1 Introduction and Statement of Results

Let \(\mathbf{X}\in \mathbb {R}^{n\times n},n\in \mathbb {Z}_{\ge 2}\) be a matrix whose entries are independent, identically distributed standard normal random variables with mean 0 and variance 1. In other words, \(\mathbf{X}\) is a matrix drawn from the real Ginibre ensemble (GinOE) [42]. This ensemble of random matrices appeared first in the 1965 paper [42] by Ginibre, the same paper which brought forth a threefold family of Gaussian random matrices (complex, real or real quaternion entries) and thereby initiated the study of eigenvalue statistics in the complex plane. At first, [42] served as a mathematical extension of Hermitian random matrix theory only but it proved to be valuable for the modeling of a wide range of physical phenomena later on: For instance, Ginibre matrices appear in the study of the fractional quantum Hall effect [29], in the stability analysis of complex biological systems [53] and neural networks [58], in quantum chaotic systems [32] and in financial market models [50]; see [48] for further applications and references. From the viewpoint of applicability and computability of the statistical properties of a given Ginibre random matrix, one commonly focuses on its extreme eigenvalues as they model extreme events in the corresponding physical system. Consequently, extreme eigenvalues correspond to events that are in general quite rare, but when they occur, then with serious consequences which makes their analysis valuable. It is precisely within the context of extreme values that the GinOE attains a special mathematical role due to its real peculiarities: the expected number of real eigenvalues of any \(\mathbf{X}\in \text {GinOE}\) is equivalent to \(\sqrt{2n/\pi }\) as \(n\rightarrow \infty \) [30] and the likelihood that all its eigenvalues are real is exactly \(2^{-n(n-1)/4}\) [31]. Both features are nowadays cornerstones of the edge behavior of real eigenvalues in the real Ginibre ensemble. In this paper, we contribute to the same field by providing original results for the limiting distribution function of the largest real eigenvalue in a thinned real Ginibre ensemble.

In order to be concrete, we first recall, cf. [10, 41, 59], that the eigenvalues \(\{z_j(\mathbf{X})\}_{j=1}^n\) of any \(\mathbf{X}\in \text {GinOE}\) form a Pfaffian point process, a fact which allows one to compute gap probabilities in the GinOE as Fredholm determinants. Of particular interest for us is the following result about the absence of real eigenvalues in \((t,\infty )\subset \mathbb {R}\), equivalently about the distribution function of the largest real eigenvalue in the finite n GinOE.

Proposition 1.1

([56, Proposition 2.2]) For every \(n\in \mathbb {Z}_{\ge 2}\),

$$\begin{aligned} \mathbb {P}\left( \max _{\begin{array}{c} j=1,\ldots ,n\\ z_j\in \mathbb {R} \end{array}}z_j(\mathbf{X})\le t\right) =\sqrt{\det _2\big (1-\chi _t\mathbf{K}_n\chi _t{\upharpoonright }_{L^2(\mathbb {R})\oplus L^2(\mathbb {R})}\big )},\ \ \ \ t\in \mathbb {R}, \end{aligned}$$
(1.1)

where \(\chi _t\) is the operator of multiplication by the characteristic function \(\chi _{[t,\infty )}\) of the interval \([t,\infty )\subset \mathbb {R}\) and \(\mathbf{K}_n\) the following Hilbert–Schmidt integral operator on \(L^2(\mathbb {R})\oplus L^2(\mathbb {R})\),

$$\begin{aligned} \mathbf{K}_n=\begin{bmatrix} \rho ^{-1}S_n\rho &{} \rho ^{-1}(DS_n^{*})\rho ^{-1}\\ -\rho (IS_n)\rho +\rho \epsilon \rho &{} \rho S_n^{*}\rho ^{-1}\end{bmatrix}. \end{aligned}$$
(1.2)

Here, \(\rho \) multiplies by any differentiable, square-integrable weight function \(\rho (x)>0\) on \(\mathbb {R}\) such that \(\rho ^{-1}(x)\equiv 1/\rho (x)\) is polynomially bounded. Moreover \(S_n\) and \(\epsilon \) are the integral operators on \(L^2(\mathbb {R})\) with kernels

$$\begin{aligned}&S_n(x,y):=\frac{1}{\sqrt{2\pi }}\mathrm {e}^{-\frac{1}{2}(x^2+y^2)}\mathfrak {e}_{n-2}(xy)+\frac{x^{n-1}\mathrm {e}^{-\frac{1}{2}x^2}}{\sqrt{2\pi }(n-2)!}\int _0^yu^{n-2}\mathrm {e}^{-\frac{1}{2}u^2}{\mathrm d}u,\\&\epsilon (x,y):=\frac{1}{2}\text {sgn}(y-x), \end{aligned}$$

