^{-\frac{1}{2}r_{1}^{2b}(M_{j,1}^{2}-2bM_{j,1}(s_{1}+s_{2})+2b^{2}(s_{1}^{2}+s_{2}^{2}))}r_{1}^{-2} \\&\quad \times \bigg \{ 1 + \frac{1}{\sqrt{n}}\bigg [ s_{1}+s_{2}+ br_{1}^{2b} \bigg ( \frac{b(2b-3)}{3}(s_{1}^{3}+s_{2}^{3}) + M_{j,1} \frac{s_{1}^{2}+s_{2}^{2}}{2} \\&\quad - M_{j,1}^{2}(s_{1}+s_{2}) - 2M_{j,1}^{3}\log (r_{1}) \bigg ) \bigg ] \\&\quad + \mathcal{O}\Big (\frac{1+M_{j,1}^{6}}{n}\Big ) \bigg \}. \end{aligned}$$

Now Eq. (7.6) follows directly from a computation using Eqs. (4.12) and (7.2). Now we prove Eq. (7.7). Using Eqs. (4.13) and (7.2), as $n \rightarrow + \infty$ with $g_{1,+} \le j \le j_{1,+}$ we obtain

$$\begin{aligned} F_{j}&= \mathcal{O}\Big ( \exp \Big \{ -\sqrt{n}(1-\delta )(j/n-br_{1}^{2b})(s_{1}+s_{2}) \Big \} \Big ) \\&= \mathcal{O}\Big ( \exp \Big \{ -br_{1}^{2b}(s_{1}+s_{2})M \Big \} \Big ). \end{aligned}$$

Since $s_{1},s_{2}>0$ are fixed and $M=M' \log n$, we can choose $M'$ large enough so that Eq. (7.7) holds. The proofs of Eqs. (7.8)–(7.13) are similar to the proof of Eq. (7.7) and are omitted. $\square$

Lemma 7.2

As $n \rightarrow + \infty$, we have

$$\begin{aligned}&K_{n}(z,w) = S_{4} + \mathcal{O}(1), \end{aligned}$$

where

$$\begin{aligned}&S_{4} := \sum _{j=g_{1,-}}^{g_{1,+}} e^{i(j-1)(\theta _{1}-\theta _{2})} \sqrt{n}\frac{\sqrt{2} \, b r_{1}^{b-2} e^{-\frac{1}{2}r_{1}^{2b}(M_{j,1}^{2}-2bM_{j,1}(s_{1}+s_{2})+2b^{2}(s_{1}^{2}+s_{2}^{2}))}}{\sqrt{\pi } {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}})} \nonumber \\&\qquad \times \bigg ( 1 + \frac{1}{\sqrt{n}}\bigg \{\frac{(5M_{j,1}^{2}r_{1}^{2b}-2)e^{-\frac{r_{1}^{2b}M_{j,1}^{2}}{2}}}{3\sqrt{2\pi } r_{1}^{b} {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}}) } - \frac{M_{j,1}}{2}+s_{1}+s_{2} + r_{1}^{2b} \bigg [ \frac{5}{6}M_{j,1}^{3} \nonumber \\&\qquad - bM_{j,1}^{2}(s_{1}+s_{2}) + bM_{j,1} \frac{s_{1}^{2}+s_{2}^{2}}{2} + \frac{b^{2}(2b-3)}{3}(s_{1}^{3}+s_{2}^{3}) \bigg ] \bigg \} \bigg ), \end{aligned}$$

(7.14)

Proof

By Lemma 7.1, as $n\rightarrow +\infty$ we have

$$\begin{aligned}&K_{n}(z,w) = S_{4} + \mathcal{O}\bigg (\sum _{j=g_{1,-}}^{g_{1,+}} \sqrt{n}\frac{e^{-\frac{1}{2}r_{1}^{2b}(M_{j,1}^{2}-2bM_{j,1}(s_{1}+s_{2}))}}{{\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}})} \frac{1+M_{j,1}^{6}}{n}\bigg )+\mathcal{O}(n^{-50}), \end{aligned}$$

