## 1 Introduction and Statement of Results

The Mittag–Leffler ensemble with parameters $$b>0$$ and $$\alpha >-1$$ is the joint probability distribution

\begin{aligned} \frac{1}{n!Z_{n}} \prod _{1 \le j < k \le n} |z_{k} -z_{j}|^{2} \prod _{j=1}^{n}|z_{j}|^{2\alpha }e^{-n |z_{j}|^{2b}}d^{2}z_{j}, \qquad z_{1},\ldots ,z_{n} \in \mathbb {C}, \end{aligned}
(1.1)

where $$Z_{n}$$ is the normalization constant. This determinantal point process can be realized as the eigenvalues of a random normal matrix M with distribution proportional to $$|\det (M)|^{2\alpha }e^{-n \, \textrm{tr}((MM^{*})^{b})}dM$$ [54]. The special case $$(b,\alpha )=(1,0)$$ also corresponds to the eigenvalue distribution of a Ginibre matrix [15, 43], i.e. an $$n \times n$$ matrix with independent complex Gaussian entries with mean 0 and variance $$\frac{1}{n}$$.

Consider the characteristic polynomial $$p_{n}(x)=\prod _{j=1}^{n}(|z_{j}|-x)$$ of the process of the moduli $$\{|z_{j}|\}_{j=1}^{n}$$. The main result of this work is a precise asymptotic formula as $$n \rightarrow + \infty$$ for

\begin{aligned} \mathbb {E}\Big [ e^{\frac{u}{\pi } \, \text {Im\,}\ln p_{n}(r)}e^{a \, \text {Re\,}\ln p_{n}(r)} \Big ], \end{aligned}
(1.2)

where $$u \in \mathbb {R}$$, $$a \in \mathbb {N}:=\{0,1,\ldots \}$$, $$r \in (0,b^{-\frac{1}{2b}})$$ are fixed, and $$\ln p_{n}(r):= \ln |p_{n}(r)| + \pi i \, \#\{z_{j}:~|z_{j}|<~r\}$$. The macroscopic large n behavior of the $$|z_{j}|$$ is described by the probability measure $$d\mu (y)=2b^{2}y^{2b-1}dy$$, whose support is $$[0,b^{-\smash {\frac{1}{2b}}}]$$ [57]; thus the condition that $$r \in (0,b^{-\frac{1}{2b}})$$ is fixed means that we focus on “the bulk regime”. By definition, the expectation (1.2) is equal to $$D_{n}/Z_{n}$$, where

\begin{aligned} D_{n}&:=\frac{1}{n!} \int _{\mathbb {C}}\ldots \int _{\mathbb {C}} \prod _{1 \le j < k \le n} |z_{k} -z_{j}|^{2} \prod _{j=1}^{n}w(z_{j})d^{2}z_{j} = \det \left( \int _{\mathbb {C}} z^{j} {\overline{z}}^{k} w(z) d^{2}z \right) _{j,k=0}^{n-1} \end{aligned}

and the weight w is given by

\begin{aligned} w(z):=|z|^{2\alpha } e^{-n |z|^{2b}} \omega (|z|), \qquad \omega (x) := |x-r|^{a} {\left\{ \begin{array}{ll} e^{u}, &{} \text{ if } x < r, \\ 1, &{} \text{ if } x \ge r. \end{array}\right. } \end{aligned}

Hence our results can also be seen as large n asymptotics for $$n \times n$$ determinants whose weight is supported on $$\mathbb {C}$$, rotation-invariant, has both jump- and root-type singularities along the circle centered at 0 of radius r (which we will call “circular" jump- and root-type singularities), and a “pointwise" root-type singularity at 0.

Since w is rotation-invariant, $$\int _{\mathbb {C}} z^{j} {\overline{z}}^{k} w(z) d^{2}z = 0$$ for $$j \ne k$$, and therefore

\begin{aligned} D_{n} = \prod _{j=0}^{n-1} \int _{\mathbb {C}} |z|^{2j} w(z) d^{2}z = (2\pi )^{n} \prod _{j=1}^{n} \int _{0}^{+\infty }v^{2j-1}w(v)dv. \end{aligned}
(1.3)

Over the past 50 years or so a lot of works have been done on structured determinants with singularities, and we briefly pause here to review the literature. In their pioneering work [38], Fisher and Hartwig made a conjecture for the asymptotics of large Toeplitz determinants when the weight is supported on the unit circle and has root- and jump-type singularities—such singularities are now called Fisher–Hartwig singularities. Many authors have contributed in proving this conjecture for certain parameter ranges, among which Lenard [53], Widom [65], Basor [7], Böttcher and Silbermann [13], and Ehrhardt [33]. A counterexample to the Fisher–Hartwig conjecture was found by Basor and Tracy in [10], and the corrected conjecture was solved for general values of the parameters by Deift, Its and Krasovsky [31]. The study of these singular determinants was motivated mainly from questions arising in the Ising model and impenetrable bosons, see [9, 32] for more historical background. In recent years, these determinants have also attracted considerable attention in the random matrix community. One reason for that is the well-known work [45] of Keating and Snaith, where numerical evidences were found of links between the characteristic polynomials of unitary and Hermitian random matrices and the zeros of the Riemann zeta function on the critical line. Expectations of powers of the absolute value of the characteristic polynomial of the Gaussian Unitary Ensemble, which are Hankel determinants with a Gaussian weight on $$\mathbb {R}$$ and root-type singularities, were investigated in [14] and their asymptotics were obtained by Garoni [42] for integer values of the parameters and by Krasovsky [46] for the general case. This result was then generalized by Berestycki, Webb and Wong [11] for one-cut regular ensembles. In a different direction, Its and Krasovsky in [44] obtained asymptotics of Hankel determinants with a jump-type singularity and a Gaussian weight. Such determinants provide information about the imaginary part of the log-characteristic polynomial of the Gaussian Unitary Ensemble. The results [11, 44] have been generalized in [19] for general Fisher–Hartwig singularities and one-cut regular ensembles. The case of one-cut regular ensembles with hard edges was then treated in [23], and the multi-cut case in [24]. Strong results on Toeplitz determinants with merging Fisher–Hartwig singularities are also available in the literature [27, 35]; these results have been useful to prove a conjecture of [41] on “the moments of the moments" of the characteristic polynomial of random unitary matrices, and [27] has also been used by Webb in [63] to establish a connection between random matrix theory and Gaussian multiplicative chaos. There exists also a vast literature on other structured determinants with Fisher–Hartwig singularities, see e.g. [28, 29] for Fredholm determinants, [8, 26, 39] for Toeplitz+Hankel determinants, and [22] for a biorthogonal generalization of Hankel determinants.

The literature on determinants associated with a singular weight supported on $$\mathbb {C}$$ is more limited. For $$a=0$$, (1.2) is the moment generating function of the disk counting function

\begin{aligned} \mathbb {E}\Big [ e^{\frac{u}{\pi } \, \text {Im\,}\ln p_{n}(r)} \Big ] = \mathbb {E}\Big [ e^{u \, \#\{z_{j}: |z_{j}|< r\}} \Big ], \end{aligned}
(1.4)

and in this case w is discontinuous along a circle but has no circular root-type singularity. Counting statistics of two-dimensional point processes have attracted a lot of interest in recent years [1, 18, 20, 25, 34, 37, 47,48,49, 59, 60]. The first two terms in the large n asymptotics of (1.4) were derived in [18, 48]Footnote 1. More precise asymptotics, including the third term of order 1, were obtained in [37] for the Ginibre ensemble (i.e. $$(b,\alpha )=(1,0)$$),Footnote 2 and the asymptotics of (1.4) including the fourth term of order $$n^{-\frac{1}{2}}$$ were then obtained in [20] for general $$b>0$$ and $$\alpha >-1$$. Several works on determinants with pointwise root-type singularities in dimension two are also available in the literature. In [5, 6, 12, 17, 50,51,52], the orthogonal polynomials for the planar Gaussian weight perturbed with a finite number of pointwise root-type singularities have been studied, see also [3] where microscopic properties of the associated point process have been analyzed. Building on [5, 50], Webb and Wong in [64] obtained a precise asymptotic formula for $$\mathbb {E} [ e^{a \, \text {Re\,}\ln q_{n}(r)} ]$$ where $$a\in \mathbb {C}$$ is fixed, $$\text {Re\,}a>-2$$, $$r<1$$ and $$q_{n}$$ is the characteristic polynomial of a Ginibre matrix, i.e. $$q_{n}(x) = \prod _{j=1}^{n}(z_{j}-x)$$ and the $$z_{j}$$ are drawn from (1.1) with $$(b,\alpha )=(1,0)$$. This expectation involves a determinant with a single pointwise root-type singularity in the bulk. The case where r is close to 1, which corresponds to the edge regime, was then investigated by Deaño and Simm in [30]. Determinants with two merging planar pointwise root-type singularities were also considered in [30], and the asymptotics were found to involve some Painlevé transcendents. The regime where a is proportional to n reveals a topological phase transition and was studied in [17]. (We also mention that determinants with regular weights supported on $$\mathbb {C}$$ have recently been studied in [2, 16].)

Determinants with circular root-type singularities have not been considered before to our knowledge. A main difficulty in the analysis of pointwise root-type singularities in dimension two stems from the fact that they break the rotation-invariance of the weight (unless of course if they are located at 0). The circular root-type singularities preserve the rotation-invariance of the weight, which makes them simpler to analyze in this respect, but they also pose a series of new challenges, which we discuss at the end of this section, and which we have been able to overcome only for integer values of a. Interestingly, these circular root-type singularities also produce some associated Hermite polynomials (see below for the definition) in the asymptotics. This came as a surprise to us and appears to be completely new. There are of course many exact formulas in random matrix theory which involve the Hermite polynomials, but we are not aware of an earlier work where these polynomials show up explicitly in the asymptotics of large determinants, let alone the associated Hermite polynomials (see however [36] where Legendre polynomials appear in the asymptotics of the sine-kernel determinant in a transition regime). For comparison, pointwise root-type singularities typically produce other kinds of ingredients in the asymptotics, such as Barnes’ G-function (as was discovered by Basor [7] in dimension one and by Webb and Wong [64] in dimension two), and circular jump-type singularities involve the error function. We also mention that ensembles with circular root-type singularities have been studied in [58, 67], and ensembles with “elliptic” root-type singularities in [55]. In [55, 58, 67], the singularities are located at the hard edge and the focus was on the leading order behavior of the kernel; in particular, the (associated) Hermite polynomials do not show up in these works.

The $$\nu$$-th associated Hermite polynomials $$\{ \textrm{He}_k^{(\nu )}:k=0,1,\dots \}$$ are defined recursively by

\begin{aligned} {\left\{ \begin{array}{ll} \textrm{He}_{k+1}^{(\nu )}(x)= x \, \textrm{He}_k^{(\nu )}(x)-(k+\nu )\textrm{He}^{(\nu )}_{k-1}(x), &{} k \ge 1, \\ \textrm{He}_0^{(\nu )}(x)=1, \quad \textrm{He}_1^{(\nu )}(x)=x, \end{array}\right. } \end{aligned}
(1.5)

and satisfy the orthogonality relations [4]

\begin{aligned} \int _{-\infty }^{+\infty } \textrm{He}_{k}^{(\nu )}(x) \textrm{He}_{\ell }^{(\nu )}(x) \frac{dx}{|D_{-\nu }(ix)|^{2}} = \sqrt{2\pi } \, (k+\nu )! \, \delta _{k\ell }, \qquad k,\ell = 0,1,\ldots \end{aligned}

where $$D_{-\nu }$$ is the parabolic cylinder function (see e.g. [56, Chapter 12]). These polynomials are explicitly given by

\begin{aligned} \textrm{He}_k^{(\nu )}(x) = {\left\{ \begin{array}{ll} \displaystyle k! \sum _{\ell =0 }^{ \lfloor k/2 \rfloor } \frac{(-1)^\ell }{\ell !(k-2\ell )!}\frac{x^{k-2\ell }}{2^\ell } = \Big [ \frac{d}{dt} \Big ]^k \Big [ e^{xt-\frac{t^2}{2}}\Big ] \Big |_{t=0}, &{} \text{ if } \nu =0, \\ \displaystyle \sum _{\ell =0}^{\lfloor k/2\rfloor } \frac{(-1)^{\ell }}{(k-2\ell )!}\bigg ( \sum _{j=0}^{\ell } \frac{(k-j)!(\nu -1+j)!}{j!(\ell -j)!(\nu -1)!2^{\ell -j}} \bigg )x^{k-2\ell }, &{} \text{ if } \nu \ge 1, \end{array}\right. } \end{aligned}
(1.6)

see [66, eq. (4.12)]. For $$\nu =0$$, they reduce to the standard Hermite polynomialsFootnote 3, i.e. $$\textrm{He}_{k}^{(0)}(x) = \textrm{He}_{k}(x)$$ for all $$k \in \mathbb {N}$$. We refer to [62] for basic properties of general associated polynomials, and to [66] for a focus on the Hermite case.

Only the polynomials $$\{\textrm{He}_{k},\textrm{He}_{k}^{(1)}:k=0,1,\ldots \}$$, corresponding to $$\nu =0$$ and $$\nu =1$$, will appear in our asymptotic formula. Our results can be presented in a unified way if we formally define $$\textrm{He}_{k}$$, $$\textrm{He}_{k}^{(1)}$$ for the first few negative k as follows:

\begin{aligned}&\textrm{He}_{-1}(x) := 0,{} & {} \textrm{He}_{-2}(x) := 1,{} & {} \textrm{He}_{-3}(x) := -\frac{x}{2}, \nonumber \\&\textrm{He}_{-1}^{(1)}(x) := 0,{} & {} \big [ k \textrm{He}_{k-2}^{(1)}(x) \big ]_{k=0} := -1,{} & {} \big [ k \textrm{He}_{k-3}^{(1)}(x) \big ]_{k=0} := x,{} & {} \nonumber \\&\big [ k \textrm{He}_{k-4}^{(1)}(x) \big ]_{k=0} := -\frac{x^{2}+1}{2}. \end{aligned}
(1.7)

These definitions are consistent with the recurrence (1.5). For general $$a \in \mathbb {N}$$, we define

\begin{aligned} p_{0,a}(x)&:= \frac{1}{i^a} \textrm{He}_a(ix) = \sum _{s=0 }^{ \lfloor a/2 \rfloor } \frac{ a! }{s!(a-2s)!} \frac{x^{a-2s}}{2^s}, \end{aligned}
(1.8)
\begin{aligned} q_{0,a}(x)&:= \!\frac{1}{i^{a-1}} \textrm{He}_{a-1}^{(1)}(ix) = \! \!\sum _{s=0 }^{ \lfloor (a-1)/2 \rfloor } \bigg ( \sum _{j=0}^s \frac{(a-1-j)!\,2^j}{(s-j)! } \bigg ) \frac{1}{(a-1-2s)!} \frac{ x^{a-1-2s}}{2^s}, \end{aligned}
(1.9)
\begin{aligned} p_{1,a}(x)&:= -\frac{a}{2} p_{0,a+1}(x) -ab \Big ( p_{0,a+1}(x)- (3a-1) \,p_{0,a-1}(x) \nonumber \\&\qquad + \frac{5}{3}(a-1)(a-2) \, p_{0,a-3}(x) \Big ), \end{aligned}
(1.10)
\begin{aligned} q_{1,a}(x)&:= -\frac{a}{2} q_{0,a+1}(x) -b \Big ( aq_{0,a+1}(x)- (3a-1) [a\,q_{0,a-1}(x)] \nonumber \\&\qquad + \frac{5}{3} [a(a-1)(a-2) \, q_{0,a-3}(x)] \Big ), \end{aligned}
(1.11)

where the brackets in (1.11) emphasize that for the first values of a, one needs to use (1.7), namely

\begin{aligned}&[a q_{0,a-1}(x)] := {\left\{ \begin{array}{ll} 1, &{} \text{ if } a=0, \\ a q_{0,a-1}(x), &{} \text{ if } a \ge 1, \end{array}\right. } \nonumber \\&[a(a-1)(a-2) q_{0,a-3}(x)] := {\left\{ \begin{array}{ll} x^{2}-1, &{} \text{ if } a = 0, \\ -x, &{} \text{ if } a = 1, \\ 2, &{} \text{ if } a = 2, \\ a(a-1)(a-2) q_{0,a-3}(x), &{} \text{ if } a \ge 3. \end{array}\right. } \nonumber \\ \end{aligned}
(1.12)

To be concrete, the first polynomials are given by

\begin{aligned} \big (p_{0,a}(x)\big )_{a=0}^{4}&= \big ( p_{0,0}(x), p_{0,1}(x), p_{0,2}(x), p_{0,3}(x), p_{0,4}(x) \big ) \\&= \Big (1, \; x, \; x^{2}+1, \; x^{3}+3x, \; x^{4}+6x^{2}+3 \Big ), \\ \big (q_{0,a}(x)\big )_{a=0}^{4}&= \Big (0, \; 1, \; x, \; x^{2}+2, \; x^{3}+5x \Big ),\\ \big (p_{1,a}(x)\big )_{a=0}^{4}&=- \big (\frac{a}{2} p_{0,a+1}(x)\big )_{a=0}^{4} \\&\quad +b\, \Big ( 0, -x^2+1, -2x^3+4x, -3x^4\!+\!6x^2\!+\!5,\!-\!4x^5\!+\!4x^3\!+\!32x \Big ),\\ \big (q_{1,a}(x)\big )_{a=0}^{4}&= - \big (\frac{a}{2} q_{0,a+1}(x)\big )_{a=0}^{4} + b \, \Big ( -\frac{5}{3}x^2+\frac{2}{3}, \frac{2}{3} x, -2x^2+\frac{8}{3}, \\&\quad -3x^3+9x , -4x^4+8x^2+16 \Big ). \end{aligned}

The large n asymptotics of $$\mathbb {E}\big [ e^{\frac{u}{\pi } \, \text {Im\,}\ln p_{n}(r)}e^{a \, \text {Re\,}\ln p_{n}(r)} \big ]$$ are naturally described in terms of the two functions

\begin{aligned} \mathcal {G}_{0}(y; u,a):=&p_{0,a}(-\sqrt{2}y)\bigg ( (-1)^{a} + \frac{e^{u}-(-1)^{a}}{2}\textrm{erfc}(y) \bigg )\nonumber \\&+ q_{0,a}(-\sqrt{2} y) (e^{u}-(-1)^{a}) \frac{e^{-y^{2}}}{\sqrt{2\pi }}, \end{aligned}
(1.13)
\begin{aligned} \mathcal {G}_{1}(y; u,a) :=&p_{1,a}(-\sqrt{2}y)\bigg ( (-1)^{a} + \frac{e^{u}-(-1)^{a}}{2}\textrm{erfc}(y) \bigg ) \nonumber \\&+ q_{1,a}(-\sqrt{2} y) (e^{u}-(-1)^{a}) \frac{e^{-y^{2}}}{\sqrt{2\pi }}, \end{aligned}
(1.14)

where $$y \in \mathbb {R}$$, $$u \in \mathbb {R}$$, $$a \in \mathbb {N}$$, and $$\textrm{erfc}$$ is the complementary error function

\begin{aligned} \textrm{erfc} (y) = \frac{2}{\sqrt{\pi }}\int _{y}^{\infty } e^{-x^{2}}dx. \end{aligned}
(1.15)

In the statement of our theorem, $$\mathcal {G}_{0}$$ appears inside a logarithm and in a denominator. It turns out that $$\mathcal {G}_{0}(y; u,a)>0$$ for all $$y \in \mathbb {R}$$, $$u \in \mathbb {R}$$ and $$a \in \mathbb {N}$$. This fact is not obvious so we defer the proof to Sect. 4, see Lemma 4.9.

### Theorem 1.1

Let $$b>0$$, $$\alpha > -1$$, $$r \in (0,b^{-\frac{1}{2b}})$$, $$u \in \mathbb {R}$$ and $$a \in \mathbb {N}$$ be fixed parameters. As $$n \rightarrow + \infty$$,

\begin{aligned} \mathbb {E}\Big [ e^{\frac{u}{\pi } \, \text {Im\,}\ln p_{n}(r)}e^{a \, \text {Re\,}\ln p_{n}(r)} \Big ] {=} \exp \bigg ( C_{1} n {+} C_{2} \sqrt{n} + C_{3} {+} \mathcal{O}\Big ( n^{-\frac{1}{2b}} + (\ln n)^{\frac{5}{8}}n^{-\frac{1}{8}} \Big )\bigg ), \end{aligned}
(1.16)

where

\begin{aligned}&C_{1} = \int _{0}^{r} \Big ( u+a \ln (r-y) \Big )d\mu (y) + \int _{r}^{b^{-\frac{1}{2b}}} a \ln (y-r) \, d\mu (y), \\&C_{2} = \sqrt{2}\, b r^{b} \int _{-\infty }^{+\infty } \Big ( \ln \mathcal {G}_{0}(y; u,a) - a \ln (\sqrt{2}|y|) - u \chi _{(-\infty ,0)}(y) \Big )\, dy,\\&C_{3} = - \bigg ( \frac{1}{2}+\alpha \bigg )u + \frac{a(1-a)}{4(1-(br^{2b})^{\frac{1}{2b}})} + \frac{a}{4}(2+a-2b+4\alpha ) \ln \Big ( (br^{2b})^{-\frac{1}{2b}}-1 \Big ) \\&\qquad \quad + \int _{-\infty }^{+\infty } \bigg \{ \frac{1}{\sqrt{2}} \,\frac{ \mathcal {G}_1( y ; u,a) }{ \mathcal {G}_0(y; u,a) } + 4b y \Big ( \ln (\mathcal {G}_{0}(y; u,a))- u \chi _{(-\infty ,0)}(y) \Big ) \\&\quad \qquad - \frac{a}{2} y \Big ( 1+2b+8b\ln (\sqrt{2}|y|) \Big ) + \frac{(2ab-a^{2})y}{4(1+y^{2})} \bigg \} \,dy, \end{aligned}

$$d\mu (y) = 2b^{2}y^{2b-1}dy$$, and $$\chi _{(-\infty ,0)}(y)= 0$$ for $$y \ge 0$$ and $$\chi _{(-\infty ,0)}(y)=1$$ for $$y<0$$.

### Remark 1.2

For $$a=0$$, the next term in (1.16) is of order $$n^{-\frac{1}{2}}$$ and was obtained explicitly in [20]. For $$a \ge 1$$, numerical simulations suggest that the $$\mathcal{O}$$-term is in fact of order $$n^{-\frac{1}{2}} + n^{-\frac{1}{2b}}$$. This also suggests that our estimate in (1.16) for the $$\mathcal{O}$$-term is optimal for $$b>4$$ and $$a \ge 1$$.