where \(\mathfrak {e}_n(z):=\sum _{k=0}^n\frac{1}{k!}z^k\) is the exponential partial sum, \(S_n^{*}\) the real adjoint of \(S_n\), D acts by differentiation on the independent variable and \(IS_n\) has kernel

$$\begin{aligned} \big (IS_n\big )(x,y):=\big (\epsilon S_n\big )(x,y),\ \ \ \ \ n\in \mathbb {Z}_{\ge 2}\ \text {even}, \end{aligned}$$

and

$$\begin{aligned} \big (IS_n\big )(x,y):=\big (\epsilon S_n\big )(x,y)+\frac{1}{2^{n/2}\Gamma (n/2)}\int _0^yu^{n-1}\mathrm {e}^{-\frac{1}{2}u^2}{\mathrm d}u,\ \ \ \ n\in \mathbb {Z}_{\ge 3}\ \text {odd}. \end{aligned}$$

Remark 1.2

The ordinary Fredholm determinant of \(\mathbf{K}_n\) is ill-defined since not all its entries vanish at \(\pm \infty \) and since \(\epsilon \) is not trace-class on \(L^2(\mathbb {R})\). This is a standard issue in random matrix theory, compare [62, Section VIII], [63, page 2199] or [27, page 79-84], and it is commonly bypassed either through the use of regularized determinants or weighted Hilbert spaces. In (1.1), we use the following regularized 2-determinant for block operators \(\text{\L} =\bigl [{\begin{matrix} L_{11} &{} L_{12}\\ L_{21} &{} L_{22}\end{matrix}}\bigr ]\) with trace class diagonal \(L_{11},L_{22}\) and Hilbert–Schmidt off-diagonal \(L_{12},L_{21}\), cf. [27, page 82],

$$\begin{aligned} \det _2(1+\text{\L} ):=\det \big ((1+\text{\L} )\mathrm {e}^{-\text{\L} }\big )\mathrm {e}^{\mathop {\mathrm {tr}}\limits (L_{11}+L_{22})}, \end{aligned}$$
(1.3)

where \(\det \) is the ordinary Fredholm determinant (which is well defined as \((1+\text{\L} )\mathrm {e}^{-\text{\L} }=1+\text {trace class}\) on \(L^2(\mathbb {R})\oplus L^2(\mathbb {R})\) for the given \(\text{\L} \), cf. [57, (3.5)]), the block operators act on \(L^2(\mathbb {R})\oplus L^2(\mathbb {R})\) and the trace in the exponent is taken in \(L^2(\mathbb {R})\). Note that (1.3) is slightly different from the Hilbert–Carleman determinant [57, Chapter 9] in that for trace class \(\text{\L} \) we have \(\det _2(1+\text{\L} )=\det (1+\text{\L} )\) and for any two of the above block operators

$$\begin{aligned} \det _2(1+\text{\L} +\mathbf{M}+\text{\L} \mathbf{M})=\det _2(1+\text{\L} )\det _2(1+\mathbf{M}). \end{aligned}$$
(1.4)

Moreover, as soon as \(\text{\L} \mathbf{M}\) and \(\mathbf{M}\text{\L} \) fit into the aforementioned class of block operators,

$$\begin{aligned} \det _2(1+\text{\L} \mathbf{M})=\det _2(1+\mathbf{M}\text{\L} ), \end{aligned}$$
(1.5)

and \(\det _2(1+\text{\L} )\ne 0\) if and only if \(1+\text{\L} \) is invertible.

The finite n GinOE result (1.1) can be used to derive a limit theorem for the largest real eigenvalue of a real Ginibre matrix which in turn quantifies the well-known saturn effect. Indeed, in order to state the corresponding limit theorem for the largest real eigenvalue we first consider the following Riemann–Hilbert problem (RHP).

Riemann-Hilbert Problem 1.3

([2, RHP 1.5]) Given \(x,\gamma \in \mathbb {R}\times [0,1]\), determine \(\mathbf{Y}(z)=\mathbf{Y}(z;x,\gamma )\in \mathbb {C}^{2\times 2}\) such that

  1. (1)

    \(\mathbf{Y}(z)\) is analytic for \(z\in \mathbb {C}\setminus \mathbb {R}\) and continuous on the closed upper and lower half-planes.