(7.15)

Since $s_{1},s_{2}>0$ are fixed, the function $x \mapsto \frac{e^{-\frac{1}{2}r_{1}^{2b}(x^{2}-2bx(s_{1}+s_{2}))}}{{\textrm{erfc}}(-xr_{1}^{b}/\sqrt{2})}$ is $\mathcal{O}(e^{-c|x|})$ as $x \rightarrow \pm \infty$ (for some $c>0$). Lemma 6.9 then implies that the $\mathcal{O}$-terms in Eq. (7.15) remain bounded as $n\rightarrow + \infty$, and the claim follows. $\square$

7.1 Proof of Theorem 3.5

Here we take $\theta _{2}=\theta _{1}$. Recall also that $s_{1},s_{2}$ in Eq. (7.1) are related to $\mathfrak {s}_{1}, \mathfrak {s}_{2}$ in Eq. (3.8) via $s_{j} = \mathfrak {s}_{j}/(\sqrt{2}br_{1}^{b})= \mathfrak {s}_{j}/(\tilde{\Delta }Q(r_{1}))^{\frac{1}{2}}$, $j=1,2$.

Using Eq. (7.14), we write $S_{4} = \sum _{j=g_{1,-}}^{g_{1,+}}(\sqrt{n} \, h_{1}(M_{j,1})+h_{2}(M_{j,1}))$, where

$$\begin{aligned}&h_{1}(x) = \frac{\sqrt{2} \, b r_{1}^{b-2} e^{-\frac{1}{2}((\frac{r_{1}^{b}x}{\sqrt{2}}-\mathfrak {s}_{1})^{2}+(\frac{r_{1}^{b}x}{\sqrt{2}}-\mathfrak {s}_{2})^{2})}}{\sqrt{\pi } {\textrm{erfc}}(-\frac{x r_{1}^{b}}{\sqrt{2}})}, \\&h_{2}(x) = h_{1}(x) \bigg \{\frac{(5x^{2}r_{1}^{2b}-2)e^{-\frac{r_{1}^{2b}x^{2}}{2}}}{3\sqrt{2\pi } r_{1}^{b} {\textrm{erfc}}(-\frac{xr_{1}^{b}}{\sqrt{2}}) } - \frac{x}{2}+\frac{\mathfrak {s}_{1}+\mathfrak {s}_{2}}{\sqrt{2} \, b r_{1}^{b}} \\&\hspace{2.3cm} + r_{1}^{2b} \bigg [ \frac{5}{6}x^{3} - x^{2}\frac{\mathfrak {s}_{1}+\mathfrak {s}_{2}}{\sqrt{2}r_{1}^{b}} + x \frac{\mathfrak {s}_{1}^{2}+\mathfrak {s}_{2}^{2}}{4br_{1}^{2b}} + \frac{2b-3}{6\sqrt{2} \, b}\frac{\mathfrak {s}_{1}^{3}+\mathfrak {s}_{2}^{3}}{r_{1}^{3b}} \bigg ] \bigg \}. \end{aligned}$$

Since $\mathfrak {s}_{1},\mathfrak {s}_{2}>0$ are fixed, $h_{1}(x),h_{2}(x) = \mathcal{O}(e^{-c |x|})$ as $x \rightarrow \pm \infty$ for some $c>0$. Since $M=M' \log n$, by Lemma 6.9, we obtain

$$\begin{aligned} S_{4}&= br_{1}^{2b}\int _{-M}^{M} h_{1}(t)dt \, n + br_{1}^{2b} \int _{-M}^{M} (h_{2}(t)-2th_{1}(t))dt \, \sqrt{n} + \mathcal{O}(1), \\&= br_{1}^{2b}\int _{-\infty }^{+\infty } h_{1}(t)dt \, n + br_{1}^{2b} \int _{-\infty }^{+\infty } (h_{2}(t)-2th_{1}(t))dt \, \sqrt{n} + \mathcal{O}(1). \end{aligned}$$

The claim now follows from a simple change of variables [55]. This finishes the proof of Theorem 3.5.