### 1.1 Outline of the proof of Theorem 1.1

Let $$\mathcal {E}_{n}:= \mathbb {E}\big [ e^{\frac{u}{\pi } \, \text {Im\,}\ln p_{n}(r)}e^{a \, \text {Re\,}\ln p_{n}(r)} \big ]$$. Our starting point is the following exact formula

\begin{aligned} \ln \mathcal {E}_{n}&= \sum _{j=1}^{n} \ln \bigg (\sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})}\nonumber \\&\quad \bigg [1+((-1)^{a}e^{u}-1)\frac{\gamma (\frac{2j+2\alpha +k}{2b},nr^{2b})}{\Gamma (\frac{2j+2\alpha +k}{2b})} \bigg ] \bigg ), \end{aligned}
(1.17)

where $$\gamma ({\tilde{a}},z)$$ is the incomplete gamma function

\begin{aligned} \gamma ({\tilde{a}},z) = \int _{0}^{z}t^{{\tilde{a}}-1}e^{-t}dt. \end{aligned}

Formula (1.17) can be derived using the facts that (1.1) is determinantal and that w is rotation-invariant, see Lemma 2.1 below. For fixed j and $$a \ge 1$$, it is easy to see that the summand in (1.17) contains a term proportional to $$n^{-\frac{1}{2b}}$$ in its large n asymptotics. This already explains why our estimate for the error term in (1.16) contains $$n^{-\frac{1}{2b}}$$.

To obtain precise asymptotics for $$\mathcal {E}_{n}$$, up to and including the term $$C_{3}$$ of order 1, we must take into account each of the n terms in the sum (1.17) (they all contribute). As can be seen from (1.17), this means that we need precise uniform asymptotics for $$\gamma ({\tilde{a}},z)$$ as both $$z \rightarrow + \infty$$ and $${\tilde{a}} \rightarrow + \infty$$ at various different relative speeds. Fortunately, these asymptotics are available in the literature [61]. Following the approach of [40] (which was further developed in [20, 21]), we will split the sum (1.17) in several parts,

\begin{aligned} \ln \mathcal {E}_{n} = S_{0} + S_{1} + S_{2} + S_{3}, \end{aligned}

where $$S_{\ell }$$, $$\ell =0,1,2,3$$ are given in (2.5–(2.8). There is a critical transition in the large $${\tilde{a}}$$ asymptotics of $$\gamma ({\tilde{a}},z)$$ when $$z \rightarrow + \infty$$ is such that $$\lambda = \frac{z}{{\tilde{a}}}\approx 1$$. The sum $$S_{2}$$ is the hardest one and precisely corresponds to this critical transition; it requires a “local analysis” involving the j-terms in (1.17) for which $$\smash {\frac{bnr^{2b}}{j}}\approx 1$$.

We found that, quite surprisingly, circular root-type singularities are significantly more involved to analyze than circular jump-type singularities. Let us highlight some of the reasons for that:

• The “global analysis” needed for $$S_{0}$$, $$S_{1}$$ and $$S_{3}$$ requires some precise Riemann sum approximations for functions with singularities. For comparison, the analogue of $$S_{0}$$, $$S_{1}$$ and $$S_{3}$$ in [20] in the case of pure circular jump-type singularities are straightforward to analyze, because the corresponding Riemann sum approximations only involve constant functions.

• Huge cancellations occur in the “local analysis” of $$S_{2}$$. In fact, to obtain $$C_{3}$$, we need to expand up to the $$(a+2)$$-th order the summand of the k-sum in (1.17). This is because, curiously, the first a terms in the expansion cancel perfectly after summing over k. To treat the general case $$a \in \mathbb {N}$$, this means that we need to expand various quantities to all orders. An important technical obstacle is that the coefficients in these various expansions are not always readily available in an explicit form; sometimes they can only be found recursively and involve heavy combinatorics, see e.g. Lemma 4.5 and the all-order expansion of $$\gamma$$ in Lemma A.2. The analysis of $$S_{2}$$ is in fact the only part in the proof where solving the problem for general $$a \in \mathbb {N}$$ is clearly harder than solving the problem for a finite number of values of a, say $$a \in \{0,1,2,3,4\}$$. This is also the only place in the proof where the (associated) Hermite polynomials arise, see Lemmas 4.6 and 4.7.

### Remark 1.3

For non-integer values of a, formula (1.17) does not hold and the connection with the incomplete gamma function is lost (and therefore the strong results from [61] cannot be used anymore). This is the main reason as to why we decided to restrict ourselves to $$a \in \mathbb {N}$$ in this work. For $$a \notin \mathbb {N}$$, the exact expression for $$\mathcal {E}_{n}$$ involves hypergeometric functions that generalize the incomplete gamma function. Also, because of the well-known relation $$D_{k}(z) = e^{-\frac{z^{2}}{4}}\textrm{He}_{k}(z)$$, $$k\in \mathbb {N}$$ (see [56, eq 12.7.2]), it is tempting to conjecture that for the general case $$a \in (-1,+\infty )$$ the large n asymptotics of $$\mathcal {E}_{n}$$ involve the parabolic cylinder function. It would be very interesting to figure that out in detail.

Outline of the paper. In Sect. 2, we prove (1.17), define the sums $$S_{j}$$, $$j=0,1,2,3$$, and establish many useful lemmas. In Sect. 3, we obtain the large n asymptotics of $$S_{0}$$, $$S_{3}$$ and $$S_{1}$$. The large n asymptotics of $$S_{2}$$ are then obtained in Sect. 4. We finish the proof of Theorem 1.1 in Sect. 5.

## 2 Preliminaries

This section contains the proof of (1.17) and the definitions of $$S_{0},\ldots ,S_{3}$$. We also establish here various preliminary lemmas that will be used in Sects. 3 and 4.

### Lemma 2.1

Formula (1.17) holds for all $$n \in \mathbb {N}_{>0}:=\{1,2,\ldots \}$$.

### Proof

The partition function $$Z_{n}$$ of the Mittag–Leffler ensemble is known to be

\begin{aligned} Z_{n} = n^{-\frac{n^{2}}{2b}}n^{-\frac{1+2\alpha }{2b}n} \frac{\pi ^{n}}{b^{n}} \prod _{j=1}^{n} \Gamma (\tfrac{j+\alpha }{b}), \end{aligned}
(2.1)

see e.g. [20, eq. (1.23)]. Since $$\mathcal {E}_{n}=D_{n}/Z_{n}$$, it only remains to find a simplified exact expression for $$D_{n}$$. Since $$a \in \mathbb {N}$$,

\begin{aligned} w(v) = v^{2\alpha }e^{-nv^{2b}} {\left\{ \begin{array}{ll} \displaystyle e^{u}\sum _{k=0}^{a} {a \atopwithdelims ()k} (-1)^{k}v^{k}r^{a-k}, &{} \text{ if } v<r, \\ \displaystyle \sum _{k=0}^{a} {a \atopwithdelims ()k} (-1)^{a-k}v^{k}r^{a-k}, &{} \text{ if } v\ge r, \end{array}\right. } \end{aligned}

and thus, by (1.3),

\begin{aligned} D_{n}&= n^{-\frac{n^{2}}{2b}}n^{-\frac{1+2\alpha }{2b}n} \frac{\pi ^{n}}{b^{n}} \prod _{j=1}^{n} \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \bigg (\Gamma (\tfrac{2j+2\alpha +k}{2b}) + ((-1)^{a}e^{u}-1)\nonumber \\&\quad \gamma (\tfrac{2j+2\alpha +k}{2b},nr^{2b}) \bigg ). \end{aligned}
(2.2)

The claim now follows directly from (2.1), (2.2) and $$\mathcal {E}_{n}=D_{n}/Z_{n}$$. $$\square$$

Throughout the paper, c and C denote positive constants which may change within a computation, and $$\ln$$ always denotes the principal branch of the logarithm.

Let $$M'$$ be a large integer independent of n, let $$\epsilon > 0$$ be a small constant independent of n, and let $$M:= n^{\frac{1}{8}}(\ln n)^{-\frac{1}{8}}$$. Define

\begin{aligned}&j_{-}:=\lceil \tfrac{bnr^{2b}}{1+\epsilon } - \alpha \rceil , \qquad j_{+} := \lfloor \tfrac{bnr^{2b}}{1-\epsilon } - \alpha \rfloor . \end{aligned}
(2.3)

We choose $$\epsilon$$ small enough so that

\begin{aligned} \frac{br^{2b}}{1-\epsilon } < \frac{1}{1+\epsilon }. \end{aligned}

Using (1.17), we divide $$\ln \mathcal {E}_{n}$$ into 4 parts

\begin{aligned} \ln \mathcal {E}_{n} = S_{0} + S_{1} + S_{2} + S_{3}, \end{aligned}
(2.4)

with

\begin{aligned}&S_{0} = \sum _{j=1}^{M'} \ln \bigg ( \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \bigg [1+((-1)^{a}e^{u}-1)\frac{\gamma (\frac{2j+2\alpha +k}{2b},nr^{2b})}{\Gamma (\frac{2j+2\alpha +k}{2b})} \bigg ] \bigg ), \end{aligned}
(2.5)
\begin{aligned}&S_{1} = \sum _{j=M'+1}^{j_{-}-1} \hspace{-0.25cm} \ln \bigg ( \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \bigg [ 1+((-1)^{a}e^{u}-1)\frac{\gamma (\frac{2j+2\alpha +k}{2b},nr^{2b})}{\Gamma (\frac{2j+2\alpha +k}{2b})} \bigg ] \bigg ), \end{aligned}
(2.6)
\begin{aligned}&S_{2} = \sum _{j=j_{-}}^{j_{+}} \ln \bigg ( \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \bigg [ 1+((-1)^{a}e^{u}-1) \frac{\gamma (\frac{2j+2\alpha +k}{2b},nr^{2b})}{\Gamma (\frac{2j+2\alpha +k}{2b})} \bigg ] \bigg ), \end{aligned}
(2.7)
\begin{aligned}&S_{3}=\sum _{j=j_{+}+1}^{n} \hspace{-0.2cm} \ln \bigg ( \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \bigg [ 1+((-1)^{a}e^{u}-1)\frac{\gamma (\frac{2j+2\alpha +k}{2b},nr^{2b})}{\Gamma (\frac{2j+2\alpha +k}{2b})} \bigg ] \bigg ). \end{aligned}
(2.8)

In Sects. 3 and 4, we will analyze these sums in order of increasing difficulty: first $$S_{0}$$, then $$S_{3}$$, then $$S_{1}$$, and finally $$S_{2}$$. The sum $$S_{0}$$ is straightforward to analyze, but $$S_{1}$$, $$S_{2}$$ and $$S_{3}$$ are more involved and require some preparation. This preparation is carried out in the next subsection.

### 2.1 Useful Lemmas

For $$\ell \in \mathbb {N}:=\{0,1,\ldots \}$$, let

\begin{aligned} {\mathfrak {g}}_\ell (x):= \sum _{k=0}^a \left( {\begin{array}{c}a\\ k\end{array}}\right) x^k k^\ell . \end{aligned}
(2.9)

If $$k=\ell =0$$ in (2.9), then $$k^{\ell }:=1$$ so that $${\mathfrak {g}}_{0}(x)=(1+x)^{a}$$.

### Remark 2.2

The sequence $$\{{\mathfrak {g}}_{\ell }\}_{\ell =0}^{+\infty }$$ satisfies $${\mathfrak {g}}_{\ell +1}(x)= x {\mathfrak {g}}_\ell '(x)$$, $$\ell \in \mathbb {N}$$. Solving this recurrence relation using the initial value $${\mathfrak {g}}_{0}(x)=(1+x)^{a}$$ yields

\begin{aligned} {\mathfrak {g}}_\ell (x)= \Big [ \frac{d}{dt}\Big ]^{\ell } \Big [ (1+e^t)^a \Big ] \Big |_{t=\ln x}, \end{aligned}

which is an interesting alternative representation of $${\mathfrak {g}}_{\ell }$$.

The next lemma establishes yet another representation of $${\mathfrak {g}}_{\ell }$$.

### Lemma 2.3

For $$\ell \in \mathbb {N}_{>0}$$, we have

\begin{aligned} {\mathfrak {g}}_{\ell }(x) = x (x+1)^{a-\min (\ell ,a)} \sum _{j=1}^{\min (\ell ,a)} S(\ell ,j) \frac{a!}{(a-j)!} x^{j-1}(x+1)^{\min (\ell ,a)-j}, \end{aligned}
(2.10)

where $$S(\ell ,j)$$ is the Stirling number of the second kind, i.e. the number of partitions of $$\{1, \dots , \ell \}$$ into exactly j nonempty subsets. Furthermore,

\begin{aligned}&{\mathfrak {g}}_{\ell }(x) = \mathcal{O}(x),{} & {} \text{ as } x \rightarrow 0, \quad \ell \in \mathbb {N}_{>0}, \end{aligned}
(2.11)
\begin{aligned}&{\mathfrak {g}}_{\ell }(x) = \mathcal{O}(\min \{1,|x+1|^{a-\ell }\}),{} & {} \text{ as } x \rightarrow -1, \quad \ell \in \mathbb {N}. \end{aligned}
(2.12)

### Remark 2.4

Since $${\mathfrak {g}}_{0}(x)=(1+x)^{a}$$, (2.10) and (2.11) do not hold for $$\ell =0$$.

### Proof

Let $$\ell \in \mathbb {N}_{>0}$$ be fixed. By [56, eq. 26.8.10], we have

\begin{aligned} k^{\ell }= \sum _{j=1}^\ell (k)_{j} S(\ell ,j), \qquad \text{ for } \text{ all } k \in \mathbb {C}, \end{aligned}
(2.13)

where $$(k)_{j}:= k(k-1)(k-2)\ldots (k-j+1)$$ is the descending factorial. Substituting (2.13) in (2.9), we obtain

\begin{aligned} {\mathfrak {g}}_\ell (x)&= \sum _{k=0}^a \sum _{j=1}^{\min (\ell ,k)} \frac{a! S(\ell ,j) x^k}{(a-k)!(k-j)!} = \sum _{j=1}^{\min (\ell ,a)}\sum _{k=j}^a \frac{a! S(\ell ,j) x^k}{(a-k)!(k-j)!} \nonumber \\&= \sum _{j=1}^{\min (\ell ,a)} S(\ell ,j) \frac{a!}{(a-j)!} \sum _{k=j}^a \left( {\begin{array}{c}a-j\\ k-j\end{array}}\right) x^k \\&= \sum _{j=1}^{\min (\ell ,a)} S(\ell ,j) \frac{a!}{(a-j)!} x^{j}(x+1)^{a-j}, \end{aligned}

which is (2.10). The expansions (2.11) and (2.12) for $$\ell \ge 1$$ directly follows from (2.10), and (2.12) for $$\ell =0$$ follows from $${\mathfrak {g}}_{0}(x)=(1+x)^{a}$$. $$\square$$

The sums $$S_{1}$$, $$S_{2}$$ and $$S_{3}$$ naturally involve the functions

\begin{aligned} \gamma _{\ell }(x) := {\left\{ \begin{array}{ll} \displaystyle \sum _{k=0}^{a} {a \atopwithdelims ()k} (-r)^{a-k} \bigg ( \frac{x}{b} \bigg )^{\frac{k}{2b}}k^{\ell }, &{} x > b r^{2b}, \\ \displaystyle \sum _{k=0}^{a} {a \atopwithdelims ()k} (-1)^{ k } r^{a-k} \bigg ( \frac{x}{b} \bigg )^{\frac{k}{2b}}k^{\ell }, &{} x \in (0,b r^{2b}). \end{array}\right. } \end{aligned}
(2.14)

The next lemma collects some properties of $$\gamma _{\ell }$$.

### Lemma 2.5

Let $$\ell \in \mathbb {N}$$. The function $$\gamma _{\ell }$$ can be written as

\begin{aligned} \gamma _{\ell }(x) = {\left\{ \begin{array}{ll} \displaystyle \Big |r-\Big (\frac{x}{b}\Big )^{\frac{1}{2b}}\Big |^{a}, &{} \text{ if } \ell =0, \\ \displaystyle \Big (\frac{x}{b}\Big )^{\frac{1}{2b}} \Big |r-\Big (\frac{x}{b}\Big )^{\frac{1}{2b}}\Big |^{a} \sum _{j=1}^{\min (\ell ,a)} \frac{a!S(\ell ,j)}{(a-j)!} \Big (\frac{x}{b}\Big )^{\frac{j-1}{2b}} \Big ( \Big (\frac{x}{b}\Big )^{\frac{1}{2b}}-r \Big )^{-j}, &{} \text{ if } \ell \ge 1. \end{array}\right. } \end{aligned}
(2.15)

In particular,

\begin{aligned}&\gamma _{\ell }(x) = \mathcal{O}(x^{\frac{1}{2b}}),{} & {} \text{ as } x \rightarrow 0, \quad \ell \in \mathbb {N}_{>0}, \end{aligned}
(2.16)
\begin{aligned}&\gamma _{\ell }(x) = \mathcal{O}(\min \{1,|x-br^{2b}|^{a-\ell }\}),{} & {} \text{ as } x \rightarrow br^{2b}, \quad \ell \in \mathbb {N}. \end{aligned}
(2.17)

### Proof

It is easily checked that

\begin{aligned} \gamma _{\ell }(x) = {\left\{ \begin{array}{ll} \displaystyle (-r)^a {\mathfrak {g}}_\ell \Big ( -\Big ( \frac{x}{br^{2b}}\Big )^{\frac{1}{2b}} \Big ) , &{} x > b r^{2b}, \\ r^a {\mathfrak {g}}_\ell \Big ( -\Big ( \dfrac{x}{br^{2b}}\Big )^{\frac{1}{2b}} \Big ) , &{} x \in (0,b r^{2b}). \end{array}\right. } \end{aligned}

The claim is now a straightforward consequence of Lemma 2.3. $$\square$$

### Lemma 2.6

Let $$k \in \mathbb {N}$$ be fixed. As $$j \rightarrow + \infty$$,

\begin{aligned} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \sim \bigg ( \frac{j}{b} \bigg )^{\frac{k}{2b}} \bigg ( 1+ \sum _{\ell =1}^{+\infty } \frac{{\mathfrak {p}}_{2\ell }(k)}{j^{\ell }}\bigg ), \end{aligned}
(2.18)

where

\begin{aligned} {\mathfrak {p}}_{2\ell }(k) := b^\ell \left( {\begin{array}{c}\frac{k}{2b}\\ \ell \end{array}}\right) B_\ell ^{(1+ \frac{k}{2b}) }\Big ( \frac{2\alpha +k}{2b}\Big ) =: \sum _{m=1}^{2\ell } {\mathfrak {p}}_{2\ell ,m}\,k^m, \end{aligned}
(2.19)

$$\left( {\begin{array}{c}\frac{k}{2b}\\ \ell \end{array}}\right) := \frac{\frac{k}{2b}(\frac{k}{2b}-1)\ldots (\frac{k}{2b}-\ell +1)}{\ell !}$$, and $$B_{\ell }^{(k)}(x)$$ is the generalized BernoulliFootnote 4 polynomial of degree $$\ell$$ defined through the generating function

\begin{aligned} \Big ( \frac{t}{e^t-1}\Big )^k \,e^{xt}= \sum _{\ell =0}^{+\infty } B_\ell ^{(k)}(x) \frac{t^n}{n!}, \qquad |t|<2\pi . \end{aligned}
(2.20)

### Remark 2.7

The degree $$2\ell$$ polynomial $${\mathfrak {p}}_{2\ell }$$ satisfies $${\mathfrak {p}}_{2\ell }(0)=0$$. This is consistent with the fact that for $$k=0$$ the left-hand side of (2.18) is 1.

### Remark 2.8

The notation “$$\sim$$" in (2.18) means that for any $$N \in \mathbb {N}$$, $$N \ge 1$$, we have

\begin{aligned} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} = \bigg ( \frac{j}{b} \bigg )^{\frac{k}{2b}} \bigg ( 1+ \sum _{\ell =1}^{N} \frac{{\mathfrak {p}}_{2\ell }(k)}{j^{\ell }} + \mathcal{O}(j^{-N-1})\bigg ), \qquad \text{ as } j \rightarrow +\infty . \end{aligned}

We will use this notation repetitively in the paper.

### Proof

The claim directly follows from [56, eq. 5.11.13]

\begin{aligned} \frac{\Gamma (v+p_{2}+p_{1})}{\Gamma (v+p_{2})} \sim v^{p_{1}} \sum _{\ell =0}^{+\infty } \left( {\begin{array}{c}p_{1}\\ \ell \end{array}}\right) \frac{B_{\ell }^{(p_{1}+1)}(p_{1}+p_{2})}{v^{\ell }} \qquad \text{ as } v \rightarrow + \infty , \; p_{1},p_{2} \text{ fixed }. \end{aligned}

$$\square$$

### Lemma 2.9

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

\begin{aligned}&\sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \sim \gamma _{0}(j/n) + \sum _{\ell =1}^{+\infty } \frac{1}{n^{\ell }} \frac{\sum _{m=1}^{2\ell }{\mathfrak {p}}_{2\ell ,m}\gamma _{m}(j/n)}{(j/n)^{\ell }}, \nonumber \\ \end{aligned}
(2.21)

uniformly for $$j \in \{j_{+}+1,\ldots ,n\}$$. Furthermore, $$M'$$ can be chosen sufficiently large (but fixed) such that the following holds: there exists $$C>0$$ such that

\begin{aligned}&\bigg |\sum _{k=0}^{a} {a \atopwithdelims ()k} \hspace{-0.07cm} \frac{r^{a-k}(-1)^{k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \hspace{-0.04cm} - \hspace{-0.04cm} \bigg \{ \hspace{-0.05cm} \gamma _{0}(j/n) \hspace{-0.03cm} + \hspace{-0.03cm} \frac{\gamma _{2}(j/n)\hspace{-0.03cm}+\hspace{-0.03cm}(4\alpha -2b)\gamma _{1}(j/n)}{8bj} \bigg \} \bigg | \nonumber \\&\quad \le C \frac{(j/n)^{\frac{1}{2b}-2}}{n^{2}}, \end{aligned}
(2.22)

for all sufficiently large n and all $$j \in \{M'+1,\ldots ,j_{-}-1\}$$.

### Proof

Since a is fixed, Lemma 2.6 implies that

\begin{aligned}&\sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \sim \sum _{k=0}^{a} {a \atopwithdelims ()k} (-r)^{a-k} \bigg ( \frac{j/n}{b} \bigg )^{\frac{k}{2b}} \bigg ( 1+ \sum _{\ell =1}^{+\infty } \frac{{\mathfrak {p}}_{2\ell }(k)}{j^{\ell }}\bigg ), \end{aligned}
(2.23)

as $$j \rightarrow + \infty$$. Since $$j/n \in (br^{2b},1]$$ for all $$j \in \{j_{+}+1,\ldots ,n\}$$, the expansion (2.21) directly follows from (2.23) and the definition (2.14) of $$\gamma _{\ell }$$. In the same way, but using now the definition (2.14) of $$\gamma _{\ell }(x)$$ for $$x \in (0,br^{2b})$$, we infer that

\begin{aligned}&\sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{r^{a-k}(-1)^{k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \sim \gamma _{0}(j/n) + \sum _{\ell =1}^{+\infty } \frac{\sum _{m=1}^{2\ell }{\mathfrak {p}}_{2\ell ,m}\gamma _{m}(j/n)}{j^{\ell }},\nonumber \\ \end{aligned}
(2.24)

as $$j \rightarrow + \infty$$ uniformly for n such that $$j/n \in (0,br^{2b})$$. The estimate (2.22) then follows from (2.24) and the behavior (2.16). $$\square$$

For the large n analysis of $$S_{3},S_{1},S_{2}$$, we will need to approximate various large sums of the form $$\sum _{j} f(\tfrac{j}{n})$$ involving some functions f with singularities and some j’s for which $$\frac{j}{n}$$ is close to these singularities; for this we will also rely on the following lemma from [21].