  2. (2)

    The boundary values \(\mathbf{Y}_{\pm }(z):=\lim _{\epsilon \downarrow 0}\mathbf{Y}(z\pm \mathrm {i}\epsilon ),z\in \mathbb {R}\) satisfy

    $$\begin{aligned}&\mathbf{Y}_+(z)=\mathbf{Y}_-(z)\begin{bmatrix}1-|r(z)|^2 &{} -\bar{r}(z)\mathrm {e}^{-2\mathrm {i}xz}\\ r(z)\mathrm {e}^{2\mathrm {i}xz} &{} 1\end{bmatrix},\\&z\in \mathbb {R}; \ \ r(z)=r(z;\gamma ):=-\mathrm {i}\sqrt{\gamma }\mathrm {e}^{-\frac{1}{4}z^2}. \end{aligned}$$
  3. (3)

    As \(z\rightarrow \infty \),

    $$\begin{aligned} \mathbf{Y}(z)=\mathbb {I}+\mathbf{Y}_1(x,\gamma )z^{-1}+\mathcal {O}\big (z^{-2}\big );\ \ \ \ \ \mathbf{Y}_1(x,\gamma )=\big [Y_1^{jk}(x,\gamma )\big ]_{j,k=1}^2. \end{aligned}$$
    (1.6)

This problem is uniquely solvable for all \((x,\gamma )\in \mathbb {R}\times [0,1]\), cf. [2, Theorem 3.9], and its solution enables us to state the limit theorem for the largest real eigenvalue as follows. Eigenvalues off the real axis are much simpler to deal with, see [56, Theorem 1.2].

Theorem 1.4

([56, Theorem 1.3], [55, Theorem 1.1], [2, Theorem 1.6]) Let \(\mathbf{X}\in \mathbb {R}^{n\times n}\) be a matrix drawn from the GinOE with eigenvalues \(\{z_j(\mathbf{X})\}_{j=1}^n\subset \mathbb {C}\). Then for every \(t\in \mathbb {R}\),

$$\begin{aligned} \lim _{n\rightarrow \infty }\mathbb {P}&\left( \max _{\begin{array}{c} j=1,\ldots ,n\\ z_j\in \mathbb {R} \end{array}}z_j(\mathbf {X})\le \sqrt{n}+t\right) =\sqrt{\det (1-\chi _tT\chi _t{\upharpoonright }_{L^2(\mathbb {R})})\Gamma _t}\\\nonumber =&\exp \left[ -\frac{1}{8}\int _t^{\infty }(x-t)\Big |y\Big (\frac{x}{2};1\Big )\Big |^2\right. \left. {\mathrm d}x +\frac{\mathrm {i}}{4}\int _t^{\infty }y\Big (\frac{x}{2};1\Big ){\mathrm d}x\right] ,\end{aligned}$$
(1.7)

where \(T:L^2(\mathbb {R})\rightarrow L^2(\mathbb {R})\) is trace class with kernel

$$\begin{aligned} T(x,y):=\frac{1}{\pi }\int _0^{\infty }\mathrm {e}^{-(x+u)^2}\mathrm {e}^{-(y+u)^2}{\mathrm d}u, \end{aligned}$$
(1.8)

and

$$\begin{aligned} \Gamma _t&:=1-\int _t^{\infty }G(x)\big ((1-T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x; \nonumber \\&g(x):=\frac{1}{\sqrt{\pi }}\mathrm {e}^{-x^2},\ \ \ \ G(x):=\int _{-\infty }^xg(y){\mathrm d}y. \end{aligned}$$
(1.9)

The function \(y=y(x;1):\mathbb {R}\times [0,1]\rightarrow \mathrm {i}\mathbb {R}\) equals \(y(x;1):=2\mathrm {i}Y_1^{12}(x,1)\), which is expressed in terms of the matrix coefficient \(\mathbf{Y}_1(x,1)\) that appeared in (1.6).

Remark 1.5

The first equality in (1.7) is due to Rider and Sinclair [56, Theorem 1.2] with a subsequent algebraic correction of the factor \(\Gamma _t\) by Poplavskyi, Tribe and Zaboronski [55, Theorem 1.1]. The second equality was derived by the authors [2, Theorem 1.6] and should be viewed as the GinOE analogue of the famous Tracy–Widom law for the largest eigenvalue in the Gaussian orthogonal ensemble (GOE), compare [62, (53)]. Indeed, as far as the largest real eigenvalue is concerned, the overall difference between GinOE and GOE stems from the appearance of the function y, i.e., the solution of a distinguished inverse scattering problem for the Zakharov–Shabat system [2, Section 1.2], rather than the more familiar Painlevé-II Hastings-McLeod transcendent.