7.2 Proof of Theorem 3.15

Here we take $\theta _{1}\ne \theta _{2} \mod 2\pi$. Let $p:=e^{-i(\theta _{1}-\theta _{2})}$. By Eq. (7.14),

$$\begin{aligned}&S_{4} = \sqrt{n} \sum _{j=g_{1,-}}^{g_{1,+}} p^{-(j-1)} c_{j} = p^{2-g_{1,-}}\sqrt{n} \sum _{\ell =1}^{N} p^{-\ell } c_{\ell }', \end{aligned}$$

where $N:=g_{1,+}-g_{1,-}+1$, and $c_{j}$, $c_{\ell }'$ are defined by

$$\begin{aligned}&c_{j} := \frac{\sqrt{2} \, b r_{1}^{b-2} e^{-\frac{1}{2}r_{1}^{2b}(M_{j,1}^{2}-2bM_{j,1}(s_{1}+s_{2})+2b^{2}(s_{1}^{2}+s_{2}^{2}))}}{\sqrt{\pi } {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}})} \\&\qquad \times \bigg ( 1 + \frac{1}{\sqrt{n}}\bigg \{\frac{(5M_{j,1}^{2}r_{1}^{2b}-2)e^{-\frac{r_{1}^{2b}M_{j,1}^{2}}{2}}}{3\sqrt{2\pi } r_{1}^{b} {\textrm{erfc}}(-\frac{M_{j,1}r_{1}^{b}}{\sqrt{2}}) } - \frac{M_{j,1}}{2}+s_{1}+s_{2} + r_{1}^{2b} \bigg [ \frac{5}{6}M_{j,1}^{3} \\&\qquad - bM_{j,1}^{2}(s_{1}+s_{2}) + bM_{j,1} \frac{s_{1}^{2}+s_{2}^{2}}{2} + \frac{b^{2}(2b-3)}{3}(s_{1}^{3}+s_{2}^{3}) \bigg ] \bigg \} \bigg ), \\&c_{\ell }' := c_{j}, \quad \hbox {where } j = \ell +g_{1,-}-1. \end{aligned}$$

Summing by parts, we get

$$\begin{aligned} \sum _{\ell =1}^{N} p^{-\ell } c_{\ell }' = c_{N}'\hat{D}_{N}-c_{1}'\hat{D}_{0}- \frac{c'_{N}-c'_{1}}{1-p^{-1}} + \sum _{\ell =1}^{N-1} (c'_{\ell +1}-c'_{\ell }) \frac{p^{-\ell -1}}{1-p^{-1}} \end{aligned}$$

where $\hat{D}_{\ell }:= \sum _{j=0}^{\ell }p^{-j}$. Note that $c_{1}' = \mathcal{O}(e^{-c M})$, $c'_{N} = \mathcal{O}(e^{-c M^{2}})$, and $c_{j+1}-c_{j} = \mathcal{O}(n^{-\frac{1}{2}})c_{j}$, $c_{j}=\mathcal{O}(e^{-c|M_{j,1}|})$. Hence, using several times summation by parts, we get

$$\begin{aligned} \sum _{\ell =1}^{N} p^{-\ell } c_{\ell }' \lesssim e^{-c M} + \frac{1}{\sqrt{n} }\sum _{\ell =1}^{N-1} p^{-\ell }c_{\ell }' \lesssim e^{-c M} + \frac{1}{n^{\frac{21}{2}}}\sum _{\ell =1}^{N-21} p^{-\ell }c_{\ell }' \lesssim \frac{M}{n^{10}}, \end{aligned}$$

where we have used that $N=\mathcal{O}(M\sqrt{n})$ for the last equality. Hence $S_{4} = \mathcal{O}(\frac{M}{n^{10}})$. By Lemma 7.2, we thus have $K_{n}(z,w) = \mathcal{O}(1)$, which is the statement of Theorem 3.15.