### Lemma 2.10

([21, 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 \text{ 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(x)dx + \frac{(1-2a_{0})f(A)+(1+2b_{0})f(B)}{2}\\&\quad + \frac{(-1+6a_{0}-6a_{0}^{2})f'(A)+(1+6b_{0}+6b_{0}^{2})f'(B)}{12n}\\&\quad + \frac{(-a_{0}+3a_{0}^{2}-2a_{0}^{3})f''(A)+(b_{0}+3b_{0}^{2} +2b_{0}^{3})f''(B)}{12n^{2}}\\&\quad + \mathcal{O}\bigg (\frac{{\mathfrak {m}}_{A}(f''')+{\mathfrak {m}}_{B}(f''')}{n^{3}} + \sum _{j=a_{n}}^{b_{n}-1} \frac{{\mathfrak {m}}_{j,n}(f'''')}{n^{4}} \bigg ), \end{aligned}

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

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

## 3 Global Analysis: Large n Asymptotics of $$S_{0}$$, $$S_{3}$$ and $$S_{1}$$

As mentioned earlier, we will analyze the sums (2.5) –(2.8) in order of increasing difficulty: first $$S_{0}$$, then $$S_{3}$$, then $$S_{1}$$, and finally $$S_{2}$$. In this section we focus on $$S_{0}$$, $$S_{3}$$ and $$S_{1}$$. We defer the analysis of $$S_{2}$$ to the next section.

### Lemma 3.1

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

\begin{aligned} S_{0} = M' \ln (r^{a}e^{u}) + \mathcal{O}(n^{-\frac{1}{2b}} ). \end{aligned}

### Proof

Using (2.5) and Lemma A.1, we obtain

\begin{aligned} S_{0}&= \sum _{j=1}^{M'} \ln \bigg ( \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \bigg [ 1+((-1)^{a}e^{u}-1)[1 + \mathcal{O}(e^{-cn})] \bigg ] \bigg ), \\&= \sum _{j=1}^{M'} \ln \bigg ( \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-1)^{k}r^{a-k}e^{u}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \bigg ) + \mathcal{O}(e^{-cn}), \qquad \text{ as } n \rightarrow + \infty . \end{aligned}

Since $$M'$$ is fixed, only the $$(k=0)$$-terms contribute to order 1 in the large n asymptotics of $$S_{0}$$; the other terms are $$\mathcal{O}(n^{-\frac{1}{2b}})$$. $$\square$$

Recall that $$S_{1},S_{3}$$ are given by (2.6) and (2.8). Following the approach of [20, 21], we define

\begin{aligned} \theta _{+}^{(n,\epsilon )}&= \bigg ( \frac{b n r^{2b}}{1-\epsilon }-\alpha \bigg )-\bigg \lfloor \frac{b n r^{2b}}{1-\epsilon }-\alpha \bigg \rfloor , \qquad \theta _{-}^{(n,\epsilon )} = \bigg \lceil \frac{b n r^{2b}}{1+\epsilon }-\alpha \bigg \rceil \\&\quad -\bigg ( \frac{b n r^{2b}}{1+\epsilon }-\alpha \bigg ), \end{aligned}

and for $$j=1,\ldots ,n$$ and $$k=0,1,\ldots ,a$$, we also define

\begin{aligned}&a_{j}:=\frac{j+\alpha }{b}, \lambda _{j} := \frac{bnr^{2b}}{j+\alpha }, \eta _{j} := (\lambda _{j}-1)\sqrt{\frac{2 (\lambda _{j}-1-\ln \lambda _{j})}{(\lambda _{j}-1)^{2}}}, \end{aligned}
(3.1)
\begin{aligned}&a_{j,k} := \frac{2j+2\alpha +k}{2b}, \lambda _{j,k} := \frac{bnr^{2b}}{j+\alpha +\frac{k}{2}}, \nonumber \\&\eta _{j,k} := (\lambda _{j,k}-1)\sqrt{\frac{2 (\lambda _{j,k}-1-\ln \lambda _{j,k})}{(\lambda _{j,k}-1)^{2}}}. \end{aligned}
(3.2)

### Lemma 3.2

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

\begin{aligned} S_{3}&= n \int _{\frac{br^{2b}}{1-\epsilon }}^{1} a \ln \bigg ( \bigg ( \frac{x}{b} \bigg )^{\frac{1}{2b}}-r \bigg )dx \\&\quad + a \frac{(2\alpha + 2\theta _{+}^{(n,\epsilon )}-1)\ln (r(1-\epsilon )^{-\frac{1}{2b}}-r)+\ln (b^{-\frac{1}{2b}}-r)}{2} \\&+ \frac{a}{4}\bigg ( \frac{1-a}{(br^{2b})^{-\frac{1}{2b}}-1}-\frac{1-a}{(1-\epsilon )^{-\frac{1}{2b}}-1}+(a-2b+4\alpha )\ln \bigg ( \frac{(br^{2b})^{-\frac{1}{2b}}-1}{(1-\epsilon )^{-\frac{1}{2b}}-1} \bigg ) \bigg ) \\&\quad + \mathcal{O}(n^{-1}). \end{aligned}

### Proof

Using (2.8) and Lemma A.2 (ii) with a and $$\lambda$$ replaced by $$a_{j,k}$$ and $$\lambda _{j,k}$$ respectively, where $$j \in \{j_{+}+1,\ldots ,n\}$$ and $$k\in \{0,\ldots ,a\}$$, we obtain

\begin{aligned} S_{3}&{=} \sum _{j=j_{+}+1}^{n} \ln \bigg ( \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j{+}2\alpha {+}k}{2b})}{\Gamma (\frac{2j{+}2\alpha }{2b})} \bigg [ 1+((-1)^{a}e^{u}-1)\mathcal{O}(e^{-\frac{a_{j,k}\eta _{j,k}^{2}}{2}}) \bigg ] \bigg ), \end{aligned}

as $$n \rightarrow + \infty$$. It is easy to check from (3.2) that $$c_{1} n \le a_{j,k} \le c_{1}'n$$, $$c_{2} \le |\lambda _{j,k}-1| \le c_{2}'$$ and $$c_{3} \le \eta _{j,k}^{2} \le c_{3}'$$ hold for some positive constants $$\{c_{j},c_{j}'\}_{j=1}^{3}$$, for all n sufficiently large, for all $$j \in \{j_{+}+1,\ldots ,n\}$$ and for all $$k \in \{0,\ldots ,a\}$$. Thus

\begin{aligned} S_{3}&= \sum _{j=j_{+}+1}^{n} \ln \bigg ( \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \bigg ) + \mathcal{O}(e^{-cn}), \qquad \text{ as } n \rightarrow + \infty . \end{aligned}
(3.3)

To complete the proof of this lemma, we need the following weaker version of (2.21):

\begin{aligned} \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})}&= \gamma _{0}(j/n) + \frac{1}{n}\frac{\gamma _{2}(j/n)+(4\alpha -2b)\gamma _{1}(j/n)}{8bj/n} \\ {}&\quad + \mathcal{O}(n^{-2}), \end{aligned}

as $$n \rightarrow +\infty$$ and simultaneously $$j \in \{j_{+}+1,\ldots ,n\}$$. Note from (2.3) that j/n lies in $$(br^{2b},1]$$ and remains bounded away from $$br^{2b}$$ as $$n \rightarrow +\infty$$ and simultaneously $$j \in \{j_{+}+1,\ldots ,n\}$$; in particular $$\gamma _{0}(j/n)$$ remains bounded away from 0. Hence, by substituting the above expansion in (3.3) and using (2.15) with $$\ell =0,1,2$$, we obtain after a computation that

\begin{aligned}&S_{3} = \Sigma _{0}+\frac{1}{n}\Sigma _{1}+\mathcal{O}(n^{-1}), \qquad \Sigma _{\ell } := \sum _{j=j_{+}+1}^{n}f_{\ell }(j/n), \; \ell =0,1, \nonumber \\&f_{0}(x) :=\ln \gamma _0(x)= a \ln \bigg ( \bigg (\frac{x}{b}\bigg )^{\frac{1}{2b}}-r \bigg ), \nonumber \\&f_{1}(x) := \frac{\gamma _2(x)+(4\alpha -2b) \gamma _1(x)}{8bx\,\gamma _0(x)} = \frac{a(\frac{x}{b})^{\frac{1}{2b}}}{8bx((\frac{x}{b})^{\frac{1}{2b}}-r)^{2}}\bigg ( (a+4\alpha -2b) \bigg (\frac{x}{b}\bigg )^{\frac{1}{2b}}\nonumber \\ {}&\qquad \qquad \quad - (1+4\alpha -2b)r \bigg ), \end{aligned}
(3.4)

where we have also used that $$\ln (1+x)=x+\mathcal{O}(x^{2})$$ as $$x \rightarrow 0$$. From Lemma 2.10 (with $$A=\frac{br^{2b}}{1-\epsilon }$$, $$a_{0} = 1-\alpha -\theta _{+}^{(n,\epsilon )}$$, $$B=1$$ and $$b_{0}=0$$), we infer that

\begin{aligned} \Sigma _{\ell } {=} n \int _{\frac{br^{2b}}{1-\epsilon }}^{1}f_{\ell }(x)dx {+} \frac{(2\alpha +2\theta _{+}^{(n,\epsilon )}-1) f_{\ell }(\frac{br^{2b}}{1-\epsilon })+f_{\ell }(1)}{2} + \mathcal{O}(n^{-1}), \qquad \ell =0,1, \end{aligned}

as $$n \rightarrow + \infty$$. We then obtain the claim for $$S_{3}$$ after a computation using the simplification

\begin{aligned} \int _{\frac{br^{2b}}{1-\epsilon }}^{1}f_{1}(x)dx&= \frac{a}{4}\bigg ( \frac{1-a}{(br^{2b})^{-\frac{1}{2b}}-1} -\frac{1-a}{(1-\epsilon )^{-\frac{1}{2b}}-1}+(a-2b+4\alpha )\\ {}&\quad \ln \bigg ( \frac{(br^{2b})^{-\frac{1}{2b}}-1}{(1-\epsilon )^{-\frac{1}{2b}}-1} \bigg ) \bigg ). \end{aligned}

$$\square$$

The large n asymptotics of $$S_{1}$$ are harder to obtain than those of $$S_{3}$$. The main reason for it is that $$S_{1}$$ involves small j’s, and for such j’s the quantities $$\gamma _{\ell }(j/n)$$ have a singular behavior, see (2.16). This is also the reason why the error term in Lemma 3.3 is more complicated than in Lemma 3.2.

### Lemma 3.3

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

\begin{aligned} S_{1}&= n \int _{0}^{\frac{br^{2b}}{1+\epsilon }} \bigg (u + a \ln \bigg ( r-\bigg ( \frac{x}{b} \bigg )^{\frac{1}{2b}} \bigg )dx - M' \ln (r^{a}e^{u}) \\&+ \frac{-(u+a\ln r)+(2\theta _{-}^{(n,\epsilon )} -1 - 2\alpha )\big (u + a \ln (r-r(1+\epsilon )^{-\frac{1}{2b}})}{2} \\&+ \frac{a}{4}\bigg ( \frac{a-1}{(1+\epsilon )^{\frac{1}{2b}}-1}+(a-2b+4\alpha )\ln \big ( 1-(1+\epsilon )^{-\frac{1}{2b}} \big ) \bigg ) + \mathcal{O}\big ( n^{-\frac{1}{2b}}+n^{-1} \big ). \end{aligned}

### Proof

Lemma A.2 (i) implies that for any $$\epsilon '>0$$ there exist $$A=A(\epsilon '),C=C(\epsilon ')>0$$ such that $$|\frac{\gamma (a,z)}{\Gamma (a)}-1| \le Ce^{-\frac{a\eta ^{2}}{2}}$$ for all $$a \ge A$$, for all $$\lambda =\frac{z}{a} \in [1+\epsilon ',+\infty ]$$, and where $$\eta$$ is given by (A.1). Let us take $$\epsilon '=\frac{\epsilon }{2}$$ and choose $$M'$$ so large that $$a_{j} = \frac{j+\alpha }{b} \ge A(\frac{\epsilon }{2})$$ for all $$j \in \{M'+1,\ldots ,j_{1,-}-1\}$$. Thus

\begin{aligned} S_{1}&= \sum _{j=M'+1}^{j_{-}-1} \ln \bigg ( \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})}\\&\bigg [ 1+((-1)^{a}e^{u}-1)(1+\mathcal{O}(e^{-\frac{a_{j,k}\eta _{j,k}^{2}}{2}})) \bigg ] \bigg ), \end{aligned}

as $$n \rightarrow + \infty$$. From a direct analysis of (3.2), we infer that $$a_{j,k}\eta _{j,k}^{2}$$ decreases as j increases from $$M'+1$$ to $$j_{-}-1$$, and $$a_{j,k}\eta _{j,k}^{2}$$ decreases also as k increases from 0 to a. Therefore

\begin{aligned} \frac{a_{j,k}\eta _{j,k}^{2}}{2} \ge \frac{a_{j_{-}-1}\eta _{j_{-}-1}^{2}}{2} \ge cn, \qquad \text{ for } \text{ all } j \in \{M'+1,\ldots ,j_{-}-1\}, \end{aligned}

for a small enough $$c>0$$. Thus

\begin{aligned} S_{1}&= \sum _{j=M'+1}^{j_{-}-1} \ln \bigg ( \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{r^{a-k}(-1)^{k}e^{u}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \bigg ) + \mathcal{O}(e^{-cn}), \qquad \text{ as } n \rightarrow + \infty . \end{aligned}
(3.5)

Substituting (2.22) in (3.5) and using (2.15) with $$\ell =0,1,2$$, we obtain

\begin{aligned}&S_{1} = {\widetilde{\Sigma }}_{0}+\frac{1}{n}{\widetilde{\Sigma }}_{1} +\mathcal{O}\bigg (n^{-2}\sum _{j=M'+1}^{j_{-}-1}(j/n)^{\frac{1}{2b}-2}\bigg ) = {\widetilde{\Sigma }}_{0}+\frac{1}{n}{\widetilde{\Sigma }}_{1} + \mathcal{O}\big ( n^{-\frac{1}{2b}}+n^{-1} \big ), \nonumber \\&{\widetilde{\Sigma }}_{0} := \sum _{j=M'+1}^{j_{-}-1} {\widetilde{f}}_{0}(j/n), \quad {\widetilde{\Sigma }}_{1} := \sum _{j=M'+1}^{j_{-}-1} f_{1}(j/n), \qquad {\widetilde{f}}_{0}(x) := u + a \ln \bigg ( r-\bigg (\frac{x}{b}\bigg )^{\frac{1}{2b}} \bigg ), \end{aligned}
(3.6)

as $$n \rightarrow + \infty$$, where $$f_{1}$$ is given by (3.4). Note that $$f_{1}(x) \sim c x^{\frac{1}{2b}-1}$$ as $$x \rightarrow 0$$; thus $$f_{1}$$ blows up at 0 if $$b > \frac{1}{2}$$. Using now Lemma 2.10 (with $$A=\frac{M'}{n}$$, $$a_{0} = 1$$, $$B=\frac{br^{2b}}{1+\epsilon }$$ and $$b_{0}=\theta _{-}^{(n,\epsilon )}-1-\alpha$$), we get

\begin{aligned}&{\widetilde{\Sigma }}_{0} = n \int _{\frac{M'}{n}}^{\frac{br^{2b}}{1+\epsilon }}{\widetilde{f}}_{0}(x)dx + \frac{-{\widetilde{f}}_{0}(\frac{M'}{n}) +(2\theta _{-}^{(n,\epsilon )}-1-2\alpha ){\widetilde{f}}_{0}(\frac{br^{2b}}{1+\epsilon })}{2} + \mathcal{O}\big ( n^{-\frac{1}{2b}}+n^{-1} \big ), \\&\frac{1}{n}{\widetilde{\Sigma }}_{1} = \int _{\frac{M'}{n}}^{\frac{br^{2b}}{1+\epsilon }}f_{1}(x)dx + \mathcal{O}\big ( n^{-\frac{1}{2b}}+n^{-1} \big ). \end{aligned}

as $$n \rightarrow + \infty$$. Furthermore, by a direct analysis of $${\widetilde{f}}_{0}$$ and $$f_{1}$$,

\begin{aligned}&{\widetilde{f}}_{0}\bigg ( \frac{M'}{n} \bigg ) = u + a \ln r + \mathcal{O}\big ( n^{-\frac{1}{2b}} \big ), \\&\int _{\frac{M'}{n}}^{\frac{br^{2b}}{1+\epsilon }}f_{1}(x)\,dx = \frac{a}{4}\bigg ( \frac{a-1}{(1+\epsilon )^{\frac{1}{2b}}-1} + (a-2b+4\alpha ) \ln \Big ( 1-(1+\epsilon )^{-\frac{1}{2b}} \Big ) \bigg )\\&\quad + \mathcal{O}\big ( n^{-\frac{1}{2b}} \big ), \\&n \int _{\frac{M'}{n}}^{\frac{br^{2b}}{1+\epsilon }}{\widetilde{f}}_{0}(x) \, dx = n \int _{0}^{\frac{br^{2b}}{1+\epsilon }}{\widetilde{f}}_{0}(x)\,dx - (u+a \ln r)M' + \mathcal{O}\big ( n^{-\frac{1}{2b}} \big ), \end{aligned}

as $$n \rightarrow + \infty$$. The claim follows after substituting the above expansions in (3.6). $$\square$$

## 4 Large n Asymptotics of $$S_{2}$$

It remains to obtain the large n asymptotics of $$S_{2}$$, which was defined in (2.7). For this, let us split $$S_{2}$$ in three pieces,

\begin{aligned}&S_{2}=S_{2}^{(1)}+S_{2}^{(2)}+S_{2}^{(3)}, \end{aligned}

where

\begin{aligned} S_{2}^{(v)}&:= \sum _{j:\lambda _{j}\in I_{v}} \ln \bigg ( \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \nonumber \\ {}&\bigg [1+((-1)^{a}e^{u}-1)\frac{\gamma (\frac{2j+2\alpha +k}{2b},nr^{2b})}{\Gamma (\frac{2j+2\alpha +k}{2b})} \bigg ] \bigg ) \end{aligned}
(4.1)

for $$v=1,2,3$$, $$\lambda _{j}$$ is given by (3.1), and

\begin{aligned} I_{1} := [1-\epsilon ,1-\tfrac{M}{\sqrt{n}}), \qquad I_{2} := [1-\tfrac{M}{\sqrt{n}},1+\tfrac{M}{\sqrt{n}}], \qquad I_{3} := (1+\tfrac{M}{\sqrt{n}},1+\epsilon ]. \end{aligned}

Equivalently, the above sums can be rewritten using

\begin{aligned}&\sum _{j:\lambda _{j}\in I_{3}} = \sum _{j=j_{-}}^{g_{-}-1}, \qquad \sum _{j:\lambda _{j}\in I_{2}} = \sum _{j= g_{-}}^{g_{+}}, \qquad \sum _{j:\lambda _{j}\in I_{1}} = \sum _{j= g_{+}+1}^{j_{+}}, \end{aligned}
(4.2)

where $$g_{-}:= \lceil \frac{bnr^{2b}}{1+\frac{M}{\sqrt{n}}}-\alpha \rceil$$, $$g_{+}:= \lfloor \frac{bnr^{2b}}{1-\frac{M}{\sqrt{n}}}-\alpha \rfloor$$. We also define $$\theta _{-}^{(n,M)},\theta _{+}^{(n,M)} \in [0,1)$$ by

\begin{aligned} \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} \end{aligned}

Note that the sums $$S_{2}^{(1)}$$ and $$S_{2}^{(3)}$$ each contain a number of elements proportional to n, while $$S_{2}^{(2)}$$ contains roughly $$M\sqrt{n}$$ elements.

### 4.1 Global Analysis: Large n Asymptotics of $$S_{2}^{(1)}$$ and $$S_{2}^{(3)}$$

We first treat $$S_{2}^{(1)}$$, $$S_{2}^{(3)}$$. These sums are delicate to analyze because they involve the asymptotics of $$\gamma (a,z)$$ in the regime $$a \rightarrow + \infty$$, $$z \rightarrow +\infty$$, when $$\lambda =\frac{z}{a}$$ is close to 1 but not very close (more precisely, $$\lambda \in [1-\epsilon ,1-\frac{M}{\sqrt{n}})\cup (1+\frac{M}{\sqrt{n}},1+\epsilon ]$$).

### Lemma 4.1

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

\begin{aligned}&S_{2}^{(1)} = n \int _{br^{2b}}^{\frac{br^{2b}}{1-\epsilon }} a \ln \bigg ( \bigg ( \frac{x}{b} \bigg )^{\frac{1}{2b}} - r \bigg )dx + \bigg \{ a b r^{2b} M \bigg ( \ln \bigg ( \frac{2b\sqrt{n}}{r M} \bigg ) + 1 \bigg ) + \frac{a(a-1)b}{2M} \\&\quad - \frac{a b (3-5a+2a^{2})}{12r^{2b}M^{3}} \bigg \} \sqrt{n} + \bigg ( -abr^{2b}M^{2}+a\frac{2\alpha +2\theta _{+}^{(n,M)}-1}{2}\\ {}&- \frac{a}{4}(a-2b+4\alpha ) \bigg ) \ln \bigg ( \frac{M}{\sqrt{n}} \bigg ) \\&\quad + \frac{ar^{2b}(2b-1+8b\ln (\frac{2b}{r}))}{8}M^{2} + \frac{2\alpha +2\theta _{+}^{(n,M)}-1}{2}a \ln \bigg ( \frac{r}{2b} \bigg ) \\&\quad + \frac{1-2\alpha -2\theta _{+}^{(n,\epsilon )}}{2}a \ln \bigg ( r(1-\epsilon )^ {-\frac{1}{2b}}-r \bigg ) \\&\quad + \frac{a}{4} \bigg \{ \frac{(1-a)(1+2b)}{2} + (a-2b+4\alpha )\ln (2b) \\&\quad + \frac{1-a}{(1-\epsilon )^{-\frac{1}{2b}}-1} + (a-2b+4\alpha )\ln \bigg ( (1-\epsilon )^{-\frac{1}{2b}}-1 \bigg ) \bigg \} + \mathcal{O}\bigg ( \frac{\sqrt{n}}{M^{5}} + \frac{M^{3}\ln n}{\sqrt{n}} \bigg ). \end{aligned}

### Remark 4.2

Since $$M=n^{\frac{1}{8}}(\ln n)^{-\frac{1}{8}}$$, the $$\mathcal{O}$$-term above is small as $$n \rightarrow + \infty$$.