Remark 1.6

We emphasize that the limit law (1.7) is not a feature of the GinOE alone. In fact, Cipolloni, Erdős and Schröder recently proved in [21, Theorem 2.3] that the edge eigenvalue statistics of a large class of real non-Hermitian random matrices with i.i.d. centered entries match those of the GinOE. Thus, in complete analogy with the Tracy–Widom law for real Wigner matrices [60], the law (1.7) is a universal limit law. The same holds true for the upcoming limit law (1.14) for thinned real non-Hermitian random matrices at their spectral edge.

1.1 Fredholm Determinant Formula

In this paper, we are concerned with the limiting (\(n\rightarrow \infty \)) distribution of the largest real eigenvalue in the following thinned real GinOE process: consider the Pfaffian point process formed by the \(m_n\le n\) real eigenvalues of some \(\mathbf{X}\in \text {GinOE}\). Fix \(\gamma \in [0,1]\) and now discard each eigenvalue \(\mathbb {R}\ni z_j(\mathbf{X}),j=1,\ldots ,m_n\) independently with likelihood \(1-\gamma \). The resulting particle system

$$\begin{aligned} \big \{z_j^{\gamma }(\mathbf{X})\big \}_{j=1}^{m_{\gamma ,n}}\ \ \ \text {with}\ \ m_{\gamma ,n}\le m_n\le n, \end{aligned}$$

forms also a random point process, see [44, Chapter 6.2.1], and most importantly for us, this process is Pfaffian as stated in our first result below.

Lemma 1.7

The above-defined thinned real GinOE process is a Pfaffian random point process with

$$\begin{aligned} \mathbb {P}\left( \max _{j=1,\ldots ,m_{\gamma ,n}}z_j^{\gamma }(\mathbf{X})\le t\right) =\sqrt{\det _2\big (1-\gamma \chi _t\mathbf{K}_n\chi _t{\upharpoonright }_{L^2(\mathbb {R})\oplus L^2(\mathbb {R})}\big )},\ \ \ \ (t,\gamma )\in \mathbb {R}\times [0,1], \end{aligned}$$
(1.10)

where the operator \(\mathbf{K}_n\) appeared in (1.2).

Identities similar to (1.10) have been derived in [13, Proposition 1.1] for the limiting GOE and the limiting Gaussian symplectic ensemble (GSE) based on Painlevé representations for the underlying eigenvalue generating functions, cf. [25, Theorem 2.1]. Our proof of Lemma 1.7 will rely on the observation that thinned Pfaffian point processes are Pfaffian with an appropriately \(\gamma \)-modified kernel, see Sect. 2, which is similar to the proof for determinantal point processes given in [51, Appendix A]. The fact that a thinned process built from a determinantal point process is also determinantal was first observed in [8]. In fact, it is the last paper [8] which re-ignited the interest in incomplete point processes in random matrix theory, simply because the incomplete or thinned matrix models allow one to transition between qualitatively different extreme behaviors. Such transitions have been studied foremost in Hermitian ensembles; compare Remarks 1.12 and 1.15 for some references; here, we are interested in the simplest thinned non-Hermitian matrix model with real entries.

Once (1.10) is established, we will then use this finite n result to derive the following limit theorem for the thinned real GinOE process, our second result. Set

$$\begin{aligned} \bar{\gamma }:=\gamma (2-\gamma ) \end{aligned}$$
(1.11)

and note that \(\bar{\gamma }\in [0,1]\) for \(\gamma \in [0,1]\). The limit is a convex combination of two simple Fredholm determinants.

Theorem 1.8

For any \((t,\gamma )\in \mathbb {R}\times [0,1]\), the limit

$$\begin{aligned} P(t;\gamma ):=\lim _{n\rightarrow \infty }\mathbb {P}\left( \max _{j=1,\ldots ,m_{\gamma ,n}}z_j^{\gamma }(\mathbf{X})\le \sqrt{n}+t\right) ,\ \ \gamma \in [0,1] \end{aligned}$$
(1.12)

exists and equals

$$\begin{aligned} P(t;\gamma )&=\sqrt{\frac{1-\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\det \big (1+\sqrt{\bar{\gamma }}\chi _tS\chi _t\upharpoonright _{L^2(\mathbb {R})}\big )\nonumber \\&\quad +\sqrt{\frac{1+\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\det \big (1-\sqrt{\bar{\gamma }}\chi _tS\chi _t\upharpoonright _{L^2(\mathbb {R})}\big ), \end{aligned}$$
(1.13)

with \(\bar{\gamma }\) defined in (1.11). Here, \(S:L^2(\mathbb {R})\rightarrow L^2(\mathbb {R})\) is the trace class integral operator with kernel

$$\begin{aligned} S(x,y)=\frac{1}{2\sqrt{\pi }}\mathrm {e}^{-\frac{1}{4}(x+y)^2}. \end{aligned}$$