Appendices

The Semi-hard 1-Point Function and Its Universality

In the following, we consider any radially symmetric potential V which is smooth near a point $r_{1}>0$ in the bulk of the droplet $S=S[V]$; we also assume that $\Delta V(r_{1})>0$. We redefine V by placing a hard wall along the circle $r=r_{1}$, say

$$\begin{aligned} Q(z)={\left\{ \begin{array}{ll} V(z),&{} |z|\le r_{1}, \\ +\infty ,&{} |z|>r_{1}. \end{array}\right. } \end{aligned}$$

Let $K_n(z,w)$ be the associated correlation kernel, i.e., the reproducing kernel of the space

$$\begin{aligned} \mathcal {W}_n^h = \{p\cdot e^{-\frac{1}{2} n Q} : p \hbox { is a holomorphic polynomial with } \hbox {deg } p \le n-1\} \subset L^2(\mathbb {C},dA). \end{aligned}$$

We write $R_n(z)=K_{n}(z,z)$ for the 1-point function.

Let us write $\tilde{K}_n(z,w)$ for the correlation kernel of the rescaled process pertaining to the blow-up map $z\mapsto \sqrt{2n\Delta V(r_{1})}\cdot (r_{1}-z)$, i.e.

$$\begin{aligned} \tilde{K}_n(z,w)=\frac{1}{2n\Delta V(r_{1})} K_{n}\Big (r_{1}-\frac{z}{\sqrt{2n\Delta V(r_{1})}},r_{1}-\frac{w}{\sqrt{2n\Delta V(r_{1})}}\Big ). \end{aligned}$$

(A.1)

Our goal is to study the asymptotics of this rescaled “semi-hard edge kernel” for z, w in a compact subset of the right half-plane: $\text {Re}\,z>0$ and $\text {Re}\,w>0$.

Remark A.1

In [16] the following slightly different rescaling was used:

$$\begin{aligned} \hat{K}_n(z,w) = \frac{1}{n\Delta V(r_{1})} K_{n}\Big (r_{1}+\frac{z}{\sqrt{n\Delta V(r_{1})}},r_{1}+\frac{w}{\sqrt{n\Delta V(r_{1})}}\Big ). \end{aligned}$$

(A.2)

Below we recall some theory for the limit of $\hat{K}_n$, and when we translate back to $\tilde{K}_n$, we use:

$$\begin{aligned} \tilde{K}_n(z,w)=\frac{1}{2} \hat{K}_n (-\frac{z}{\sqrt{2}},-\frac{w}{\sqrt{2}}), \qquad \hat{K}_n (z,w) = 2\tilde{K}_n(-\sqrt{2}z,-\sqrt{2}w). \end{aligned}$$

(A.3)

By [16] we know that there is a sequence of cocycles $c_n(z,w)$ such that the kernels $c_n(z,w)\hat{K}_{n}(z,w)$ converge locally uniformly for z, w in the left half-plane to a limit kernel $\hat{K}(z,w)$ of the form

$$\begin{aligned} \hat{K}(z,w)=G(z,w)\Psi (z,w) \qquad \text {Re}\,z<0,\text {Re}\,w<0, \end{aligned}$$

(A.4)

where $G(z,w)=e^{ z\bar{w}-\frac{1}{2}|z|^2-\frac{1}{2}|w|^2}$ is the “Ginibre kernel” while $\Psi (z,w)$ is a function analytic in z and $\bar{w}$ and well-defined for $\text {Re}\,z<0$, $\text {Re}\,w<0$. Also, the limit kernel gives rise to a solution $\hat{R}(z)$ of the Ward equation

$$\begin{aligned} \overline{\partial } \hat{C}(z) = \hat{R}(z)-1-\Delta \log \hat{R}(z),\qquad \text {Re}\,z<0, \end{aligned}$$

(A.5)

where

$$\begin{aligned} \hat{R}(z)=\hat{K}(z,z)=\Psi (z,z),\qquad \hat{C}(z)=\int _{\text {Re}\,w<0}\frac{|\hat{K}(z,w)|^2}{\hat{K}(z,z)}\frac{1}{z-w}\, dA(w). \end{aligned}$$

(If $\hat{R}(z)>0$ for one z, then $\hat{R}(z)>0$ everywhere in the left half-plane, see [16] for details.)