### Remark 4.3

The above asymptotics are of the form

\begin{aligned} S_{2}^{(1)} = E_{1}^{(\epsilon )}n + {\tilde{E}}_{2}^{(M)} \sqrt{n} \ln n + E_{2}^{(M)}\sqrt{n} + {\tilde{E}}_{3}^{(n,M)} \ln n + E_{3}^{(n,\epsilon ,M)} + o(1). \end{aligned}

This may seem a bit counter intuitive as the asymptotics of Theorem 1.1 only contain terms proportional to n, $$\sqrt{n}$$ and 1. In fact, remarkable cancellations will occur in the asymptotics of $$S_{2}^{(1)}+S_{2}^{(2)}+S_{2}^{(3)}$$; in particular $${\tilde{E}}_{2}^{(M)}$$ and $${\tilde{E}}_{3}^{(n,M)}$$ will get perfectly canceled by other terms in the large n asymptotics of $$S_{2}^{(2)}$$ and $$S_{2}^{(3)}$$, and we will show in Lemma 4.12 below that the large n asymptotics of $$S_{2} = S_{2}^{(1)}+S_{2}^{(2)}+S_{2}^{(3)}$$ are of the form $$S_{2} = {\widehat{C}}_{1}^{(\epsilon )} n + {\widehat{C}}_{2} \sqrt{n} + {\widehat{C}}_{3}^{(n,\epsilon )} + o(1)$$.

### Proof

By (4.1) and Lemma A.2, $$S_{2}^{(1)}$$ admits the following exact formula

\begin{aligned} S_{2}^{(1)} =&\sum _{j:\lambda _{j}\in I_{1}} \ln \bigg (\sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \\&\times \bigg [ 1+((-1)^{a}e^{u}-1)\bigg ( \frac{1}{2}\textrm{erfc}\Big (\hspace{-0.1cm}-\hspace{-0.05cm}\eta _{j,k} \sqrt{\tfrac{a_{j,k}}{2}}\Big ) \\&\times - R_{a_{j,k}}(\eta _{j,k}) \bigg ) \bigg ]\bigg ), \end{aligned}

where $$\eta _{j,k}$$ and $$a_{j,k}$$ are given by (3.2). Since $$I_{1}=[1-\epsilon ,1-\frac{M}{\sqrt{n}})$$,

\begin{aligned}&\eta _{j,k} = \lambda _{j,k}-1 +\mathcal{O}((\lambda _{j,k}-1)^{2}) \le -\tfrac{M}{\sqrt{n}} + \mathcal{O}(\tfrac{M^{2}}{n}),{} & {} \text{ as } n \rightarrow +\infty , \\&-\eta _{j,k} \sqrt{a_{j,k}/2} =-\eta _{j,k} \sqrt{ \tfrac{n r^{2b}}{ 2\lambda _{j,k} } } \ge \tfrac{Mr^{b}}{\sqrt{2}} + \mathcal{O}(\tfrac{M^{2}}{\sqrt{n}}),{} & {} \text{ as } n \rightarrow +\infty , \end{aligned}

uniformly for $$j\in \{j: \lambda _{j} \in I_{1}\}$$ and $$k \in \{0,1,\ldots ,a\}$$. Also, $$M = n^{\frac{1}{8}}(\ln n)^{-\frac{1}{8}}$$, and thus, by (A.2),

\begin{aligned}&R_{a_{j,k}}(\eta _{j,k}) = \mathcal{O}(e^{-\frac{r^{2b}M^{2}}{ 4 }}) = \mathcal{O}(e^{-n^{c}}),{} & {} \frac{1}{2}\textrm{erfc}\Big (\hspace{-0.1cm}-\hspace{-0.05cm}\eta _{j,k} \sqrt{\tfrac{a_{j,k}}{2}}\Big ) = \mathcal{O}(e^{-n^{c}}), \end{aligned}

as $$n \rightarrow + \infty$$ uniformly for $$j\in \{j: \lambda _{j} \in I_{1}\}$$ and $$k \in \{0,1,\ldots ,a\}$$, and (using also (4.2))

\begin{aligned} S_{2}^{(1)}&= \sum _{j=g_{+}+1}^{j_{+}} \ln \bigg (\sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \bigg ) + \mathcal{O}(e^{-n^{c}}), \qquad \text{ as } n \rightarrow + \infty . \end{aligned}
(4.3)

In the same way as for (2.21), we obtain

\begin{aligned}&\sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \sim \gamma _{0}(j/n) + \sum _{\ell =1}^{+\infty } \frac{\sum _{m=1}^{2\ell }{\mathfrak {p}}_{2\ell ,m}\gamma _{m}(j/n)}{j^{\ell }} \end{aligned}
(4.4)

as $$n \rightarrow + \infty$$ uniformly for $$j \in \{g_{+}+1,\ldots ,j_{+}\}$$. However, unlike for $$S_{3}$$, the first subleading term (corresponding to $$\ell =1$$) is not sufficient for our purpose; we also need the coefficients of $${\mathfrak {p}}_{4}$$ which are given by (see (2.19))

\begin{aligned} {\mathfrak {p}}_{4}(k)&= \frac{k(k-2b)}{8}B_2^{(1+\frac{k}{2b})}\Big ( \frac{2\alpha +k}{2b}\Big ) \nonumber \\&= \frac{k(k-2b)(8b^{2}+3(k+4\alpha )^{2}-2b(7k+24\alpha ))}{384b^{2}}. \end{aligned}
(4.5)

The reason as to why we need $${\mathfrak {p}}_{4}$$ can be seen as follows. The sum on the left-hand side of (4.4) appears inside $$\ln$$ in (4.3). Since $$\ln (\gamma _{0}+B)=\ln \gamma _{0} + \ln (1+B/\gamma _{0})$$, what is relevant is to estimate the asymptotics series of (4.4) divided by $$\gamma _{0}(j/n)$$. Lemma 2.5 implies

\begin{aligned} \frac{\gamma _{\ell }(x)}{\gamma _{0}(x)}{} & {} = \mathcal{O}(\min \{|x-br^{2b}|^{-a},|x-br^{2b}|^{-\ell }\}) = \mathcal{O}(|x-br^{2b}|^{-\min \{a,\ell \}}), \qquad \nonumber \\{} & {} \quad \text{ as } x \rightarrow br^{2b}. \end{aligned}
(4.6)

Using the definitions of $$j_{+}$$ and $$g_{+}$$ (see 2.3 and (4.2)), we see that for $$j \in \{g_{+}+1,\ldots ,j_{+}\}$$, j/n lies in $$(br^{2b},1]$$ and $$g_{+}/n-br^{2b}$$ is of order $$\frac{M}{\sqrt{n}}$$ as $$n \rightarrow +\infty$$. Therefore, for $$\ell =1,2,\ldots$$,

\begin{aligned} \sum _{j=g_{+}+1}^{j_{+}}\frac{\sum _{m=1}^{2\ell } {\mathfrak {p}}_{2\ell ,m}\gamma _{m}(j/n)}{j^{\ell }\gamma _{0}(j/n)} = \mathcal{O}\bigg ( \sum _{j=g_{+}+1}^{j_{+}}\frac{|j/n - br^{2b}|^{-\min \{a,2\ell \}}}{n^{\ell }} \bigg ) = \mathcal{O}\bigg ( \frac{\sqrt{n}}{M^{2\ell }} \bigg ), \end{aligned}
(4.7)

as $$n \rightarrow + \infty$$ uniformly for $$j \in \{g_{+}+1,\ldots ,j_{+}\}$$. Since $$M=n^{\frac{1}{8}}(\ln n)^{-\frac{1}{8}}$$, only the terms corresponding to $$\ell \ge 3$$ in (4.4) (i.e. the terms associated with $${\mathfrak {p}}_{6},{\mathfrak {p}}_{8},\ldots$$) will give an error in the asymptotics of $$S_{2}^{(1)}$$. Substituting (4.4)–(4.5) in (4.3) and using (2.15) (with $$\ell =0,1,2,3,4)$$, (2.17) and (4.7), we obtain

\begin{aligned}&S_{2}^{(1)} = \Sigma _{0}^{(1)} + \frac{1}{n}\Sigma _{1}^{(1)} + \frac{1}{n^{2}}\Sigma _{2}^{(1)} + \mathcal{O}\bigg ( \frac{\sqrt{n}}{M^{5}} \bigg ) \end{aligned}
(4.8)

as $$n \rightarrow + \infty$$, where

\begin{aligned} \Sigma _{\ell }^{(1)} := \sum _{j=g_{+}+1}^{j_{+}} f_{\ell }(j/n), \qquad \ell =0,1,2, \end{aligned}

$$f_{0}$$ and $$f_{1}$$ are as in (3.4), and $$f_{2}$$ is defined by

\begin{aligned} f_{2}(x)&: = \frac{1}{384b^{2}x^{2}}\bigg \{ -16b(b^{2}-6b\alpha +6\alpha ^{2})\frac{a\big (\frac{x}{b}\big )^ {\frac{1}{2b}}}{(\frac{x}{b}\big )^{\frac{1}{2b}}-r} + 12(3b^{2}-8b\alpha +4\alpha ^{2})\nonumber \\ {}&\quad \frac{a\big (\frac{x}{b}\big )^{\frac{1}{2b}}\big (a\big (\frac{x}{b}\big ) ^{\frac{1}{2b}}-r\big )}{\big (a\big (\frac{x}{b}\big )^{\frac{1}{2b}}-r\big )^{2}} \nonumber \\&\quad +4(6\alpha -5b) \frac{a\big (\frac{x}{b}\big )^{\frac{1}{2b}}\big (a^{2}\big (\frac{x}{b}\big ) ^{\frac{1}{b}}+(1-3a)r\big (\frac{x}{b}\big )^{\frac{1}{2b}}+r^{2}\big )}{\big (a\big (\frac{x}{b}\big )^{\frac{1}{2b}}-r\big )^{3}} \nonumber \\&\quad + \frac{3a\big (\frac{x}{b}\big )^{\frac{1}{2b}} \big [ a^{3}\big (\frac{x}{b}\big )^{\frac{3}{2b}} + (4a-6a^{2}-1)r \big (\frac{x}{b}\big )^{\frac{1}{b}} + (7a-4)r^{2}\big (\frac{x}{b}\big )^{\frac{1}{2b}} -r^{3} \big ]}{\big (a\big (\frac{x}{b}\big )^{\frac{1}{2b}}-r\big )^{4}} \bigg \} \nonumber \\&\quad - \frac{f_{1}(x)^{2}}{2}. \end{aligned}
(4.9)

Using Lemma 2.10 (with $$A=\frac{br^{2b}}{1-\frac{M}{\sqrt{n}}}$$, $$a_{0}=1-\alpha -\theta _{+}^{(n,M)}$$, $$B = \frac{br^{2b}}{1-\epsilon }$$ and $$b_{0}=-\alpha -\theta _{+}^{(n,\epsilon )}$$), we get

\begin{aligned} \Sigma _{0}^{(1)}&= n \int _{\frac{br^{2b}}{1-\frac{M}{\sqrt{n}}}}^ {\frac{br^{2b}}{1-\epsilon }}f_{0}(x)dx + \frac{(2\alpha +2\theta _{+} ^{(n,M)}-1)f_{0}(\frac{br^{2b}}{1-\frac{M}{\sqrt{n}}}) +(1-2\alpha -2\theta _{+}^{(n,\epsilon )})f_{0}(\frac{br^{2b}}{1-\epsilon })}{2} \nonumber \\&\quad + \mathcal{O}\bigg ( \frac{1}{M\sqrt{n}} \bigg ), \nonumber \\ \frac{1}{n}\Sigma _{1}^{(1)}&= \int _{\frac{br^{2b}}{1-\frac{M}{\sqrt{n}}}}^{\frac{br^{2b}}{1-\epsilon }}f_{1}(x)dx + \mathcal{O}\bigg ( \frac{1}{M^{2}} \bigg ), \qquad \frac{1}{n^{2}}\Sigma _{2}^{(1)} = \frac{1}{n}\int _{\frac{br^{2b}}{1-\frac{M}{\sqrt{n}}}}^{\frac{br^{2b}}{1-\epsilon }}f_{2}(x)dx + \mathcal{O}\bigg ( \frac{1}{M^{4}} \bigg ), \end{aligned}
(4.10)

as $$n \rightarrow + \infty$$. To obtain the above error terms, we also used (4.6), (2.15), and in particular that

\begin{aligned} f_{0}'\bigg ( \frac{br^{2b}}{1-\frac{M}{\sqrt{n}}} \bigg )&= \mathcal{O}\bigg ( \frac{\sqrt{n}}{M} \bigg ), \qquad f_{1}\bigg ( \frac{br^{2b}}{1-\frac{M}{\sqrt{n}}} \bigg ) = \mathcal{O}\bigg ( \frac{n}{M^{2}} \bigg ), \qquad f_{2}\bigg ( \frac{br^{2b}}{1-\frac{M}{\sqrt{n}}} \bigg ) \\&= \mathcal{O}\bigg ( \frac{n^{2}}{M^{4}} \bigg ) \end{aligned}

as $$n \rightarrow + \infty$$. Using the asymptotics of $$f_{0}, f_{1}$$ and $$f_{2}$$ near $$x=br^{2b}$$ with $$x>br^{2b}$$, namely

\begin{aligned}&f_{0}(x) = a \log (x-br^{2b}) + a \log (\tfrac{r^{1-2b}}{2b^{2}}) + a \frac{(1-2b)(x-br^{2b})}{4b^{2}r^{2b}} + \mathcal{O}\big ((x-br^{2b})^{2}\big ), \\&f_{1}(x) = \frac{(a-1)ab^{2}r^{2b}}{2(x-br^{2b})^{2}} + \frac{a(a-2b+4\alpha )}{4(x-br^{2b})} + \mathcal{O}(1), \\&f_{2}(x) = - \frac{a(3-5a+2a^{2})b^{4}r^{4b}}{4(x-br^{2b})^{4}} + \mathcal{O}\big ((x-br^{2b})^{-3}\big ), \end{aligned}

we obtain

\begin{aligned} f_{0}\bigg ( \frac{br^{2b}}{1-\frac{M}{\sqrt{n}}} \bigg )&= a \ln \bigg ( \frac{Mr}{2b\sqrt{n}} \bigg ) + \mathcal{O}\bigg ( \frac{M}{\sqrt{n}} \bigg ), \end{aligned}
(4.11)
\begin{aligned} n \int _{\frac{br^{2b}}{1-\frac{M}{\sqrt{n}}}}^{\frac{br^{2b}}{1-\epsilon }}f_{0}(x)dx&= n \int _{br^{2b}}^{\frac{br^{2b}}{1-\epsilon }}f_{0}(x)dx + \sqrt{n} \; a b r^{2b} M \bigg ( 1+\ln \bigg ( \frac{2b\sqrt{n}}{r M} \bigg ) \bigg ) \nonumber \\&\quad + a b r^{2b} M^{2} \ln \bigg ( \frac{\sqrt{n}}{M} \bigg ) \nonumber \\&\quad + \frac{a r^{2b} M^{2}(-1+2b+8b \ln (\frac{2b}{r}))}{8} + \mathcal{O}\bigg ( \frac{M^{3}\ln n}{\sqrt{n}} \bigg ), \end{aligned}
(4.12)
\begin{aligned} \int _{\frac{br^{2b}}{1-\frac{M}{\sqrt{n}}}}^{\frac{br^{2b}}{1-\epsilon }}f_{1}(x)dx&= \frac{a(a-1)b\sqrt{n}}{2M} + \frac{a}{4}\bigg \{ \frac{(1-a)(1+2b)}{2} + \frac{1-a}{(\frac{1}{1-\epsilon })^{\frac{1}{2b}}-1} \nonumber \\&\quad - (a-2b+4\alpha )\ln \bigg ( \frac{M}{2b\sqrt{n}} \bigg ) \nonumber \\&\quad + (a-2b+4\alpha ) \ln \bigg ( (1-\epsilon )^{-\frac{1}{2b}}-1 \bigg ) \bigg \} + \mathcal{O}\bigg ( \frac{M}{\sqrt{n}} \bigg ), \end{aligned}
(4.13)
\begin{aligned} \frac{1}{n}\int _{\frac{br^{2b}}{1-\frac{M}{\sqrt{n}}}}^{\frac{br^{2b}}{1-\epsilon }}f_{2}(x)dx&= - \frac{a b (3-5a+2a^{2})\sqrt{n}}{12 r^{2b} M^{3}} + \mathcal{O}\bigg ( \frac{1}{M^{2}} \bigg ), \end{aligned}
(4.14)

as $$n \rightarrow + \infty$$. Substituting (4.11)–(4.14) in (4.10) and then in (4.8), we obtain the claim after another long but direct computation. $$\square$$

### Lemma 4.4

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

\begin{aligned} S_{2}^{(3)} =&\; n \int _{\frac{br^{2b}}{1+\epsilon }}^{br^{2b}}\bigg [ u + a \ln \bigg ( r - \bigg ( \frac{x}{b} \bigg )^{\frac{1}{2b}} \bigg ) \bigg ] dx + \bigg \{ b r^{2b} M \bigg ( a\ln \bigg ( \frac{2b\sqrt{n}}{r M} \bigg ) + a - u \bigg ) \\&+ \frac{a(a-1)b}{2M}- \frac{a b (3-5a+2a^{2})}{12r^{2b}M^{3}} \bigg \} \sqrt{n} + \bigg ( abr^{2b}M^{2}+a\frac{2\theta _{-}^{(n,M)}-1-2\alpha }{2}\\ {}&+ \frac{a}{4}(a-2b+4\alpha ) \bigg ) \ln \bigg ( \frac{M}{\sqrt{n}} \bigg )+ \frac{r^{2b}(a-2ab+8bu+8ab\ln (\frac{r}{2b}))}{8}M^{2}\\ {}&+ \frac{2\theta _{-}^{(n,M)}-1-2\alpha }{2}\bigg ( u + a \ln \bigg ( \frac{r}{2b} \bigg )\bigg ) + \frac{1+2\alpha -2\theta _{-}^{(n,\epsilon )}}{2}\bigg ( u + a \ln \bigg ( r-r(1+\epsilon )^{-\frac{1}{2b}} \bigg ) \bigg ) \\&+ \frac{a}{4} \bigg \{ \frac{(a-1)(1+2b)}{2} - (a-2b+4\alpha )\ln (2b) \\&+ \frac{1-a}{1-(1+\epsilon )^{-\frac{1}{2b}}} - (a-2b+4\alpha )\ln \bigg ( 1-(1+\epsilon )^{-\frac{1}{2b}} \bigg ) \bigg \} \\&+ \mathcal{O}\bigg ( \frac{\sqrt{n}}{M^{5}} + \frac{M^{3}\ln n}{\sqrt{n}} \bigg ). \end{aligned}

### Proof

By (4.1) and Lemma A.2,

\begin{aligned} S_{2}^{(3)}&= \sum _{j:\lambda _{j}\in I_{3}} \ln \bigg (\sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \\&\times \bigg [ 1+((-1)^{a}e^{u}-1)\bigg ( \frac{1}{2}\textrm{erfc}\Big (\hspace{-0.1cm}-\hspace{-0.05cm}\eta _{j,k} \sqrt{\tfrac{a_{j,k}}{2}}\Big ) - R_{a_{j,k}}(\eta _{j,k}) \bigg ) \bigg ]\bigg ). \end{aligned}

Because $$I_{3} = (1+\frac{M}{\sqrt{n}},1+\epsilon ]$$, we have

\begin{aligned} \eta _{j,k} =&\lambda _{j,k}-1 +\mathcal{O}((\lambda _{j,k}-1)^{2}) \ge \tfrac{M}{\sqrt{n}} + \mathcal{O}(\tfrac{M^{2}}{n}),{} & {} \text{ as } n \rightarrow \infty , \\&-\eta _{j,k} \sqrt{a_{j,k}/2} \le - \tfrac{M r^{b}}{\sqrt{2}} + \mathcal{O}(\tfrac{M^{2}}{\sqrt{n}}),{} & {} \text{ as } n \rightarrow \infty , \end{aligned}

uniformly for $$j\in \{j:\lambda _{j}\in I_{3}\}$$ and $$k \in \{1,\ldots ,a\}$$. Since $$M = n^{\frac{1}{8}}(\ln n)^{-\frac{1}{8}}$$, we get

\begin{aligned}&R_{a_{j,k}}(\eta _{j,k}) = \mathcal{O}(e^{-\frac{r^{2b}M^{2}}{4} }) = \mathcal{O}(e^{-n^{c}}),{} & {} \frac{1}{2}\textrm{erfc}\Big (\hspace{-0.1cm}-\hspace{-0.05cm}\eta _{j,k} \sqrt{\tfrac{a_{j,k}}{2}}\Big ) = 1-\mathcal{O}(e^{-\frac{r^{2b}M^{2}}{4} }) \\ {}&= 1-\mathcal{O}(e^{-n^{c}}), \end{aligned}

as $$n \rightarrow + \infty$$ uniformly for $$j\in \{j:\lambda _{j}\in I_{3}\}$$ and $$k \in \{1,\ldots ,a\}$$, and thus

\begin{aligned} S_{2}^{(3)}&= \sum _{j=j_{-}}^{g_{-}-1} \ln \bigg ( \sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{r^{a-k}(-1)^{k}e^{u}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \bigg ) + \mathcal{O}(e^{-n^{c}}), \end{aligned}
(4.15)

where we have also used (4.2). By (2.24),

\begin{aligned}&\sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{r^{a-k}(-1)^{k}e^{u}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \sim e^{u} \bigg \{\gamma _{0}(j/n) + \sum _{\ell =1}^{+\infty } \frac{\sum _{m=1}^{2\ell }{\mathfrak {p}}_{2\ell ,m}\gamma _{m}(j/n)}{j^{\ell }} \bigg \}, \end{aligned}
(4.16)

as $$n \rightarrow + \infty$$ uniformly for $$j \in \{j_{-},\ldots ,g_{-}-1\}$$. Using Lemma 2.5, (4.6) and the definitions (2.3), (4.2) of $$j_-$$ and $$g_-$$, we infer that (4.7) holds also as $$n \rightarrow + \infty$$ uniformly for $$j\in \{j_{-},\ldots ,g_{-}-1\}$$. Hence, substituting (4.16) in (4.15) and using (2.15) (with $$\ell =0,1,2,3,4$$), (2.17) and (4.7), we get

\begin{aligned}&S_{2}^{(3)} = \Sigma _{0}^{(3)} + \frac{1}{n}\Sigma _{1}^{(3)} + \frac{1}{n^{2}}\Sigma _{2}^{(3)} + \mathcal{O}\bigg ( \frac{\sqrt{n}}{M^{5}} \bigg ) \end{aligned}
(4.17)

as $$n \rightarrow + \infty$$, where

\begin{aligned} \Sigma _{0}^{(3)} := \sum _{j=j_{-}}^{g_{-}-1} {\widetilde{f}}_{0}(j/n), \qquad \Sigma _{\ell }^{(3)} := \sum _{j=j_{-}}^{g_{-}-1} f_{\ell }(j/n), \qquad \ell =1,2, \end{aligned}

$${\widetilde{f}}_{0}$$ is given by (3.6), $$f_{1}$$ is given by (3.4), and $$f_{2}$$ is given by (4.9).