By standard balayage arguments [11], we know that $\hat{R}$ blows up at the hard edge (this is not surprising since $K_{n}(z,z)$ grows like $n^2$ in the hard edge regime), i.e.

$$\begin{aligned} \lim _{\text {Re}\,z\rightarrow 0} \hat{R}(z)=+\infty . \end{aligned}$$

(A.6)

It is also useful to note that when $\text {Re}\,z\rightarrow -\infty$ we have, by well-known bulk asymptotics [8],

$$\begin{aligned} \lim _{\text {Re}\,z\rightarrow -\infty }\hat{R}(z)=1. \end{aligned}$$

(A.7)

The rotational symmetry of the ensemble also easily implies that $\hat{R}$ is vertical translation invariant (see [15] for a detailed proof of this)

$$\begin{aligned} \hat{R}(z+it)=\hat{R}(z),\qquad \forall t\in \mathbb {R}. \end{aligned}$$

(A.8)

This is equivalent to saying that the analytic function $\Psi$ in Eq. (A.4) takes the form $\Psi (z,w)=\Phi (z+\bar{w})$ (see e.g. [15]) where $\Phi (z)$ is an analytic function of one variable, defined for $\text {Re}\,z<0$.

Thus we have

$$\begin{aligned} \hat{K}(z,w)=G(z,w)\Phi (z+\bar{w}),\qquad \hat{R}(z)=\Phi (z+\bar{z}). \end{aligned}$$

(A.9)

We now have the following result, which is implicit in the analysis in [16].

Proposition A.2

There is a unique analytic function $\Phi (z)$ for $\text {Re}\,z < 0$ which satisfies the Ward equation (A.5) with boundary conditions Eqs. (A.6) and (A.7). This function is given by

$$\begin{aligned} \Phi (z)=\frac{1}{\sqrt{2\pi }} \int _{-\infty }^{+\infty }\frac{e^{-\frac{1}{2} (z-t)^2}}{\frac{1}{2} {\textrm{erfc}} \frac{t}{\sqrt{2}}}\, dt,\qquad (\text {Re}\,z<0). \end{aligned}$$

(A.10)

Proof

That Eq. (A.10) solves the Ward equation is shown in [16, Theorem 8], and Eqs. (A.6) and (A.7) are likewise clear for the solution Eq. (A.10). In the other direction, given an arbitrary solution $\Phi$ there exists, by [16, Theorem 8], an interval $I\subset \mathbb {R}$ such that $\Phi$ has the structure

$$\begin{aligned} \Phi (z)=\frac{1}{\sqrt{2\pi }}\int _I \frac{e^{-\frac{1}{2} (z-t)^2}}{\frac{1}{2} {\textrm{erfc}} \frac{t}{\sqrt{2}}}\, dt. \end{aligned}$$

The conditions $\Phi (x)\rightarrow 1$ as $x\rightarrow -\infty$ and $\Phi (x)\rightarrow +\infty$ as $x\rightarrow 0$ easily implies that we must have $I=\mathbb {R}$. $\square$

It follows that the limit kernel $\hat{K}(z,w)$ is uniquely given by

$$\begin{aligned} \hat{K}(z,w)=e^{z\bar{w}-\frac{1}{2}|z|^2-\frac{1}{2}|w|^2}\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{+\infty }\frac{e^{-\frac{1}{2} (z+\bar{w}-t)^2}}{\frac{1}{2} {\textrm{erfc}} \frac{t}{\sqrt{2}}}\, dt,\qquad \text {Re}\,z<0,\, \text {Re}\,w<0. \end{aligned}$$

Using Eq. (A.3), we then get for z and w in the right half-plane $\text {Re}\,z>0$, $\text {Re}\,w>0$ that

$$\begin{aligned} \tilde{K}(z,w)&=\frac{1}{2} e^{\frac{1}{2} z\bar{w}-\frac{1}{4}|z|^2-\frac{1}{4}|w|^2}\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{+\infty }\frac{e^{-\frac{1}{2} (\frac{z+\bar{w}}{\sqrt{2}}+t)^2}}{\frac{1}{2} {\textrm{erfc}}\frac{t}{\sqrt{2}}}\, dt\nonumber \\&=e^{\frac{1}{2} z\bar{w}-\frac{1}{4}|z|^2-\frac{1}{4}|w|^2}\frac{1}{\sqrt{\pi }}\int _{-\infty }^{+\infty }\frac{e^{-\frac{1}{4}(z+\bar{w}+2t)^2}}{{\textrm{erfc}}\,t}\, dt. \end{aligned}$$

(A.11)

If we choose $\mathfrak {s}_1,\mathfrak {s}_2$ real and positive this gives

$$\begin{aligned} \tilde{K}(\mathfrak {s}_1,\mathfrak {s}_2) = \int _{-\infty }^{+\infty } \frac{e^{-\frac{(t+\mathfrak {s}_{1})^{2} + (t+\mathfrak {s}_{2})^{2}}{2}}}{\sqrt{\pi } {\textrm{erfc}}\,t}dt. \end{aligned}$$

(A.12)

Universality for non-rotation invariant potentials. Note that the above arguments work not only for rotationally symmetric potentials, but for arbitrary potentials and scaling limits $\tilde{K}$ which are vertical translation invariant.

Indeed, consider a smooth potential V (not necessarily rotation invariant), fix any point $r_{1}$ and define Q as follows:

$$\begin{aligned} Q(z)={\left\{ \begin{array}{ll} V(z),&{} z \in D, \\ +\infty ,&{} z \notin D, \end{array}\right. } \end{aligned}$$

where $D \subset \hbox {int }S[V]$ is such that $\partial D$ is smooth and passes through $r_{1}$. Assume that $\Delta V(r_{1})>0$. Let $K_{n}$ be the correlation kernel associated to Q, and define $\tilde{K}_{n},\hat{K}_{n}$ as in Eqs. (A.1) and (A.2). By [16] we know that each subsequential limit $\hat{K}$ of the rescaled kernels $\hat{K}_n$ satisfy Ward’s equation (A.5) as well as the boundary conditions Eqs. (A.6) and (A.7). Hence if $\hat{K}$ is vertical translation invariant, i.e., $\hat{K}(z+it,z+it)=\hat{K}(z,z)$ for all $t\in \mathbb {R}$, [16, Theorem 8] implies that $\hat{K}$ is uniquely given by Eqs. (A.9), (A.10). In other words, the kernel $\tilde{K}$ in Eq. (A.11) is universal among translation invariant scaling limits.

Uniform Asymptotics of the Incomplete Gamma Function

To analyze the large n behavior of $K_{n}(z,w)$, we will use the asymptotics of $\gamma (a,z)$ in various regimes of the parameters a and z. These asymptotics are available in the literature and are summarized in the following lemmas.

Lemma B.1

([57, formula 8.11.2]). Let a be fixed. As $z \rightarrow \infty$,

$$\begin{aligned} \gamma (a,z) = \Gamma (a) + \mathcal{O}(z^{a-1}e^{-z}). \end{aligned}$$

Lemma B.2

([64, Sect. 11.2.4]). The following hold:

$$\begin{aligned}&\frac{\gamma (a,z)}{\Gamma (a)} = \frac{1}{2}{\textrm{erfc}}(-\eta \sqrt{a/2}) - R_{a}(\eta ), \qquad R_{a}(\eta ) := \frac{e^{-\frac{1}{2}a \eta ^{2}}}{2\pi i}\int _{-\infty }^{\infty }e^{-\frac{1}{2}a u^{2}}g(u)du, \end{aligned}$$