Using Lemma 2.10 (with $$A=\frac{br^{2b}}{1+\epsilon }$$, $$a_{0}=\theta _{-}^{(n,\epsilon )}-\alpha$$, $$B = \frac{br^{2b}}{1+\frac{M}{\sqrt{n}}}$$ and $$b_{0} = \theta _{-}^{(n,M)}-1-\alpha$$), we get

\begin{aligned} \Sigma _{0}^{(3)} =&n \int _{\frac{br^{2b}}{1+\epsilon }}^ {\frac{br^{2b}}{1+\frac{M}{\sqrt{n}}}}{\widetilde{f}}_{0}(x)dx + \frac{(1+2\alpha -2\theta _{-}^{(n,\epsilon )}){\widetilde{f}}_{0} (\frac{br^{2b}}{1+\epsilon })+(2\theta _{-}^{(n,M)}-1-2\alpha ) {\widetilde{f}}_{0}(\frac{br^{2b}}{1+\frac{M}{\sqrt{n}}})}{2} \nonumber \\&+ \mathcal{O}\bigg ( \frac{1}{M\sqrt{n}} \bigg ), \nonumber \\ \frac{1}{n}\Sigma _{1}^{(3)} =&\int _{\frac{br^{2b}}{1+\epsilon }}^{\frac{br^{2b}}{1+\frac{M}{\sqrt{n}}}} f_{1}(x)dx + \mathcal{O}\bigg ( \frac{1}{M^{2}} \bigg ), \qquad \frac{1}{n^{2}}\Sigma _{2}^{(3)} = \frac{1}{n}\int _{\frac{br^{2b}}{1+\epsilon }}^{\frac{br^{2b}}{1+\frac{M}{\sqrt{n}}}} f_{2}(x)dx + \mathcal{O}\bigg ( \frac{1}{M^{4}} \bigg ),\quad \end{aligned}
(4.18)

as $$n \rightarrow + \infty$$. Furthermore, a long but straightforward analysis of $${\widetilde{f}}_{0}$$, $$f_{1}$$ and $$f_{2}$$ shows that

\begin{aligned} {\widetilde{f}}_{0}\bigg ( \frac{br^{2b}}{1+\frac{M}{\sqrt{n}}} \bigg )&= u+ a \ln \bigg ( \frac{Mr}{2b\sqrt{n}} \bigg ) + \mathcal{O}\bigg ( \frac{M}{\sqrt{n}} \bigg ), \end{aligned}
(4.19)
\begin{aligned} n \int _{\frac{br^{2b}}{1+\epsilon }}^{\frac{br^{2b}}{1+\frac{M}{\sqrt{n}}}}{\widetilde{f}}_{0}(x)dx&= n \int _{\frac{br^{2b}}{1+\epsilon }}^{br^{2b}}{\widetilde{f}}_{0}(x)dx + \sqrt{n} \; b r^{2b} \nonumber \\&\quad M \bigg ( a-u+a\ln \bigg ( \frac{2b\sqrt{n}}{r M} \bigg ) \bigg ) + a b r^{2b} M^{2} \ln \bigg ( \frac{M}{\sqrt{n}} \bigg ) \nonumber \\&\quad + \frac{r^{2b} M^{2}(a-2ab+8bu+8ab \ln (\frac{r}{2b}))}{8} + \mathcal{O}\bigg ( \frac{M^{3}\ln n}{\sqrt{n}} \bigg ), \end{aligned}
(4.20)
\begin{aligned} \int _{\frac{br^{2b}}{1+\epsilon }}^{\frac{br^{2b}}{1+\frac{M}{\sqrt{n}}}}f_{1}(x)dx&= \frac{a(a-1)b\sqrt{n}}{2M} + \frac{a}{4}\bigg \{ \frac{(a-1)(1+2b)}{2} + \frac{1-a}{1-(\frac{1}{1+\epsilon })^{\frac{1}{2b}}} \nonumber \\ {}&\quad + (a-2b+4\alpha )\ln \bigg ( \frac{M}{2b\sqrt{n}} \bigg ) \nonumber \\&\quad - (a-2b+4\alpha ) \ln \bigg ( 1-(1+\epsilon )^{-\frac{1}{2b}} \bigg ) \bigg \} + \mathcal{O}\bigg ( \frac{M}{\sqrt{n}} \bigg ), \nonumber \\ \end{aligned}
(4.21)
\begin{aligned} \frac{1}{n}\int _{\frac{br^{2b}}{1+\epsilon }}^{\frac{br^{2b}}{1+\frac{M}{\sqrt{n}}}}f_{2}(x)dx&= - \frac{a b (3-5a+2a^{2})\sqrt{n}}{12 r^{2b} M^{3}} + \mathcal{O}\bigg ( \frac{1}{M^{2}} \bigg ). \end{aligned}
(4.22)

Substituting (4.19)–(4.22) in (4.18) and then in (4.17), we obtain the claim after another long but direct computation. $$\square$$

### 4.2 Local Analysis: Large n Asymptotics of $$S_{2}^{(2)}$$

Our next goal is to obtain the large n asymptotics of $$S_{2}^{(2)}$$. This is the most technical part of the proof of Theorem 1.1. As mentioned in the introduction, a major obstacle in the asymptotic analysis of $$S_{2}^{(2)}$$ is that, in order to treat the general case $$a \in \mathbb {N}$$, we need to expand various quantities to all orders. Lemma 4.5 below provides a general scheme to compute the coefficients appearing in these expansions in a recursive way. These coefficients are not all readily available in explicit forms, however only a few of those will really matter for us. Lemmas 4.6 and 4.7 establish some non-trivial identities between those relevant coefficients and the (associated) Hermite polynomials. The large n asymptotics of $$S_{2}^{(2)}$$ are then obtained in Lemma 4.11.

Let us define

\begin{aligned}&M_{j,k} := \sqrt{n}(\lambda _{j,k}-1),{} & {} k \in \{1,\ldots ,m\}, \; j \in \{j: \lambda _{j} \in I_{2}\}=\{g_{-},\ldots ,g_{+}\}, \end{aligned}
(4.23)
\begin{aligned}&M_{j} := \sqrt{n}(\lambda _{j}-1),{} & {} j \in \{j: \lambda _{j} \in I_{2}\}=\{g_{-},\ldots ,g_{+}\}. \end{aligned}
(4.24)

Since $$I_{2}=[1-\tfrac{M}{\sqrt{n}},1+\tfrac{M}{\sqrt{n}}]$$ and $$\lambda _{j}$$ is decreasing in j, we have $$-M \le M_{g_{+}}< \cdots < M_{g_{-}} \le M$$. Note from (3.1)–(3.2) that $$\lambda _{j,k}$$ is close to $$\lambda _{j}$$ for large n, and from (4.23)–(4.24) that $$M_{j,k}$$ is close to $$M_{j}$$ for large n. For convenience, for $$j \in \{g_{-},\ldots ,g_{+}\}$$, we also define

\begin{aligned} \Xi _{j,k} := \frac{M_j r^b}{\sqrt{2}} - \frac{\sqrt{a_{j,k}}}{\sqrt{2}} \eta _{j,k}, \end{aligned}
(4.25)

where $$a_{j,k}$$ and $$\eta _{j,k}$$ are given in (3.2). Also, by (4.1) and Lemma A.2, we have

\begin{aligned} S_{2}^{(2)} =&\sum _{j:\lambda _{j}\in I_{2}} \ln \bigg (\sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \nonumber \\&\times \bigg [ 1+((-1)^{a}e^{u}-1)\bigg ( \textrm{erfc}\Big (\hspace{-0.1cm}-\hspace{-0.05cm}\eta _{j,k} \sqrt{\tfrac{a_{j,k}}{2}}\Big ) - R_{a_{j,k}}(\eta _{j,k}) \bigg ) \bigg ]\bigg ). \end{aligned}
(4.26)

The next lemma provides the asymptotics of $$\Xi _{j,k}$$ and of the summand of the k-sum in (4.26) as $$n\rightarrow +\infty$$ uniformly for $$j\in \{j:\lambda _{j}\in I_{2}\}=\{g_{-},\ldots ,g_{+}\}$$.

### Lemma 4.5

Let $$k \in \{0,1,\ldots ,a\}$$ be fixed. As $$n \rightarrow + \infty$$ and uniformly for $$j \in \{g_{-},\ldots ,g_{+}\}$$, we have

\begin{aligned}&\frac{1}{n^{\frac{k}{2b}}}\frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \sim r^{k} \sum _{\ell =0}^{+\infty } \frac{\sum _{m=0}^{\ell }\sum _{p=0}^{\ell } {\mathfrak {q}}_{\ell ,m,p}^{(1)}k^{m}M_{j}^{p}}{n^{\frac{\ell }{2}}}, \end{aligned}
(4.27)
\begin{aligned}&\Xi _{j,k} \sim \Xi _{j,k}^{\textrm{formal}} := \sum _{\ell =1}^ {+\infty } \frac{\sum _{m=0}^{ \lfloor (\ell +1)/2 \rfloor } d_{\ell ,m} \, k^m M_j^{\ell +1-2m}}{ n^{\ell /2} }, \end{aligned}
(4.28)
\begin{aligned}&\frac{1}{2}\textrm{erfc}\Big (\hspace{-0.1cm}-\hspace{-0.05cm}\eta _{j,k} \sqrt{\tfrac{a_{j,k}}{2}}\Big ) \sim \frac{1}{2}\textrm{erfc}\Big ( \hspace{-0.1cm}-\hspace{-0.05cm}\frac{M_{j}r^{b}}{\sqrt{2}} \Big )\nonumber \\&-\frac{e^{-\frac{M_{j}^{2}r^{2b}}{2}}}{\sqrt{2\pi }} \sum _{\ell =1}^{+\infty } \frac{\sum _{m=0}^{\ell }\sum _{p=0}^{3\ell -1-2m} {\mathfrak {q}}_{\ell ,m,p}^{(2)}k^{m}M_{j}^{p}}{n^{\frac{\ell }{2}}}, \end{aligned}
(4.29)
\begin{aligned}&R_{a_{j,k}}(\eta _{j,k}) \sim -\frac{e^{-\frac{M_{j}^{2}r^{2b}}{2}}}{3\sqrt{2\pi } \; r^{b} \sqrt{n}} \sum _{\ell =0}^{+\infty } \frac{\sum _{m=0}^{\ell }\sum _{p=0}^{3\ell -2m} {\mathfrak {q}}_{\ell ,m,p}^{(3)}k^{m}M_{j}^{p}}{n^{\frac{\ell }{2}}}, \end{aligned}
(4.30)
\begin{aligned}&\frac{1}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \bigg [ 1+((-1)^{a}e^{u}-1)\bigg ( \frac{1}{2}\textrm{erfc}\Big (\hspace{-0.1cm}-\hspace{-0.05cm}\eta _{j,k} \sqrt{\tfrac{a_{j,k}}{2}}\Big ) - R_{a_{j,k}}(\eta _{j,k}) \bigg ) \bigg ]\nonumber \\ {}&\sim r^{k}\sum _{\ell =0}^{+\infty } \frac{\mathcal {A}_{\ell }(M_{j};k)}{\sqrt{n^{\ell } }}, \end{aligned}
(4.31)

for some $${\mathfrak {q}}_{\ell ,m,p}^{(1)},{\mathfrak {q}}_{\ell ,m,p}^{(2)}, {\mathfrak {q}}_{\ell ,m,p}^{(3)},{\mathfrak {q}}_{\ell ,m,p}^{(4)},d_{\ell ,m} \in \mathbb {C},$$ $${\mathfrak {q}}_{0,0,0}^{(1)}:=1$$, $${\mathfrak {q}}_{0,0,0}^{(3)}:=1$$, where $$a_{j,k}$$, $$\eta _{j,k}$$ are given in (3.2), and the $$\mathcal {A}_{\ell }$$’s are defined by

\begin{aligned} \mathcal {A}_{0}(x;k)&= 1+\frac{(-1)^{a}e^{u}-1}{2}\textrm{erfc}\bigg ( -\frac{r^{b}x}{\sqrt{2}} \bigg ), \end{aligned}
(4.32)
\begin{aligned} \mathcal {A}_{\ell }(x;k)&= \bigg ( 1+\frac{(-1)^{a}e^{u}-1}{2}\textrm{erfc}\bigg ( -\frac{r^{b}x}{\sqrt{2}} \bigg ) \bigg ) \sum _{m=1}^{\ell }\sum _{p=0}^{\ell }{\mathfrak {q}}_{\ell ,m,p}^{(1)}k^{m}x^{p} \nonumber \\&\quad + ((-1)^{a}e^{u}-1)\frac{e^{-\frac{r^{2b}x^{2}}{2}}}{\sqrt{2\pi } } \sum _{m=0}^{\ell }\sum _{p=0}^{ 3\ell -1-2m } {\mathfrak {q}}_{\ell ,m,p}^{(4)}k^{m}x^{p}, \quad \ell \ge 1. \end{aligned}
(4.33)

For $$\ell \ge 1$$, $$m=0$$ and $$p=0,\ldots ,\ell$$, $${\mathfrak {q}}_{\ell ,m,p}^{(1)}:=0$$. The other coefficients $$\{{\mathfrak {q}}_{\ell ,m,p}^{(1)}\}$$ are given by

\begin{aligned}&\sum _{m=1}^{\ell }\sum _{p=0}^{\ell } {\mathfrak {q}}_{\ell ,m,p}^{(1)}k^{m}x^{p}= \sum _{s=0 }^{ \lfloor \ell /2 \rfloor } \frac{1}{ ( r^{2b} )^{s} } \left( {\begin{array}{c}\frac{k}{2b}\\ s\end{array}}\right) B_s^{(1+ \frac{k}{2b}) }\Big ( \frac{k}{2b}\Big ) \left( {\begin{array}{c} s-\frac{k}{2b} \\ \ell -2s \end{array}}\right) x^{\ell -2s}, \end{aligned}
(4.34)

and the coefficients $$\{{\mathfrak {q}}_{\ell ,m,p}^{(j)}\}_{j=2}^{4}$$ can be found by equaling the terms of the same order in n in the following formal power series

\begin{aligned}&\sum _{\ell =1}^{+\infty } \frac{\sum _{m=0}^{\ell }\sum _{p=0}^{3\ell -1-2m} {\mathfrak {q}}_{\ell ,m,p}^{(2)}k^{m}M_{j}^{p}}{n^{\frac{\ell }{2}}} = \sum _{\ell =1}^{+\infty } \frac{ 2^{ \ell /2 } }{\ell !} \textrm{He}_{\ell -1} ( M_j r^b ) (\Xi _{j,k}^{\textrm{formal}})^\ell , \end{aligned}
(4.35)
\begin{aligned}&\sum _{\ell =0}^{+\infty } \frac{\sum _{m=0}^{\ell }\sum _{p=0}^{3\ell -2m} {\mathfrak {q}}_{\ell ,m,p}^{(3)}k^{m}M_{j}^{p}}{n^{\frac{\ell }{2}}}\nonumber \\ {}&= -3 r^{b}\sqrt{n} \bigg ( \sum _{\ell =0}^{+\infty } \frac{ 2^{ \ell /2 } }{\ell !}\, \textrm{He}_\ell ( M_j r^b ) (\Xi _{j,k}^{\textrm{formal}})^\ell \bigg ) \bigg (\sum _{\ell =0}^{+\infty } \frac{c_{\ell }(\eta _{j,k})}{a_{j,k}^{\ell +\frac{1}{2}}} \bigg ), \end{aligned}
(4.36)
\begin{aligned}&\sum _{\ell =1}^{+\infty } \frac{1}{n^{ \frac{\ell }{2} }} \sum _{m=0}^{\ell }\sum _{p=0}^{3\ell -1-2m} {\mathfrak {q}}_{\ell ,m,p}^{(4)}k^{m}M_{j}^{p} = -\bigg ( 1+ \sum _{\ell =1}^{+\infty } \frac{\sum _{m=1}^{\ell }\sum _{p=0}^{\ell } {\mathfrak {q}}_{\ell ,m,p}^{(1)}k^{m}M_{j}^{p}}{n^{\frac{\ell }{2}}}\bigg ) \nonumber \\&\qquad \times \bigg \{ \sum _{\ell =1}^{+\infty } \frac{\sum _{m=0}^{\ell }\sum _{p=0}^{ 3\ell -1-2m } {\mathfrak {q}}_{\ell ,m,p}^{(2)}k^{m}M_{j}^{p}}{n^{\frac{\ell }{2}}} \nonumber \\ {}&\qquad -\frac{1}{3r^{b}\sqrt{n}}\sum _{\ell =0}^{+\infty } \frac{\sum _{m=0}^{\ell }\sum _{p=0}^{3\ell -2m} {\mathfrak {q}}_{\ell ,m,p}^{(3)}k^{m}M_{j}^{p}}{n^{\frac{\ell }{2}}} \bigg \}, \end{aligned}
(4.37)

where the $$c_{\ell }$$’s are defined recursively as in (A.3).

### Proof

By (3.1) and (4.24), we have $$j+\alpha = \frac{bnr^{2b}}{1+\frac{M_{j}}{\sqrt{n}}}$$. We conclude that the left-hand side of (4.27) is independent of $$\alpha$$, and thus the coefficients $${\mathfrak {q}}_{\ell ,m,p}^{(1)}$$ do not depend on $$\alpha$$. For $$\alpha =0$$, using (2.18), we get

\begin{aligned} \frac{1}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+k}{2b})}{\Gamma (\frac{2j}{2b})}&\sim r^k \sum _{\ell =0}^{+\infty } \frac{1}{ ( r^{2b} )^{\ell } } \left( {\begin{array}{c}\frac{k}{2b}\\ \ell \end{array}}\right) B_\ell ^{(1+ \frac{k}{2b}) }\Big ( \frac{k}{2b}\Big ) \frac{1}{n^{\ell }}\bigg ( 1+\frac{M_j}{\sqrt{n}} \bigg )^{\ell -\frac{k}{2b}} \\&= r^k \sum _{\ell =0}^{+\infty } \frac{1}{ ( r^{2b} )^{\ell } } \left( {\begin{array}{c}\frac{k}{2b}\\ \ell \end{array}}\right) B_\ell ^{(1+ \frac{k}{2b}) }\Big ( \frac{k}{2b}\Big ) \frac{1}{n^{\ell }} \sum _{s=0}^{+\infty } \left( {\begin{array}{c}\ell -\frac{k}{2b}\\ s\end{array}}\right) \frac{M_j^{s}}{n^{\frac{s}{2}}} \\&= r^k \sum _{\ell =0}^{+\infty } \bigg [ \sum _{s=0 }^{ \lfloor \ell /2 \rfloor } \frac{1}{ ( r^{2b} )^{s} } \left( {\begin{array}{c}\frac{k}{2b}\\ s\end{array}}\right) B_s^{(1+ \frac{k}{2b}) }\Big ( \frac{k}{2b}\Big ) \left( {\begin{array}{c} s-\frac{k}{2b} \\ \ell -2s \end{array}}\right) M_j^{\ell -2s} \bigg ] \frac{1}{n^{\frac{\ell }{2}}}, \end{aligned}

and (4.27), (4.34) follow. By (4.25),

\begin{aligned} \Xi _{j,k}-\frac{ M_j r^b }{\sqrt{2}}&= - \sqrt{nr^{2b}} (\lambda _{j,k}-1) \sqrt{ \frac{ \lambda _{j,k}-1-\ln \lambda _{j,k} }{ \lambda _{j,k} (\lambda _{j,k}-1)^{2} } }\\&\quad \sim - \sqrt{nr^{2b}} \sum _{s=1}^{+\infty } {\mathfrak {a}}_{s}^{(1)} (\lambda _{j,k}-1)^{s} \end{aligned}

as $$n \rightarrow + \infty$$ uniformly for $$j \in \{g_{-},\ldots ,g_{+}\}$$, for some $$\{{\mathfrak {a}}_{s}^{(1)}\}_{s=1}^{+\infty }\subset \mathbb {C}$$ that are independent of j and k. The all-order expansion (4.28) now follows from

\begin{aligned} \lambda _{j,k}&= \Big ( 1+\frac{M_j}{\sqrt{n}} \Big )\bigg ( 1+\frac{k}{2bnr^{2b}} \Big (1+\frac{M_j}{\sqrt{n}}\Big )\bigg )^{-1} \sim \sum _{\ell =0}^{+\infty } \Big ( \frac{-k}{2bnr^{2b}} \Big )^\ell \Big (1+\frac{M_j}{\sqrt{n}}\Big )^{\ell +1}\\&= \sum _{\ell =0}^{+\infty } \frac{ [\lambda _{j,k}]_\ell }{n^{\ell /2}}, \end{aligned}

where $$[\lambda _{j,k}]_\ell$$ is given by

\begin{aligned}{}[\lambda _{j,k}]_\ell = \sum _{s=\lceil (\ell -1)/3 \rceil }^{\lfloor \ell /2 \rfloor } \Big ( \frac{-1}{2br^{2b}} \Big )^s \left( {\begin{array}{c}s+1\\ \ell -2s \end{array}}\right) \, k^s M_j^{\ell -2s} . \end{aligned}

Using that

\begin{aligned} \frac{1}{2}\frac{d^{\ell }}{dz^{\ell }} \textrm{erfc}(z)= -\frac{1}{\sqrt{\pi }} \frac{d^{\ell -1}}{dz^{\ell -1}} e^{-z^2} = \frac{(-1)^{\ell }}{\sqrt{2\pi }} \, 2^{\frac{\ell }{2}} \textrm{He}_{\ell -1}(\sqrt{2}z) \,e^{-z^2}, \quad \ell \in \mathbb {N},\qquad \quad \end{aligned}
(4.38)

and $$\textrm{He}_\ell (z)=(-1)^{\ell }\textrm{He}_\ell (-z)$$, we infer that

\begin{aligned} \frac{1}{2} \textrm{erfc} \Big ( -\frac{ M_j r^b }{\sqrt{2}}+ \Xi _{j,k}\Big )&\sim \frac{1}{2} \textrm{erfc} \Big ( -\frac{ M_j r^b }{\sqrt{2}}\Big )-\frac{e^{ -\frac{M_j^2 r^{2b}}{2} }}{\sqrt{2\pi }} \sum _{\ell =1}^{+\infty } \frac{ 2^{ \ell /2 } }{\ell !} \textrm{He}_{\ell -1} ( M_j r^b ) \Xi _{j,k}^{\ell } \end{aligned}

as $$n \rightarrow + \infty$$ uniformly for $$j \in \{g_{-},\ldots ,g_{+}\}$$. To verify (4.29), it is thus enough to show (4.35). For this purpose, note that by (4.28), we have

\begin{aligned} (\Xi _{j,k}^{\textrm{formal}})^{\ell }&= \bigg (\sum _{s=1}^{+\infty } \frac{\sum _{m=0}^{ \lfloor (s+1)/2 \rfloor } d_{s,m} \, k^m M_j^{s+1-2m}}{ n^{s/2} } \bigg )^{\ell } \nonumber \\&= \sum _{s=\ell }^{+\infty } \frac{\sum _{m=0}^{ \lfloor (s+\ell )/2 \rfloor } d_{s,m}^{(\ell )} \, k^m M_j^{s+\ell -2m}}{ n^{s/2} } \end{aligned}
(4.39)

for some coefficients $$\{d_{s,m}^{(\ell )}\} \subset \mathbb {C}$$. Inserting this expansion and (1.6) in the right-hand side of (4.35), we obtain

\begin{aligned} \sum _{\ell =1}^{+\infty }&\frac{ 2^{ \ell /2 } }{\ell !} \textrm{He}_{\ell -1} ( M_j r^b ) (\Xi _{j,k}^{\textrm{formal}})^{\ell } \sim \sum _{\ell =1}^{+\infty } \sum _{s=\ell }^{+\infty } \frac{ 2^{ \ell /2 } }{\ell !} \textrm{He}_{\ell -1} ( M_j r^b ) \\&\frac{\sum _{m=0}^{ \lfloor (s+\ell )/2 \rfloor } d_{s,m}^{(\ell )} \, k^m M_j^{s+\ell -2m}}{ n^{s/2} } \\&\sim \sum _{\ell =1}^{+\infty } \sum _{s=\ell }^{+\infty } \frac{ 2^{ \ell /2 } }{\ell !} \bigg ((\ell -1)! \sum _{q=0 }^{ \lfloor (\ell -1)/2 \rfloor } \frac{(-1)^q}{q!(\ell -1-2q)!}\frac{(M_{j}r^{b})^{\ell -1-2q}}{2^q} \bigg ) \\&\frac{\sum _{m=0}^{ \lfloor (s+\ell )/2 \rfloor } d_{s,m}^{(\ell )} \, k^m M_j^{s+\ell -2m}}{ n^{s/2} }. \end{aligned}

Rearranging the summations in the right-hand side of this equation, we obtain (4.35).

We now turn to the proof of (4.30) and (4.36). By (A.2), we have

\begin{aligned}&R_{a_{j,k}}(\eta _{j,k}) \sim \frac{e^{-\frac{1}{2}a_{j,k} \eta _{j,k}^{2}}}{\sqrt{2\pi a_{j,k} }}\sum _{\ell =0}^{+\infty } \frac{c_{\ell }(\eta _{j,k})}{a_{j,k}^{\ell }}, \qquad \text{ as } n \rightarrow + \infty , \end{aligned}

uniformly for $$j \in \{g_{-},\ldots ,g_{+}\}$$. Using again (4.38), we get

\begin{aligned} e^{-\frac{1}{2}a_{j,k} \eta _{j,k}^{2}} \sim e^{-\frac{M_{j}^{2}r^{2b}}{2}} \sum _{\ell =0}^{+\infty } 2^{ \ell /2 }\, \textrm{He}_\ell ( M_j r^b ) \frac{ \Xi _{j,k}^\ell }{ \ell ! }. \end{aligned}

Combining these expansions yields

\begin{aligned} R_{a_{j,k}}(\eta _{j,k}) \sim \frac{ 1 }{\sqrt{2\pi } } e^{-\frac{M_{j}^{2}r^{2b}}{2}} \bigg ( \sum _{\ell =0}^{+\infty } \frac{ 2^{ \ell /2 } }{ \ell ! } \, \textrm{He}_\ell ( M_j r^b ) \Xi _{j,k}^{\ell } \bigg ) \bigg ( \sum _{\ell =0}^{+\infty } \frac{c_{\ell }(\eta _{j,k})}{a_{j,k}^{\ell +\frac{1}{2} }} \bigg ). \end{aligned}

Therefore, by (4.28), to prove (4.30) it is enough to show (4.36). The first term (inside the left parenthesis) on the right-hand side of (4.36) can be expanded in a similar way as (4.35). On the other hand, the second term (inside the right parenthesis) can be expanded using that the $$c_{\ell }$$’s (which are defined below (A.2)) are smooth, and the expansions

\begin{aligned}&a_{j,k}= \frac{n r^{2b}}{1+\frac{M_j}{\sqrt{n}} }+\frac{k}{2b} \sim \frac{k}{2b} + nr^{2b} \sum _{\ell =0}^{+\infty }\frac{(-1)^{\ell }M_{j}^{\ell }}{n^{\ell /2}}, \end{aligned}
(4.40)
\begin{aligned}&\eta _{j,k} \sim \sum _{s=1}^{+\infty } {\mathfrak {b}}_{s} (\lambda _{j,k}-1)^{s}= \sum _{s=1}^{+\infty } {\mathfrak {b}}_{s} \bigg ( \sum _{\ell =1}^{+\infty } \frac{ [\lambda _{j,k}]_\ell }{n^{\ell /2}} \bigg )^{s}, \end{aligned}
(4.41)

where the coefficients $$\{{\mathfrak {b}}_{s}\}_{s=1}^{+\infty }\subset \mathbb {C}$$ are independent of j and k. Combining all of the above, we obtain the desired asymptotic expansion (4.36).

Finally, (4.31), (4.32), (4.33) and (4.37) are direct consequences of (4.27)–(4.30). $$\square$$

### Lemma 4.6

The following relations hold

\begin{aligned}&(2b)^a r^{ab} (-1)^a a! \sum _{p=0}^a {\mathfrak {q}}_{a,a,p}^{(1)} \Big ( \frac{x}{r^b}\Big )^{p} =p_{0,a}(x), \end{aligned}
(4.42)
\begin{aligned}&(2b)^{a+1} r^{(a+1)b} (-1)^a a! \sum _{p=0}^{a+1} {\mathfrak {q}}_{a+1,a,p}^{(1)} \Big ( \frac{x}{r^b}\Big )^{p} \nonumber \\&\quad = -ab \Big ( p_{0,a+1}(x) + (1-3a) p_{0,a-1}(x)\nonumber \\ {}&\qquad + \frac{5}{3}(a-1)(a-2) p_{0,a-3}(x) \Big ), \end{aligned}
(4.43)

where $$p_{0,a}(x)$$ is given by (1.8).

### Proof

It follows from (4.34) that

\begin{aligned}&(2b)^a r^{ab} (-1)^a a! \sum _{p=0}^a {\mathfrak {q}}_{a,a,p}^{(1)} \Big ( \frac{x}{r^b}\Big )^{p} \\&=(2b)^a (-1)^a \sum _{s=0 }^{ \lfloor a/2 \rfloor } \Big [\frac{d}{dk}\Big ]^a \bigg [ B_s^{(1+ \frac{k}{2b}) }\Big ( \frac{k}{2b}\Big ) \left( {\begin{array}{c}\frac{k}{2b}\\ s\end{array}}\right) \left( {\begin{array}{c} s-\frac{k}{2b} \\ a-2s \end{array}}\right) \bigg ] \bigg |_{k=0} x^{a-2s}. \end{aligned}

For each $$s \in \{0,\ldots ,\lfloor \frac{a}{2} \rfloor \}$$, the polynomial $$k \mapsto B_s^{(1+ \frac{k}{2b}) }( \frac{k}{2b}) \left( {\begin{array}{c}\frac{k}{2b}\\ s\end{array}}\right) \left( {\begin{array}{c} s-\frac{k}{2b} \\ a-2\,s \end{array}}\right)$$ is of the form $$\sum _{\ell =0}^{a} {\tilde{c}}_{\ell }(k/b)^{a}$$, where $${\tilde{c}}_{0},\ldots ,{\tilde{c}}_{a} \in \mathbb {C}$$ are independent of b. Thanks to the prefactor $$(2b)^a$$, the above expression is thus independent of b. Replacing b by $$\frac{1}{2}$$ yields

\begin{aligned} (2b)^a r^{ab} (-1)^a a! \sum _{p=0}^a {\mathfrak {q}}_{a,a,p}^{(1)} \Big ( \frac{x}{r^b}\Big )^{p}&= (-1)^a \sum _{s=0 }^{ \lfloor a/2 \rfloor } \Big [\frac{d}{dk}\Big ]^a \bigg [ B_s^{(1+ k) }( k ) \left( {\begin{array}{c}k\\ s\end{array}}\right) \left( {\begin{array}{c} s-k \\ a-2s \end{array}}\right) \bigg ] \bigg |_{k=0} x^{a-2s}. \end{aligned}
(4.44)

By (2.20),

\begin{aligned} B_s^{(1+ k) }( k ) = \Big ( \frac{d}{dt}\Big )^s \Big [ \Big ( \frac{t}{e^t-1}\Big )^{k+1} e^{ kt } \Big ]\Big |_{t=0} = \frac{1}{2^s}\Big ( k^s-\frac{s(s+5)}{6} k^{s-1} + \cdots \Big ) \end{aligned}
(4.45)

is a polynomial in k of degree s, and

\begin{aligned} \left( {\begin{array}{c}k\\ s\end{array}}\right) \left( {\begin{array}{c} s-k \\ a-2s \end{array}}\right) = \frac{ (-1)^{a} }{s! (a-2s)!} \Big ( k^{a-s}+\frac{ 7s^2-(6a-3)s+a^2-a}{2} \, k^{a-1-s}+\cdots \Big ) \end{aligned}
(4.46)

is a polynomial in k of degree $$a-s$$. Combining (4.45) and (4.46), we obtain

\begin{aligned} \Big [\frac{d}{dk}\Big ]^a \bigg [ B_s^{(1+ k) }( k ) \left( {\begin{array}{c}k\\ s\end{array}}\right) \left( {\begin{array}{c} s-k \\ a-2s \end{array}}\right) \bigg ] \bigg |_{k=0} = \frac{ (-1)^{a} a!}{s! (a-2s)! 2^{s}}, \end{aligned}

and now (4.42) follows directly from the right-most expression of $$p_{0,a}$$ in (1.8). Now we turn to the proof of (4.43). In a similar way as (4.44), using (4.34),

\begin{aligned}{} & {} (2b)^{a+1} r^{(a+1)b} (-1)^a a! \sum _{p=0}^{a+1} {\mathfrak {q}}_{a+1,a,p}^{(1)} \Big ( \frac{x}{r^b}\Big )^{p}\\{} & {} \quad = 2b\, (-1)^a \sum _{s=0 }^{ \lfloor (a+1)/2 \rfloor } \Big [\frac{d}{dk}\Big ]^a \bigg [ B_s^{(1+k) }(k) \left( {\begin{array}{c}k\\ s\end{array}}\right) \left( {\begin{array}{c} s-k \\ a+1-2s \end{array}}\right) \bigg ] \bigg |_{k=0} x^{a+1-2s}. \end{aligned}

From (4.45) and (4.46) with a replaced by $$a+1$$, we get

\begin{aligned}&\Big [\frac{d}{dk}\Big ]^a \bigg [ B_s^{(1+k) }(k) \left( {\begin{array}{c}k\\ s\end{array}}\right) \left( {\begin{array}{c} s-k \\ a+1-2s \end{array}}\right) \bigg ] \bigg |_{k=0} \\&= \frac{ (-1)^{a+1} a!}{s! (a+1-2s)! 2^{s}} \frac{20s^{2}-2(9a+7)s+3a(a+1)}{6}, \end{aligned}

and thus

\begin{aligned}{} & {} (2b)^{a+1} r^{(a+1)b} (-1)^a a! \sum _{p=0}^{a+1} {\mathfrak {q}}_{a+1,a,p}^{(1)} \Big ( \frac{x}{r^b}\Big )^{p} \\ {}{} & {} = -\frac{b}{3} \sum _{s=0 }^{ \lfloor (a+1)/2 \rfloor } \frac{a!\, (20s^2-2(9a+7)s+3a(a+1)) }{ s! (a+1-2s)!} \frac{ x^{a+1-2s} }{ 2^s}. \end{aligned}

Finally, the expression (4.43) follows from the manipulation

\begin{aligned}&\sum _{s=0 }^{ \lfloor (a+1)/2 \rfloor } \frac{a!\, (20s^2-2(9a+7)s+3a(a+1)) }{ s! (a+1-2s)!} \frac{ x^{a+1-2s} }{ 2^s}\\&\quad = 3a \,p_{0,a+1}(x)+ a! \sum _{s=0 }^{ \lfloor (a-1)/2 \rfloor } \frac{ 10s+3(1-3a) }{ s! (a-1-2s)!} \frac{ x^{a-1-2s} }{ 2^s }\\&\quad = 3a \,p_{0,a+1}(x)+ 3a(1-3a) \,p_{0,a-1}(x) \\&\qquad + a! \sum _{s=1 }^{ \lfloor (a-1)/2 \rfloor } \frac{ 10 }{ (s-1)! (a-1-2s)!} \frac{ x^{a-1-2s} }{ 2^s }\\&\quad = 3a \,p_{0,a+1}(x)+ 3a(1-3a) \,p_{0,a-1}(x) + 5 a(a-1)(a-2) \, p_{0,a-3}(x). \end{aligned}

$$\square$$

### Lemma 4.7

The following relations hold

\begin{aligned}&(2br^{b})^{a+1} (-1)^{a+1} (a+1)! \sum _{p=0}^{a} {\mathfrak {q}}_{a+1,a+1,p}^{(4)} \Big (\frac{x}{r^b}\Big )^{p} = q_{0,a+1}(x), \end{aligned}
(4.47)
\begin{aligned}&(2br^{b})^{a+1} (-1)^a a! \sum _{p=0}^{a+2} {\mathfrak {q}}_{a+1,a,p}^{(4)} \Big ( \frac{x}{r^b}\Big )^{p} \nonumber \\&\quad = -b \Big ( a \, q_{0,a+1}(x) + (1-3a) [a q_{0,a-1}(x)] \nonumber \\&\qquad + \frac{5}{3} [a(a-1)(a-2) q_{0,a-3}(x)] \Big ), \end{aligned}
(4.48)

where $$q_{0,a}(x)$$, $$[a q_{0,a-1}(x)]$$ and $$[a(a-1)(a-2) q_{0,a-3}(x)]$$ are given by (1.9) and (1.12).

### Remark 4.8

The degree of the polynomial in the right-hand side of (4.48) is given by

\begin{aligned} {\left\{ \begin{array}{ll} 2 &{} if a=0, \\ a &{} if a\ge 1. \end{array}\right. } \end{aligned}

In particular, $${\mathfrak {q}}_{a+1,a,a+1}^{(4)} ={\mathfrak {q}}_{a+1,a,a+2}^{(4)}=0.$$

### Proof

Let us first rewrite the sums on the left-hand sides of (4.47) and (4.48) in terms of the coefficients $$\{{\mathfrak {q}}_{\ell ,m,p}^{(1)},{\mathfrak {q}}_{\ell ,m,p}^{(2)}, {\mathfrak {q}}_{\ell ,m,p}^{(3)}\}$$. By (4.37),

\begin{aligned}&\sum _{m=0}^{a+1}\sum _{p=0}^{3a+2-2m} {\mathfrak {q}}_{a+1,m,p}^{(4)}k^{m}x^{p} = - \sum _{s=0}^{a} \Bigg \{ \sum _{m=0}^{s}\sum _{p=0}^{s} {\mathfrak {q}}_{s,m,p}^{(1)}k^{m}x^{p} \nonumber \\&\qquad \times \bigg \{ \sum _{m=0}^{a+1-s}\sum _{p=0}^{ 3(a+1-s)-1-2m } \hspace{-0.3cm} {\mathfrak {q}}_{a+1-s,m,p}^{(2)}k^{m}x^{p} -\frac{1}{3r^{b}} \sum _{m=0}^{a-s}\sum _{p=0}^{3(a-s)-2m} {\mathfrak {q}}_{a-s,m,p}^{(3)}k^{m}x^{p} \bigg \} \Bigg \}. \end{aligned}

Equaling the coefficients of $$k^{a+1}$$ and of $$k^{a}$$ gives the identities

\begin{aligned}&\sum _{p=0}^{a} {\mathfrak {q}}_{a+1,a+1,p}^{(4)}x^{p} = - \sum _{s=0}^{a} \Bigg \{ \sum _{p=0}^{s} {\mathfrak {q}}_{s,s,p}^{(1)}x^{p} \times \bigg \{ \sum _{p=0}^{ a-s } {\mathfrak {q}}_{a+1-s,a+1-s,p}^{(2)}x^{p} \bigg \} \Bigg \} \end{aligned}
(4.49)

and

\begin{aligned}&\sum _{p=0}^{a+2} {\mathfrak {q}}_{a+1,a,p}^{(4)}x^{p} \hspace{-0.05cm} = \hspace{-0.05cm} - \hspace{-0.05cm} \sum _{s=1}^{a} \hspace{-0.05cm} \Bigg \{ \hspace{-0.05cm} \sum _{p=0}^{s} {\mathfrak {q}}_{s,s-1,p}^{(1)}x^{p} \hspace{-0.05cm} \times \hspace{-0.05cm} \bigg \{ \hspace{-0.05cm} \sum _{p=0}^{ a-s } {\mathfrak {q}}_{a+1-s,a+1-s,p}^{(2)}x^{p} \bigg \} \hspace{-0.05cm}\Bigg \} \nonumber \\&\quad - \sum _{s=0}^{a} \Bigg \{ \sum _{p=0}^{s} {\mathfrak {q}}_{s,s,p}^{(1)}x^{p} \times \bigg \{ \sum _{p=0}^{ a-s+2 } {\mathfrak {q}}_{a+1-s,a-s,p}^{(2)}x^{p} -\frac{1}{3r^{b}} \sum _{p=0}^{a-s} {\mathfrak {q}}_{a-s,a-s,p}^{(3)}x^{p} \bigg \} \Bigg \}. \end{aligned}
(4.50)

It remains to simplify the right-hand sides of (4.49) and (4.50). Sums of the form $$\sum _{p=0}^{s} {\mathfrak {q}}_{s,s,p}^{(1)}x^{p}$$ and $$\sum _{p=0}^{s} {\mathfrak {q}}_{s,s-1,p}^{(1)}x^{p}$$ were already simplified in (4.42) and (4.43). Also, by (4.35),

\begin{aligned} \sum _{p=0}^{\ell -1} {\mathfrak {q}}_{\ell ,\ell ,p}^{(2)} \Big ( \frac{x}{r^b}\Big )^{p} = \sum _{s=1}^{\ell } \frac{ 2^{ s/2 } }{s!} \textrm{He}_{s-1} ( x ) \frac{1}{\ell !}\Big [\frac{d}{dk}\Big ]^{\ell }\Big [[(\Xi _{j,k}^ {\textrm{formal}})^s]_{\ell }\Big ]\Big |_{k=0,M_{j}\rightarrow \frac{x}{r^{b}}}\nonumber \\ \end{aligned}
(4.51)

where $$[(\Xi _{j,k}^{\textrm{formal}})^s]_{\ell }$$ is the coefficient of the term of order $$n^{-\frac{\ell }{2}}$$ in the asymptotic series $$(\Xi _{j,k}^{\textrm{formal}})^s$$. Using (4.28), we infer that

\begin{aligned} \frac{1}{\ell !}\Big [\frac{d}{dk}\Big ]^{\ell } \Big [[(\Xi _{j,k}^{\textrm{formal}})^s]_{\ell }\Big ]\Big |_{k=0} = \delta _{s,\ell } d_{1,1}^{\ell }, \qquad s=1,\ldots ,\ell . \end{aligned}
(4.52)

Also, a direct computation using (4.25) shows that

\begin{aligned} \Xi _{j,k} = \frac{1}{2\sqrt{2} b r^{b} \sqrt{n}}\bigg \{ k + \frac{5 (r^{b}M_{j})^{2}}{3}b + \frac{1}{\sqrt{n}} \bigg ( \frac{k M_{j}}{3} - \frac{53}{36}bM_{j}^{3}r^{2b} \bigg ) + \mathcal{O}(n^{-1}) \bigg \} \end{aligned}
(4.53)

as $$n \rightarrow + \infty$$ uniformly for $$j \in \{g_{-},\ldots ,g_{+}\}$$. In particular, $$d_{1,1} = \frac{1}{2\sqrt{2} b r^{b}}$$ and thus, by (4.51)-(4.52),

\begin{aligned} \sum _{p=0}^{\ell -1} {\mathfrak {q}}_{\ell ,\ell ,p}^{(2)} \Big ( \frac{x}{r^b}\Big )^{p} = \frac{1}{ (2b r^b)^{\ell } } \frac{1}{ \ell ! } \textrm{He}_{\ell -1}(x), \qquad \ell \ge 1. \end{aligned}
(4.54)

Combining (4.49) with (1.8), (4.42) and (4.54) yields

\begin{aligned}&(2br^{b})^{a+1} (-1)^{a+1} (a+1)! \sum _{p=0}^{a} {\mathfrak {q}}_{a+1,a+1,p}^{(4)} \Big (\frac{x}{r^b}\Big )^{p} = (-1)^a \sum _{s=0}^a \left( {\begin{array}{c}a+1\\ s\end{array}}\right) i^s\\&\quad \textrm{He}_s(ix) \textrm{He}_{a-s}(x). \end{aligned}

Using the functional equation (see e.g. [66, eq.(9.6)])

\begin{aligned} \sum _{s=0}^{a} \left( {\begin{array}{c}a+1\\ s\end{array}}\right) i^s \textrm{He}_s(ix) \textrm{He}_{a-s}(x) = i^a\, \textrm{He}_a^{(1)}(ix), \qquad a \in \mathbb {N}, \end{aligned}
(4.55)

we obtain the desired identity (4.47). It remains to simplify the right-hand side of (4.50) (with x replaced by $$\frac{x}{r^{b}}$$) and to prove (4.48). In view of (4.42), (4.43) and (4.54), it only remains to evaluate explicitly sums of the forms

\begin{aligned} \sum _{p=0}^{\ell +2} {\mathfrak {q}}_{\ell +1,\ell ,p}^{(2)} \Big ( \frac{x}{r^b}\Big )^{p} \quad \text{ and } \quad \frac{1}{3 r^{b} }\sum _{p=0}^{\ell } {\mathfrak {q}}_{\ell ,\ell ,p}^{(3)} \Big ( \frac{x}{r^b}\Big )^{p}, \qquad \ell \ge 0. \end{aligned}
(4.56)

For the first sum, we use (4.35) to get

\begin{aligned}&\sum _{p=0}^{\ell +2} {\mathfrak {q}}_{\ell +1,\ell ,p}^{(2)} \Big ( \frac{x}{r^b}\Big )^{p} = \sum _{s=1}^{\ell +1} \frac{ 2^{ s/2 } }{s!} \textrm{He}_{s-1} ( x )\frac{1}{\ell !}\Big [\frac{d}{dk}\Big ]^{\ell } \Big [[(\Xi _{j,k}^{\textrm{formal}})^s]_{\ell +1}\Big ]\Big |_{k=0,M_{j}\rightarrow \frac{x}{r^{b}}}, \nonumber \\ {}&\qquad \ell \ge 0. \end{aligned}
(4.57)

A direct computation using (4.28) shows that

\begin{aligned} \frac{1}{\ell !}\Big [\frac{d}{dk}\Big ]^{\ell } \Big [[(\Xi _{j,k}^{\textrm{formal}})^s]_{\ell +1}\Big ]\Big |_{k=0} = {\left\{ \begin{array}{ll} 0, &{} \text{ if } 1 \le s \le \ell -2, \\ (\ell -1) d_{1,1}^{\ell -2}d_{3,2}, &{} \text{ if } s = \ell -1, \\ \ell \frac{x}{r^{b}} d_{1,1}^{\ell -1}d_{2,1}, &{} \text{ if } s = \ell , \\ (\ell +1) \frac{x^{2}}{r^{2b}} d_{1,1}^{\ell }d_{1,0}, &{} \text{ if } s = \ell +1, \end{array}\right. } \end{aligned}

and by (4.25) and (4.53),

\begin{aligned} d_{1,1} = \frac{1}{2\sqrt{2} b r^{b}}, \qquad d_{1,0} = \frac{5r^{b}}{6\sqrt{2}}, \qquad d_{2,1} = \frac{1}{6\sqrt{2} br^{b}}, \qquad d_{3,2} = \frac{-1}{24\sqrt{2}b^{2}r^{3b}}. \end{aligned}

Substituting the above in (4.57), for $$\ell \ge 0$$ we get

\begin{aligned} \sum _{p=0}^{\ell +2} {\mathfrak {q}}_{\ell +1,\ell ,p}^{(2)} \Big ( \frac{x}{r^b}\Big )^{p}&= \frac{1}{(2b)^\ell r^{b(\ell +1) } }\frac{1}{ \ell ! } \Big ( \frac{5}{6} \, x^2 \, \textrm{He}_{\ell }(x) + \frac{\ell }{3} \, x\, \textrm{He}_{\ell -1}(x) -\frac{\ell (\ell -1)}{6} \textrm{He}_{\ell -2}(x) \Big ), \nonumber \\&= \frac{1}{(2b)^\ell r^{b(\ell +1) } }\frac{1}{ \ell ! } \Big [ \frac{5x^{2}+2\ell }{6} \, \textrm{He}_{\ell }(x) + \frac{\ell (\ell -1)}{6} \textrm{He}_{\ell -2}(x) \Big ], \end{aligned}
(4.58)

where $$\textrm{He}_{-2}(x) \equiv \textrm{He}_{-1}(x) \equiv 0$$, and for the second line we have used the three-term recurrence relation (1.5) of $$\textrm{He}_\ell$$.