(B.1)

where ${\textrm{erfc}}$ is given by Eq. (3.2), $g(u):= \frac{dt}{du}\frac{1}{\lambda -t}+\frac{1}{u+i \eta }$,

$$\begin{aligned}&\lambda = \frac{z}{a}, \quad \eta = (\lambda -1)\sqrt{\frac{2 (\lambda -1-\log \lambda )}{(\lambda -1)^{2}}}, \quad u=-i(t-1)\sqrt{\frac{2(t-1-\log t)}{(t-1)^{2}}}, \end{aligned}$$

(B.2)

and the principal branch is used for the roots. In particular, $t \in \mathcal {L}:=\{\frac{\theta }{\sin \theta } e^{i\theta }: -\pi< \theta < \pi \}$ for $u \in \mathbb {R}$ and $\eta \in \mathbb {R}$ for $\lambda >0$. Furthermore, as $a \rightarrow + \infty$, uniformly for $z \in [0,\infty )$,

$$\begin{aligned}&R_{a}(\eta ) \sim \frac{e^{-\frac{1}{2}a \eta ^{2}}}{\sqrt{2\pi a}}\sum _{j=0}^{\infty } \frac{c_{j}(\eta )}{a^{j}}, \end{aligned}$$

(B.3)

where all coefficients $c_{j}(\eta )$ are bounded functions of $\eta \in \mathbb {R}$ (i.e. bounded for $\lambda \in (0,\infty )$). The coefficients $c_{0}(\eta )$ and $c_{1}(\eta )$ are explicitly given by (see [64, p. 312])

$$\begin{aligned} c_{0}(\eta ) = \frac{1}{\lambda -1}-\frac{1}{\eta }, \qquad c_{1}(\eta ) = \frac{1}{\eta ^{3}}-\frac{1}{(\lambda -1)^{3}}-\frac{1}{(\lambda -1)^{2}}-\frac{1}{12(\lambda -1)}. \end{aligned}$$

(B.4)

Recall that ${\textrm{erfc}}$ satisfies [57, 7.12.1]

$$\begin{aligned}&{\textrm{erfc}}(y) = \frac{e^{-y^{2}}}{\sqrt{\pi }}\bigg ( \frac{1}{y}-\frac{1}{2y^{3}}+\frac{3}{4y^{5}}-\frac{15}{8y^{7}} + \mathcal{O}(y^{-9}) \bigg ),{} & {} \hbox {as } y \rightarrow + \infty , \end{aligned}$$

(B.5)

and ${\textrm{erfc}}(-y) = 2-{\textrm{erfc}}(y)$. By combining Lemma B.2 with the above large y asymptotics of ${\textrm{erfc}}(y)$, we get the following.

Lemma B.3

Let $\eta$ be as in Eq. (B.2).

• (i) Let $\delta >0$ be fixed. As $a \rightarrow +\infty$, uniformly for $\lambda \ge 1+\delta$,

  $$\begin{aligned} \frac{\gamma (a,\lambda a)}{\Gamma (a)} = 1 + \frac{e^{-\frac{a\eta ^{2}}{2}}}{\sqrt{2\pi }} \bigg ( \frac{-1}{\lambda -1}\frac{1}{\sqrt{a}}+\frac{1+10\lambda +\lambda ^{2}}{12(\lambda -1)^{3}} \frac{1}{a^{3/2}} + \mathcal{O}(a^{-5/2}) \bigg ). \end{aligned}$$

• (ii) As $a \rightarrow +\infty$, uniformly for $\lambda$ in compact subsets of (0, 1),

  $$\begin{aligned} \frac{\gamma (a,\lambda a)}{\Gamma (a)} = \frac{e^{-\frac{a\eta ^{2}}{2}}}{\sqrt{2\pi }} \bigg ( \frac{-1}{\lambda -1}\frac{1}{\sqrt{a}}+\frac{1+10\lambda +\lambda ^{2}}{12(\lambda -1)^{3}} \frac{1}{a^{3/2}} + \mathcal{O}(a^{-5/2}) \bigg ). \end{aligned}$$