Now we turn to the problem of simplifying the second sum in (4.56). Let us write

\begin{aligned} \sqrt{n}\bigg (\sum _{\ell =0}^{+\infty } \frac{c_{\ell }(\eta _{j,k})}{a_{j,k}^{\ell +\frac{1}{2}}} \bigg ) \sim \sum _{\ell =0}^{+\infty } \frac{e_{\ell }}{n^{\frac{\ell }{2}}}, \qquad \text{ as } n \rightarrow + \infty \end{aligned}

for some $$\{e_{\ell }\}_{\ell =0}^{+\infty } \subset \mathbb {C}$$. By (4.36),

\begin{aligned}&\frac{1}{3r^{b}}\sum _{p=0}^{\ell } {\mathfrak {q}}_{\ell ,\ell ,p}^{(3)} = - \frac{1}{\ell !}\Big [\frac{d}{dk}\Big ]^{\ell } \bigg \{ e_{\ell } + \sum _{s=1}^{\ell } e_{\ell -s} \sum _{m=1}^{s} \frac{ 2^{ m/2 } }{m!} \textrm{He}_{m} ( x ) \Big [[(\Xi _{j,k}^{\textrm{formal}})^m]_{s}\Big ] \bigg \}\Big |_{k=0,M_{j}\rightarrow \frac{x}{r^{b}}} \nonumber \\&= - \frac{1}{\ell !} \bigg \{ \Big [\frac{d}{dk}\Big ]^{\ell }e_{\ell } + \sum _{s=1}^{\ell } \sum _{m=1}^{s} \frac{ 2^{ m/2 } }{m!} \textrm{He}_{m} ( x ) \sum _{q=0}^{\ell }\left( {\begin{array}{c}\ell \\ q\end{array}}\right) \nonumber \\&\Big [\frac{d}{dk}\Big ]^{q}e_{\ell -s} \; \Big [\frac{d}{dk}\Big ]^{\ell -q}\Big [[(\Xi _{j,k}^{\textrm{formal}})^m]_{s}\Big ] \bigg \}\Big |_{k=0,M_{j}\rightarrow \frac{x}{r^{b}}}. \end{aligned}
(4.59)

Long but direct calculations using (4.39), (4.40) and (4.41) show that

\begin{aligned}&\Big [\frac{d}{dk}\Big ]^{q}e_{\ell -s}\Big |_{k=0} = 0,{} & {} \text{ for } \; \ell \ge 0, \; 0 \le s \le \ell , \; q \ge 1+\Big \lfloor \frac{\ell -s}{2} \Big \rfloor , \\&\Big [\frac{d}{dk}\Big ]^{\ell -q}\Big [[(\Xi _{j,k}^{\textrm{formal}})^m] _{s}\Big ]\Big |_{k=0} = 0,{} & {} \text{ for } \; \ell \ge 1, \; 1 \le s \le \ell , \; 1 \le m \le s, \; q \\&\le \Big \lfloor \ell -\frac{s+m+1}{2} \Big \rfloor . \end{aligned}

This means that for $$\ell \ge 1$$, the only term that contributes in (4.59) corresponds to $$m=s=\ell$$ and $$q=0$$. Thus, for any $$\ell \ge 0$$, we have

\begin{aligned} \frac{1}{3r^{b}}\sum _{p=0}^{\ell } {\mathfrak {q}}_{\ell ,\ell ,p}^{(3)} = - \bigg \{ e_{0}\frac{ 2^{ \ell /2 } }{\ell !} \textrm{He}_{\ell } ( x ) \; \frac{1}{\ell !}\Big [\frac{d}{dk}\Big ]^{\ell } \Big [[(\Xi _{j,k}^{\textrm{formal}})^\ell ]_{\ell }\Big ] \bigg \}\Big |_{k=0,M_{j}\rightarrow \frac{x}{r^{b}}}. \end{aligned}

A direct computation shows that $$e_{0}=-\frac{1}{3r^{b}}$$. Using also (4.52), we get

\begin{aligned} \frac{1}{3r^{b}}\sum _{p=0}^{\ell } {\mathfrak {q}}_{\ell ,\ell ,p}^{(3)} \Big ( \frac{x}{r^b}\Big ) ^{p}= \frac{1}{3} \frac{1}{(2b)^{\ell } r^{b(\ell +1) } }\frac{1}{ \ell ! } \textrm{He}_{\ell }(x), \qquad \ell \ge 0. \end{aligned}
(4.60)

Using (4.58) and (4.60), for $$\ell \ge 0$$ we obtain

\begin{aligned}&\sum _{p=0}^{\ell +2} {\mathfrak {q}}_{\ell +1,\ell ,p}^{(2)} \Big ( \frac{x}{r^b}\Big ) ^{p} -\frac{1}{3r^{b}} \sum _{p=0}^{\ell } {\mathfrak {q}}_{\ell ,\ell ,p}^{(3)} \Big ( \frac{x}{r^b}\Big ) ^{p} \nonumber \\&= \frac{1}{(2b)^{\ell } r^{b(\ell +1) } }\frac{1}{\ell ! } \Big [ \frac{5x^{2}+2(\ell -1)}{6} \, \textrm{He}_{\ell }(x) + \frac{\ell (\ell -1)}{6} \textrm{He}_{\ell -2}(x) \Big ]. \end{aligned}
(4.61)

Combining (4.50) with (4.42), (4.43) and (4.61), we obtain after some calculations that

\begin{aligned}{} & {} (2br^{b})^{a+1} (-1)^a a! \sum _{p=0}^{a+2} {\mathfrak {q}}_{a+1,a,p}^{(4)} \Big ( \frac{x}{r^b}\Big )^{p}\nonumber \\ {}{} & {} =- \frac{b}{3}\bigg \{ \sum _{s=1}^a (-1)^{a-s} \left( {\begin{array}{c}a\\ s-1\end{array}}\right) {\widehat{p}}_{0,s}(x) \textrm{He}_{a-s}(x) +{\mathfrak {s}}_a(x) \bigg \}, \end{aligned}
(4.62)

where

\begin{aligned} {\widehat{p}}_{0,a+1}(x)&:= 3 a\,p_{0,a+1}(x)+ 3a(1-3a) \,p_{0,a-1}(x) + 5a (a-1)(a-2) \, p_{0,a-3}(x), \end{aligned}
(4.63)
\begin{aligned} {\mathfrak {s}}_a(x)&:= (-1)^{a}\sum _{s=0}^a \left( {\begin{array}{c}a\\ s\end{array}}\right) i^s\textrm{He}_{s}(ix) \Big [ ( 5 x^2+2(a-s-1)) \, \textrm{He}_{a-s}(x) \nonumber \\ {}&\quad + (a-s)(a-s-1) \textrm{He}_{a-s-2}(x) \Big ] . \end{aligned}
(4.64)

The polynomial $${\mathfrak {s}}_a$$ can actually be considerably simplified. Indeed, since

\begin{aligned} \textrm{He}_s(x)= \Big [ \frac{d}{dt} \Big ]^s \Big [ e^{xt-\frac{t^2}{2}}\Big ] \Big |_{t=0}, \qquad i^s \textrm{He}_s(ix)=\Big [ \frac{d}{dt} \Big ]^s \Big [ e^{-xt+\frac{t^2}{2}}\Big ] \Big |_{t=0}, \end{aligned}

we have

\begin{aligned} \sum _{s=0}^a \left( {\begin{array}{c}a\\ s\end{array}}\right) i^s\textrm{He}_s(ix) \textrm{He}_{a-s}(x)= \Big [ \frac{d}{dt} \Big ]^a \Big [ 1 \Big ] \Big |_{t=0}= {\left\{ \begin{array}{ll} 1 &{} if a=0, \\ 0 &{} if a \ge 1. \end{array}\right. } \end{aligned}
(4.65)

Hence,

\begin{aligned}{} & {} \sum _{s=0}^a \left( {\begin{array}{c}a\\ s\end{array}}\right) (a-s)(a-s-1) i^s\textrm{He}_{s}(ix) \textrm{He}_{a-s-2}(x) \nonumber \\{} & {} \quad =a(a-1) \sum _{s=0}^{a-2} \left( {\begin{array}{c}a-2\\ s\end{array}}\right) i^s\textrm{He}_{s}(ix) \textrm{He}_{a-s-2}(x) = {\left\{ \begin{array}{ll} 2 &{} if a=2, \\ 0 &{}if a\not =2, \end{array}\right. } \end{aligned}
(4.66)

and using also the recurrence relation (1.5) we get

\begin{aligned}{} & {} \sum _{s=0}^a \left( {\begin{array}{c}a\\ s\end{array}}\right) (a-s) i^s\textrm{He}_{s}(ix) \textrm{He}_{a-s}(x) \nonumber \\{} & {} \qquad =a \sum _{s=0}^{a-1} i^s \left( {\begin{array}{c}a-1\\ s\end{array}}\right) \textrm{He}_{s}(ix) \Big ( x \textrm{He}_{a-s-1}(x)-(a-s-1)\textrm{He}_{a-s-2}(x) \Big ) \nonumber \\{} & {} \qquad = {\left\{ \begin{array}{ll} x &{} if a=1, \\ -2 &{}if a=2, \\ 0 &{}otherwise . \end{array}\right. } \end{aligned}
(4.67)

Using (4.65), (4.66) and (4.67) to simplify $${\mathfrak {s}}_{a}$$, we finally obtain

\begin{aligned} {\mathfrak {s}}_a(x)= {\left\{ \begin{array}{ll} 5x^2-2 &{} if a=0, \\ -2x &{} if a=1, \\ -2 &{} if a=2, \\ 0 &{} if a\ge 3. \end{array}\right. } \end{aligned}

Let us now simplify the sum in (4.62). First, substituting the definitions (4.63) and (1.8), we rewrite it as

\begin{aligned}&\sum _{s=1}^a (-1)^{a-s} \left( {\begin{array}{c}a\\ s-1\end{array}}\right) {\widehat{p}}_{0,s}(x) \textrm{He}_{a-s}(x) = A_{1} + A_{2} + A_{3}, \end{aligned}
(4.68)

where

\begin{aligned}&A_{1} := 3(-1)^{a} \sum _{s=2}^a \left( {\begin{array}{c}a\\ s-1\end{array}}\right) (s-1) i^s \textrm{He}_{s}(ix)\textrm{He}_{a-s}(x), \\&A_{2} := -3(-1)^a \sum _{s=2}^a \left( {\begin{array}{c}a\\ s-1\end{array}}\right) (s-1)\Big ( 3(s-2)+2\Big ) i^{s-2} \textrm{He}_{s-2}(ix)\textrm{He}_{a-s}(x), \\&A_{3} := 5(-1)^a \sum _{s=4}^a \left( {\begin{array}{c}a\\ s-1\end{array}}\right) (s-1)(s-2)(s-3) i^{s} \textrm{He}_{s-4}(ix)\textrm{He}_{a-s}(x). \end{aligned}

To simplify $$A_{1}$$, we first establish two formulas. Using (4.65) and $$\left( {\begin{array}{c}a+1\\ s\end{array}}\right) =\left( {\begin{array}{c}a\\ s\end{array}}\right) +\left( {\begin{array}{c}a\\ s-1\end{array}}\right)$$ in (4.55), we infer that

\begin{aligned} \sum _{s=1}^a \left( {\begin{array}{c}a\\ s-1\end{array}}\right) i^s\textrm{He}_s(ix) \textrm{He}_{a-s}(x) = {\left\{ \begin{array}{ll} i^a\, \textrm{He}_a^{(1)}(ix), &{} \text{ if } a \ge 1, \\ 0, &{} \text{ if } a=0. \end{array}\right. } \end{aligned}
(4.69)

Also, using the recurrence (1.5) together with (4.65),

\begin{aligned}&a \sum _{s=1}^{a} \left( {\begin{array}{c}a-1\\ s-1\end{array}}\right) i^{s} \textrm{He}_{s}(ix) \textrm{He}_{a-s}(x) \nonumber \\ {}&\quad = a\sum _{s=1}^{a} \left( {\begin{array}{c}a-1\\ s-1\end{array}}\right) i^{s} \Big ( i x \textrm{He}_{s-1}(ix)-(s-1)\textrm{He}_{s-2}(ix) \Big ) \textrm{He}_{a-s}(x) \nonumber \\&\quad = {\left\{ \begin{array}{ll} -x, &{} \text{ if } a = 1, \\ 2, &{} \text{ if } a=2, \\ 0, &{} \text{ otherwise. } \end{array}\right. } \end{aligned}
(4.70)

For $$A_{1}$$, we use $$\left( {\begin{array}{c}a-1\\ s-2\end{array}}\right) =\left( {\begin{array}{c}a\\ s-1\end{array}}\right) -\left( {\begin{array}{c}a-1\\ s-1\end{array}}\right)$$ together with (4.69) and (4.70), and find

\begin{aligned} A_{1}= 3(-1)^{a} a \sum _{s=2}^a \left( {\begin{array}{c}a-1\\ s-2\end{array}}\right) i^s \textrm{He}_{s}(ix)\textrm{He}_{a-s}(x) = 3 a \,q_{0,a+1}(x) + {\left\{ \begin{array}{ll} -3x &{} if a=1, \\ -6 &{} if a=2, \\ 0 &{} otherwise . \end{array}\right. } \end{aligned}
(4.71)

Similarly, by (4.55) and (4.69),

\begin{aligned} A_{2}&= 3(-1)^a \sum _{s=2}^a \Big [3a(a-1)\left( {\begin{array}{c}a-2\\ s-3\end{array}}\right) + 2a\left( {\begin{array}{c}a-1\\ s-2\end{array}}\right) \Big ] i^{s-2} \textrm{He}_{s-2}(ix)\textrm{He}_{a-s}(x) \nonumber \\&= 3(1-3a) \,[a q_{0,a-1}(x)]+ {\left\{ \begin{array}{ll} -3 &{}if a=0, \\ 18 &{}if a=2, \\ 0 &{} otherwise. \end{array}\right. } \end{aligned}
(4.72)

Simplifying $$A_{3}$$ is a simpler task as it only relies on (4.55), namely

\begin{aligned} A_{3}&= 5(-1)^a a(a-1)(a-2)\sum _{s=0}^{a-4} \left( {\begin{array}{c}a-3\\ s\end{array}}\right) i^{s} \textrm{He}_{s}(ix)\textrm{He}_{a-4-s}(x) \nonumber \\&= {\left\{ \begin{array}{ll} 0, &{} \hspace{-1cm} \text{ if } a\in \{0,1,2\} \\ 5a(a-1)(a-2)q_{0,a-3}(x), &{} \hspace{-0.1cm} \text{ if } a \ge 3 \end{array}\right. } \nonumber \\&= 5[a(a-1)(a-2)q_{0,a-3}(x)] + {\left\{ \begin{array}{ll} 5(1-x^{2}), &{} \text{ if } a=0, \\ 5x, &{} \text{ if } a = 1, \\ -10, &{} \text{ if } a = 2, \\ 0, &{} \text{ otherwise. } \end{array}\right. } \end{aligned}
(4.73)

Therefore, by (4.68), (4.71), (4.72) and (4.73), we have

\begin{aligned}&\sum _{s=1}^a (-1)^{a-s} \left( {\begin{array}{c}a\\ s-1\end{array}}\right) {\widehat{p}}_{0,s}(x) \textrm{He}_{a-s}(x) \nonumber \\&= 3 a \,q_{0,a+1}(x)+3 (1-3a) [a\,q_{0,a-1}(x)]+5[a(a-1)(a-2)\,q_{0,a-3}(x)]-{\mathfrak {s}}_a(x). \end{aligned}
(4.74)

Now (4.48) directly follows from (4.62) and (4.74), which completes the proof. $$\square$$

In the large n asymptotics of $$S_{2}^{(2)}$$, which are obtained in Lemma 4.11 below, the function $$\mathcal {G}_{0}(y; u,a)$$, defined in (1.13), will appear inside a logarithm and in a denominator. The next lemma ensures that this function is positive for relevant values of the parameters.

### Lemma 4.9

The function $$\mathcal {G}_{0}(y; u,a)$$ given in (1.13) is positive for all $$y \in \mathbb {R}$$, $$u \in \mathbb {R}$$, and $$a \in \mathbb {N}$$.

### Proof

Let us write

\begin{aligned} \mathcal {G}_{0,0}(y,a)&:= (-1)^{a} \bigg [ p_{0,a}(-\sqrt{2}y)\frac{2-\textrm{erfc}(y)}{2} - q_{0,a}(-\sqrt{2} y) \frac{e^{-y^{2}}}{\sqrt{2\pi }} \bigg ],\\ \mathcal {G}_{0,1}(y,a)&:= p_{0,a}(-\sqrt{2}y) \textrm{erfc}(y) + 2\,q_{0,a}(-\sqrt{2} y) \frac{e^{-y^{2}}}{\sqrt{2\pi }} . \end{aligned}

Then by (1.13), we have

\begin{aligned} \mathcal {G}_{0}(y; u,a)=\mathcal {G}_{0,0}(y,a)+\frac{e^{u}}{2}\mathcal {G}_{0,1}(y,a). \end{aligned}
(4.75)

Using the definitions (1.8), (1.9), we obtain

\begin{aligned} p_{0,a+1}'(x){} & {} =(a+1) p_{0,a}(x), \qquad q_{0,a+1}'(x)=(a+1) q_{0,a}(x)+x q_{0,a+1}(x)\nonumber \\{} & {} \quad -p_{0,a+1}(x). \end{aligned}
(4.76)

The first identity in (4.76) is well-known and easy to prove. The second one follows from two known identities, namely the differentiation rule $$(\textrm{He}_{k}^{(\nu )})'(x) = (k+\nu )\textrm{He}_{k-1}^{(\nu )}(x)-\nu \textrm{He}_{k-1}^{(\nu +1)}(x)$$ and the recurrence relation $$\textrm{He}_{k+1}^{(\nu -1)}(x) = x \textrm{He}_{k}^{(\nu )}(x)-\nu \textrm{He}_{k-1}^{(\nu +1)}(x)$$, see e.g. [66, eqs (6.3) and (6.6)]. It follows from (4.76) that

\begin{aligned} \frac{d}{dy} \mathcal {G}_{0,0}(y,a+1){} & {} = \sqrt{2}(a+1) \mathcal {G}_{0,0}(y,a),\qquad \frac{d}{dy} \mathcal {G}_{0,1}(y,a+1) \nonumber \\ {}{} & {} = - \sqrt{2}(a+1) \mathcal {G}_{0,1}(y,a). \end{aligned}
(4.77)

Let us now show that

\begin{aligned}&\mathcal {G}_{0,1}(y,a) >0,{} & {} y\in \mathbb {R}, \end{aligned}
(4.78)
\begin{aligned}&\mathcal {G}_{0,1}(y,a) = \sqrt{ \frac{2}{\pi } } e^{-y^2} \frac{a!}{ (\sqrt{2}y)^{a+1} } \Big (1+O(y^{-2}) \Big ),{} & {} \text{ as } y \rightarrow +\infty . \end{aligned}
(4.79)

We shall prove (4.78) and (4.79) by induction on a. For $$a=0$$, we have $$\mathcal {G}_{0,1}(y,0)=\textrm{erfc}(y)$$ and (4.78), (4.79) follow. Assume now that (4.78) and (4.79) hold for a given a. By combining (4.77) with (4.79), we infer that (4.79) also holds with a replaced by $$a+1$$; in particular $$\mathcal {G}_{0,1}(y,a+1)>0$$ for all sufficiently large $$y>0$$. Also, by (4.77) and (4.78), the function $$\mathcal {G}_{0,1}(y,a+1)$$ is decreasing for $$y \in \mathbb {R}$$. Since $$\mathcal {G}_{0,1}(y,a+1)>0$$ for all sufficiently large $$y>0$$, we conclude that $$\mathcal {G}_{0,1}(y,a+1) >0$$ for all $$y \in \mathbb {R}$$.

The proof that $$\mathcal {G}_{0,0}(y,a) >0$$ for all $$y\in \mathbb {R}$$ is similar, so we omit it.

The statement of the lemma now readily follows from (4.75). $$\square$$

Recall from (3.1) and (4.24) that $$M_{j}=\sqrt{n}(\lambda _{j}-1)$$ with $$\lambda _{j}:= bnr^{2b}/(j+\alpha )$$. To obtain the large n asymptotics of $$\smash {S_{2}^{(2)}}$$, we need the following lemma from [21] to approximate sums of the form $$\sum _{j}h(M_{j})$$ (by contrast, Lemma 2.10 approximates sums of the form $$\sum _{j}f(j/n)$$).

### Lemma 4.10

([21, Lemma 3.11]) Let $$h \in C^{3}(\mathbb {R})$$. As $$n \rightarrow + \infty$$, we have

\begin{aligned} \sum _{j=g_{-}}^{g_{+}}h(M_{j})&= br^{2b} \int _{-M}^{M} h(t) dt \; \sqrt{n} - 2 b r^{2b} \int _{-M}^{M} th(t) dt \\ {}&\quad + \bigg ( \frac{1}{2}-\theta _{-}^{(n,M)} \bigg )h(M)+ \bigg ( \frac{1}{2}-\theta _{+}^{(n,M)} \bigg )h(-M) \\&\quad + \frac{1}{\sqrt{n}}\bigg [ 3br^{2b} \int _{-M}^{M}t^{2}h(t)dt + \bigg (\frac{1}{12}+\frac{\theta _{-}^{(n,M)}(\theta _{-}^{(n,M)}-1)}{2} \bigg )\frac{h'(M)}{br^{2b}}\\&\quad - \bigg ( \frac{1}{12}+\frac{\theta _{+}^{(n,M)}(\theta _{+}^{(n,M)}-1)}{2} \bigg )\frac{h'(-M)}{br^{2b}} \bigg ] \\&\quad + \mathcal{O}\Bigg ( \frac{1}{n^{3/2}} \sum _{j=g_{-}+1}^{g_{+}} \bigg ( (1+|M_{j}|^{3}) \tilde{{\mathfrak {m}}}_{j,n}(h) + (1+M_{j}^{2})\tilde{{\mathfrak {m}}}_{j,n}(h')\\&\quad + (1+|M_{j}|) \tilde{{\mathfrak {m}}}_{j,n}(h'') + \tilde{{\mathfrak {m}}}_{j,n}(h''') \bigg ) \Bigg ), \end{aligned}

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

### Lemma 4.11

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

\begin{aligned} S_{2}^{(2)}&= -abr^{2b}M\sqrt{n} \ln n + br^{2b} \sqrt{n} \int _{-M}^{M}h_{0}(t)dt + a\frac{\theta _{-}^{(n,M)}+\theta _{+}^{(n,M)}-1}{2}\ln n \\&\quad + br^{2b} \int _{-M}^{M} \big ( h_{1}(t)-2th_{0}(t) \big )dt + \bigg ( \frac{1}{2}-\theta _{-}^{(n,M)} \bigg )h_{0}(M) + \bigg ( \frac{1}{2}-\theta _{+}^{(n,M)} \bigg ) \\&\quad h_{0}(-M)+ \mathcal{O}\bigg ( \frac{M^{3}}{\sqrt{n}}\ln n \bigg ), \end{aligned}

where

\begin{aligned} \mathcal {H}_{0}(x)&:=\frac{r^{a-ab}}{(2b)^{a}} \, \mathcal {G}_0\Big ( \hspace{-0.1cm}-\frac{r^bx}{\sqrt{2}};u,a\Big ), \end{aligned}
(4.80)
\begin{aligned} h_{0}(x)&:= \ln \big (\mathcal {H}_{0}(x)\big ), \end{aligned}
(4.81)
\begin{aligned} h_1(x)&: = \frac{r^{a-(1+a)b}}{(2b)^{a+1}} \,\frac{1}{\mathcal {H}_0(x)}\, \mathcal {G}_1\Big ( \hspace{-0.1cm}-\frac{r^bx}{\sqrt{2}};u,a\Big ), \end{aligned}
(4.82)

and the functions $$\mathcal {G}_0$$ and $$\mathcal {G}_1$$ are given by (1.13) and (1.14).

### Proof

Recall the formula (4.26) for $$S_{2}^{(2)}$$, namely

\begin{aligned} S_{2}^{(2)}&= \sum _{j:\lambda _{j}\in I_{2}} \ln \bigg (\sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \nonumber \\&\quad \times \bigg [ 1+((-1)^{a}e^{u}-1)\bigg ( \textrm{erfc}\Big (\hspace{-0.1cm}-\hspace{-0.05cm}\eta _{j,k} \sqrt{\tfrac{a_{j,k}}{2}}\Big ) - R_{a_{j,k}}(\eta _{j,k}) \bigg ) \bigg ]\bigg ). \end{aligned}
(4.83)

Recall also that for all $$j \in \{j:\lambda _{j}\in I_{2}\}$$, we have $$1-\frac{M}{\sqrt{n}} \le \lambda _{j} = \frac{bnr^{2b}}{j+\alpha } \le 1+\frac{M}{\sqrt{n}}$$, and $$-M \le M_{j} \le M$$.

The expansion (4.31) implies that

\begin{aligned}&\sum _{k=0}^{a} {a \atopwithdelims ()k} \frac{(-r)^{a-k}}{n^{\frac{k}{2b}}} \frac{\Gamma (\frac{2j+2\alpha +k}{2b})}{\Gamma (\frac{2j+2\alpha }{2b})} \bigg [ 1+((-1)^{a}e^{u}-1)\bigg ( \frac{1}{2}\textrm{erfc}\Big (\hspace{-0.1cm}-\hspace{-0.05cm}\eta _{j,k} \sqrt{\tfrac{a_{j,k}}{2}}\Big ) \hspace{-0.1cm}\nonumber \\&\quad - R_{a_{j,k}}(\eta _{j,k}) \bigg ) \bigg ] \sim \sum _{\ell =0}^{+\infty } \frac{\mathcal {B}_{\ell }(M_{j};a)}{\sqrt{n^{\ell } }}, \end{aligned}
(4.84)

where

\begin{aligned} \mathcal {B}_{\ell }(x;a) := \sum _{k=0}^{a} {a \atopwithdelims ()k} (-1)^{a-k}r^{a}\mathcal {A}_{\ell }(x;k). \end{aligned}
(4.85)

It turns out that the first a terms in the expansion (4.84) are 0, i.e.

\begin{aligned} \mathcal {B}_{0}(\cdot ;a) \equiv \mathcal {B}_{1}(\cdot ;a) \equiv \cdots \equiv \mathcal {B}_{a-1}(\cdot ;a) \equiv 0. \end{aligned}
(4.86)

To prove this, we use [56, eq 26.8.6], i.e.

\begin{aligned} \sum _{k=0}^{a}{a \atopwithdelims ()k}(-1)^{a-k}k^{\ell } = a! \,S(\ell ,a), \qquad \ell ,a \in \mathbb {N}, \end{aligned}

where we recall that $$S(\ell ,a)$$ is the Stirling number of the second kind. In particular,

\begin{aligned} \sum _{k=0}^{a}{a \atopwithdelims ()k}(-1)^{a-k}k^{\ell } = {\left\{ \begin{array}{ll} 0 &{}\text{ if } \ell < a, \\ a! &{}\text{ if } \ell =a, \\ (a+1)!\,a/2 &{}\text{ if } \ell =a+1. \end{array}\right. } \end{aligned}
(4.87)

Since $$k\mapsto \mathcal {A}_{\ell }(x;k)$$ is a polynomial of degree $$\ell$$ by (4.33), the identities (4.86) directly follow from (4.85).

Let us now compute $$\mathcal {B}_a(x;a)$$ and $$\mathcal {B}_{a+1}(x;a)$$. By (4.33), (4.85) and (4.87), we have

\begin{aligned} \mathcal {B}_a(x;a)&=r^a \bigg ( (-1)^a+\frac{e^{u}-(-1)^a}{2}\textrm{erfc}\bigg ( -\frac{r^{b}x}{\sqrt{2}} \bigg ) \bigg ) (-1)^a a! \sum _{p=0}^{a}{\mathfrak {q}}_{a,a,p}^{(1)} x^{p} \nonumber \\&\qquad +\, r^a (e^{u}-(-1)^{a})\frac{e^{-\frac{r^{2b}x^{2}}{2}}}{\sqrt{2\pi } }(-1)^{a} a! \sum _{p=0}^{a-1} {\mathfrak {q}}_{a,a,p}^{(4)} x^{p} \end{aligned}
(4.88)

and

\begin{aligned} \mathcal {B}_{a+1}(x;a)&= r^a \bigg ( (-1)^a+\frac{e^{u}-(-1)^a}{2}\textrm{erfc}\bigg ( -\frac{r^{b}x}{\sqrt{2}} \bigg ) \bigg ) (-1)^a a! \nonumber \\&\quad \sum _{p=0}^{a+1} \Big ( {\mathfrak {q}}_{a+1,a,p}^{(1)}+\frac{a(a+1)}{2} {\mathfrak {q}}_{a+1,a+1,p}^{(1)} \Big ) x^{p} \nonumber \\&\quad + r^a (e^{u}-(-1)^{a})\frac{e^{-\frac{r^{2b}x^{2}}{2}}}{\sqrt{2\pi } }(-1)^{a} a! \bigg ( \sum _{p=0}^{a+2} {\mathfrak {q}}_{a+1,a,p}^{(4)} x^{p} + \frac{a(a+1)}{2}\nonumber \\&\quad \sum _{p=0}^{a}{\mathfrak {q}}_{a+1,a+1,p}^{(4)} x^{p} \bigg ) . \end{aligned}
(4.89)

More generally, for $$\ell \ge 1$$, we have

\begin{aligned} \mathcal {B}_{a+\ell }(x;a)&= r^{a}\bigg ( (-1)^{a}+\frac{e^{u}-(-1)^{a}}{2}\textrm{erfc}\bigg ( -\frac{r^{b}x}{\sqrt{2}} \bigg ) \bigg )(-1)^{a}a!\nonumber \\&\quad \sum _{m=\max (a,1)}^{a+\ell }\sum _{p=0}^{a+\ell } {\mathfrak {q}}_{a+\ell ,m,p}^{(1)}S(m,a)x^{p} \nonumber \\&\quad + r^{a}(e^{u}-(-1)^{a})\frac{e^{-\frac{r^{2b}x^{2}}{2}}}{\sqrt{2\pi } }(-1)^{a}a! \sum _{m=a}^{a+\ell }\sum _{p=0}^{ 3(a+\ell )-1-2m } {\mathfrak {q}}_{a+\ell ,m,p}^{(4)}S(m,a)x^{p}. \end{aligned}
(4.90)

Let us define

\begin{aligned} \mathcal {H}_{0}(x):=\mathcal {B}_{a}(x;a), \qquad h_{0}(x):=\ln (\mathcal {H}_{0}(x)), \qquad h_{1}(x):=\frac{\mathcal {B}_{a+1}(x;a)}{\mathcal {B}_{a}(x;a)}. \end{aligned}

By combining (4.88) and (4.89) with Lemmas 4.6 and 4.7, we obtain after simplifications that the functions $$\mathcal {H}_{0}(x)$$ and $$h_{1}(x)$$ can be written as in (4.80) and (4.82).

Since the case $$a=0$$ was already done in [19], from here on we focus on the more complicated case $$a\ge 1$$. We note the following important facts:

1. (1)

By Lemma 4.9, $$\mathcal {H}_{0}(x)>0$$ for all $$x \in \mathbb {R}$$.

2. (2)

By (4.80) and (4.81), $$h_{0}(x)$$ grows logarithmically at $$\pm \infty$$.

3. (3)

By (4.80) and (4.82), $$h_{1}(x)$$ grows linearly at $$\pm \infty$$.

4. (4)

By (4.90) and (4.88), $$\frac{\mathcal {B}_{a+\ell }(x;a)}{\mathcal {B}_{a}(x;a)}=\mathcal{O}(x^{\ell })$$ as $$x \rightarrow \pm \infty$$.

(Another reason as to why the case $$a=0$$ is significantly simpler than the case $$a \ge 1$$ stems from the fact that for $$a=0$$, the function $$h_{0}(x)$$ remains bounded, and furthermore $$h_{1}(x)$$ becomes exponentially small as $$x\rightarrow \pm \infty$$.)

After substituting (4.84) in (4.83), we obtain

\begin{aligned} S_{2}^{(2)} = -\frac{a}{2}\ln n \; \sum _{j=g_{-}}^{g_{+}}1 + \Sigma _{0}^{(2)} + \frac{1}{\sqrt{n}}\Sigma _{1}^{(2)} + \mathcal{O}\bigg ( \frac{1}{n}\sum _{j=g_{k,-}}^{g_{k,+}}M_{j}^{2} \bigg ), \end{aligned}
(4.91)

where

\begin{aligned}&\Sigma _{0}^{(2)} := \sum _{j=g_{-}}^{g_{+}} h_{0}(M_{j}), \qquad \Sigma _{1}^{(2)} := \sum _{j=g_{-}}^{g_{+}} h_{1}(M_{j}). \end{aligned}

The $$\mathcal{O}$$-term in (4.91) is $$\mathcal{O}(\frac{M^{3}}{\sqrt{n}})$$. Also, by Lemma 4.10,

\begin{aligned}&-\frac{a}{2}\ln n \; \sum _{j=g_{-}}^{g_{+}}1 = - a b r^{2b}M\sqrt{n} \ln n + a\frac{\theta _{-}^{(n,M)}+\theta _{+}^{(n,M)}-1}{2} \ln n + \mathcal{O}\bigg ( \frac{M^{3}}{\sqrt{n}}\ln n \bigg ), \\&\Sigma _{0}^{(2)} = br^{2b}\sqrt{n}\int _{-M}^{M}h_{0}(t) \, dt -2br^{2b}\int _{-M}^{M}t \, h_{0}(t) \, dt \\&\hspace{1cm} + \bigg ( \frac{1}{2}-\theta _{-}^{(n,M)} \bigg )h_{0}(M) + \bigg ( \frac{1}{2}-\theta _{+}^{(n,M)} \bigg )h_{0}(-M) + \mathcal{O}\bigg ( \frac{M^{3}\ln n}{\sqrt{n}} \bigg ), \\&\frac{1}{\sqrt{n}}\Sigma _{1}^{(2)} = br^{2b} \int _{-M}^{M} h_{1}(t)dt + \mathcal{O}\bigg ( \frac{M^{3}}{\sqrt{n}} \bigg ), \qquad \frac{1}{n}\sum _{j=g_{k,-}}^{g_{k,+}}M_{j}^{2} = \mathcal{O}\bigg ( \frac{M^{3}}{\sqrt{n}} \bigg ), \end{aligned}

as $$n \rightarrow + \infty$$. We now obtain the claim after a computation. $$\square$$

### 4.3 Asymptotics of $$S_{2}$$

We are now in a position to compute the large n asymptotics of $$S_{2}$$.

### Lemma 4.12

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

\begin{aligned}&S_{2} = {\widehat{C}}_{1}^{(\epsilon )} n + {\widehat{C}}_{2} \sqrt{n} + {\widehat{C}}_{3}^{(n,\epsilon )} + \mathcal{O}\bigg ( \frac{M^{3}}{\sqrt{n}}\ln n + \frac{\sqrt{n}}{M^{5}} \bigg ), \end{aligned}

where

\begin{aligned} {\widehat{C}}_{1}^{(\epsilon )}&= \int _{\frac{br^{2b}}{1+\epsilon }}^{br^{2b}} \bigg ( u + a \ln \bigg ( r-\bigg ( \frac{x}{b} \bigg )^{\frac{1}{2b}} \bigg ) \bigg )dx + \int _{br^{2b}}^{\frac{br^{2b}}{1-\epsilon }} a \ln \bigg ( \bigg ( \frac{x}{b} \bigg )^{\frac{1}{2b}} - r \bigg )dx, \end{aligned}
(4.92)
\begin{aligned} {\widehat{C}}_{2}&= b r^{2b} \int _{-\infty }^{+\infty } {\hat{h}}_{0}(t) \, dt, \nonumber \\ {\widehat{C}}_{3}^{(n,\epsilon )}&= -\alpha u + \frac{1+2\alpha - 2\theta _{-}^{(n,\epsilon )}}{2}\bigg ( u + a \ln \Big ( r-r(1+\epsilon )^{-\frac{1}{2b}} \Big ) \bigg ) \nonumber \\&\quad + \frac{1-2\alpha -2\theta _{+}^{(n,\epsilon )}}{2}a \ln \Big ( r(1-\epsilon )^{-\frac{1}{2b}}-r \Big ) \nonumber \\&\quad + \frac{a}{4}\bigg \{ \frac{1-a}{1-(1+\epsilon )^{-\frac{1}{2b}}} + \frac{1-a}{(1-\epsilon )^{-\frac{1}{2b}}-1} + (a-2b+4\alpha ) \ln \bigg ( \frac{(1-\epsilon )^{-\frac{1}{2b}}-1}{1-(1+\epsilon )^{-\frac{1}{2b}}} \bigg ) \bigg \} \nonumber \\&\quad + br^{2b} \int _{-\infty }^{+\infty } \bigg \{ h_{1}(t)-2\,th_{0}(t) + \frac{a}{4b} \bigg ( 1+2b+8b\ln \bigg ( \frac{r|t|}{2b} \bigg ) \bigg ) t + 2u t \chi _{(0,\infty )} (t)\nonumber \\ {}&\quad - \frac{(2ab-a^{2})t}{4br^{2b}(1+t^{2})} \bigg \} \,dt . \end{aligned}
(4.93)

Here

\begin{aligned} {\hat{h}}_{0}(x) = h_{0}(x) - a \ln \bigg ( \frac{r|x|}{2b} \bigg ) -u \,\chi _{(0,\infty )}(x) \end{aligned}

and $$h_{0}, h_{1}$$ are given in the statement of Lemma 4.11.

### Proof

Combining Lemmas 4.1, 4.4 and 4.11 yields

\begin{aligned}&S_{2} = {\widehat{C}}_{1}^{(\epsilon )} + {\widehat{C}}_{2}'^{(M)} \sqrt{n} \ln n + {\widehat{C}}_{2}^{(M)} \sqrt{n} + {\widehat{C}}_{3}'^{(n,M)} \ln n + {\widehat{C}}_{3}^{(n,\epsilon ,M)}\nonumber \\ {}&\quad + \mathcal{O}\bigg ( \frac{M^{3}}{\sqrt{n}}\ln n + \frac{\sqrt{n}}{M^{5}} \bigg ), \end{aligned}
(4.94)

as $$n \rightarrow +\infty$$, where $${\widehat{C}}_{1}^{(\epsilon )}$$ is given by (4.92). After short (but remarkable) simplifications, we obtain $${\widehat{C}}'^{(M)}_{2} = {\widehat{C}}_{3}'^{(\epsilon ,M)} = 0$$. The quantity $${\widehat{C}}_{2}^{(M)}$$ is given by

\begin{aligned} {\widehat{C}}_{2}^{(M)}&= br^{2b}M\bigg ( 2a \ln \bigg ( \frac{2b}{r M} \bigg ) + 2a -u \bigg ) + \frac{a(a-1)b}{M} - \frac{ab (a-1)(2a-3) }{6r^{2b}M^{3}} \\&\quad + br^{2b}\int _{-M}^{M}h_{0}(t)dt. \end{aligned}

Using the definition (4.81) of $$h_{0}$$, we verify that

\begin{aligned} h_{0}(t)&= a \ln \Big ( \frac{r|t|}{2b} \Big )+ u\chi _{(0,\infty )}(t) + \frac{a(a-1)}{2r^{2b}t^{2}} - \frac{a(a-1)(2a-3)}{4r^{4b}t^{4}} + \mathcal{O}(t^{-6}),\nonumber \\ {}&\quad \text{ as } t \rightarrow \pm \infty , \end{aligned}
(4.95)

from which we conclude

\begin{aligned} {\widehat{C}}_{2}^{(M)} \sqrt{n} = {\widehat{C}}_{2} \sqrt{n} + \mathcal{O}\bigg ( \frac{\sqrt{n}}{M^{5}} \bigg ), \qquad \text{ as } n \rightarrow + \infty , \end{aligned}

where $${\widehat{C}}_{2}$$ is given by (4.93). The quantity $${\widehat{C}}_{3}^{(n,\epsilon ,M)}$$ in (4.94) is given by

\begin{aligned} {\widehat{C}}_{3}^{(n,\epsilon ,M)} = \bigg ( \frac{1}{2}-\theta _{-}^{(n,M)} \bigg ) {\widehat{C}}_{3,1}^{(M)} + \bigg ( \frac{1}{2}-\theta _{+}^{(n,M)} \bigg ) {\widehat{C}}_{3,2}^{(M)} + {\widehat{C}}_{3,3}^{(n,\epsilon )} + {\widehat{C}}_{3,4}^{(M)}, \end{aligned}

where

\begin{aligned} {\widehat{C}}_{3,1}^{(M)}&:= h_{0}(M) - u - a \ln \bigg ( \frac{rM}{2b} \bigg ), \qquad {\widehat{C}}_{3,2}^{(M)} := h_{0}(-M) - a \ln \bigg ( \frac{rM}{2b} \bigg ), \\ {\widehat{C}}_{3,3}^{(n,\epsilon )}&:= \frac{1+2\alpha - 2\theta _{-}^{(n,\epsilon )}}{2}\bigg ( u + a \ln \Big ( r-r(1+\epsilon )^{-\frac{1}{2b}} \Big ) \bigg ) \\&\quad + \frac{1-2\alpha -2\theta _{+}^{(n,\epsilon )}}{2}a \ln \Big ( r(1-\epsilon )^{-\frac{1}{2b}}-r \Big ) \\&\quad -\alpha u + \frac{a}{4}\bigg \{ \frac{1-a}{1-(1+\epsilon )^{-\frac{1}{2b}}} + \frac{1-a}{(1-\epsilon )^{-\frac{1}{2b}}-1} \\&\quad + (a-2b+4\alpha ) \ln \bigg ( \frac{(1-\epsilon )^{-\frac{1}{2b}}-1}{1-(1+\epsilon )^{-\frac{1}{2b}}} \bigg ) \bigg \}, \\ {\widehat{C}}_{3,4}^{(M)}&= br^{2b} \bigg \{ u M^{2} + \int _{-M}^{M} (h_{1}(t)-2th_{0}(t))dt \bigg \}. \end{aligned}

It readily follows from (4.95) that

\begin{aligned} {\widehat{C}}_{3,1}^{(M)} = \mathcal{O}\bigg ( \frac{1}{M^{2}} \bigg ), \qquad {\widehat{C}}_{3,2}^{(M)} = \mathcal{O}\bigg ( \frac{1}{M^{2}} \bigg ), \qquad \text{ as } n \rightarrow + \infty . \end{aligned}

We also verify from (4.81), (4.82), (4.95), and

\begin{aligned} \frac{p_{1,a}(x)}{p_{0,a}(x)}=-\frac{a}{2}(1+2b)x-\frac{a}{2}\frac{a+2b(1-2a)}{x}+O(x^{-3}), \qquad \text{ as } x \rightarrow \pm \infty \end{aligned}

that

\begin{aligned} h_{1}(t)-2th_{0}(t)&= - \frac{a}{4b} \bigg ( 1+2b+8b\ln \bigg ( \frac{r|t|}{2b} \bigg ) \bigg ) t -2ut \chi _{(0,\infty )}(t) + \frac{2ab-a^{2}}{4br^{2b}t} \\&\quad + \mathcal{O}(t^{-3}) \end{aligned}

as $$t \rightarrow \pm \infty$$. We conclude that

\begin{aligned} {\widehat{C}}_{3,4}^{(M)}&{=} br^{2b} \int _{-\infty }^{{+}\infty } \bigg \{ h_{1}(t)-2\,th_{0}(t) {+} \frac{a}{4b} \bigg ( 1+2b+8b\ln \bigg ( \frac{r|t|}{2b} \bigg ) \bigg ) t {+} 2u t \chi _{(0,\infty )} (t) \\&\quad - \frac{(2ab-a^{2})t}{4br^{2b}(1+t^{2})} \bigg \} \,dt + \mathcal{O}(M^{-2}), \qquad \text{ as } n \rightarrow + \infty , \end{aligned}

and the claim follows. $$\square$$

## 5 Proof of Theorem 1.1

By combining (2.4) with Lemmas 3.1, 3.2, 3.3 and 4.12, we obtain

\begin{aligned} \ln \mathcal {E}_{n} = S_{0}+S_{1}+S_{2}+S_{3} = C_{1} n + C_{2} \sqrt{n} + C_{3} + \mathcal{O}\bigg ( \frac{M^{3}}{\sqrt{n}}\ln n + \frac{\sqrt{n}}{M^{5}} + n^{-\frac{1}{2b}} \bigg ), \end{aligned}

as $$n \rightarrow +\infty$$. Here we obtain the constants $$C_{1}$$, $$C_{2}$$ and $$C_{3}$$ of (1.16) after a long computation using (1.13)–(1.14), a change of variables, and simplifying. Since $$M = n^{\frac{1}{8}}(\ln n)^{-\frac{1}{8}}$$, the error term is $$\mathcal{O}\Big ( n^{-\frac{1}{2b}} + (\ln n)^{\frac{5}{8}}n^{-\frac{1}{8}} \Big )$$, which finishes the proof of Theorem 1.1.