The special value \(\gamma =1\) reduces (1.13) to

$$\begin{aligned} P(t;1)=\det \big (1-\chi _tS\chi _t{\upharpoonright }_{L^2(\mathbb {R})}\big ), \end{aligned}$$

which was first proven by the authors in [2, Theorem 1.11]. Note that the formula for P(t; 1) is the analogue of the Ferrari–Spohn formula [35] in the GOE, generalized to the thinned GOE by Forrester in [39, Corollary 1]. Comparing (1.13) to the last reference (modulo the typo correction \(\xi \mapsto \bar{\xi }\) in the determinants in the first line of [39, (1.22)] and after completing squares), we spot a striking resemblance between the thinned GOE and the thinned GinOE: up to the kernel replacement

$$\begin{aligned} S(x,y)\mapsto \text {Ai}(x+y), \end{aligned}$$

with the Airy function \(w=\text {Ai}(z)\), see [54, 9.2.2], the formulæare exactly the same.

1.2 Integrability of the Thinned Real GinOE Process

In our third result, we express the limiting distribution function \(P(t;\gamma )\) in (1.12) in terms of the solution of RHP 1.3 and thus in terms of the solution to an inverse scattering problem for the Zakharov–Shabat system. Here are the details:

Theorem 1.9

For any \((t,\gamma )\in \mathbb {R}\times [0,1]\),

$$\begin{aligned} P(t;\gamma )&=\exp \left[ -\frac{1}{8}\int _t^{\infty }(x-t)\Big |y\Big (\frac{x}{2};\bar{\gamma }\Big )\Big |^2{\mathrm d}x\right] \nonumber \\&\quad \left( \sqrt{\frac{1-\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\mathrm {e}^{\frac{1}{2}\mu (t;\bar{\gamma })}+\sqrt{\frac{1+\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\mathrm {e}^{-\frac{1}{2}\mu (t;\bar{\gamma })}\right) \end{aligned}$$
(1.14)

where the function \(y=y(x;\gamma ):\mathbb {R}\times [0,1]\rightarrow \mathrm {i}\mathbb {R}\) is given by \(y(x;\gamma ):=2\mathrm {i}Y_1^{12}(x,\gamma )\) in terms of (1.6) and

$$\begin{aligned} \mu (t;\gamma ):=-\frac{\mathrm {i}}{2}\int _t^{\infty }y\Big (\frac{x}{2};\gamma \Big ){\mathrm d}x. \end{aligned}$$
(1.15)

Remark 1.10

Note that for every \((t,\gamma )\in \mathbb {R}\times [0,1]\),

$$\begin{aligned} \sqrt{\frac{1-\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\mathrm {e}^{\frac{1}{2}\mu (t;\bar{\gamma })}+\sqrt{\frac{1+\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\mathrm {e}^{-\frac{1}{2}\mu (t;\bar{\gamma })}=\sqrt{\frac{\gamma -1-\cosh \mu (t;\bar{\gamma })+\sqrt{\bar{\gamma }}\sinh \mu (t;\bar{\gamma })}{\gamma -2}}. \end{aligned}$$
(1.16)

We emphasize that the structure in the right-hand side of (1.14), (1.16) is completely similar to the one in the limiting distribution function for the largest eigenvalue in the thinned GOE ensemble, cf. [13, (1.6)]. It is only the appearance of the solution to the Zakharov–Shabat inverse scattering problem which sets the thinned GinOE apart from the thinned GOE—at least as far as the largest real eigenvalue is concerned; compare Remark 1.5 for the special case \(\gamma =1\). We further emphasize this point with our fourth result, a simple corollary to Theorem 1.13: let \(E(m,(t,\infty ))\) denote the limiting (as \(n\rightarrow \infty \)) probability that there are \(m\in \mathbb {Z}_{\ge 0}\) edge scaled real eigenvalues \(\mu _j(\mathbf{X}):=z_j(\mathbf{X})-\sqrt{n}\in \mathbb {R}\) of a matrix \(\mathbf{X}\in \text {GinOE}\) in the interval \((t,\infty )\subset \mathbb {R}\). Now define the associated generating function

$$\begin{aligned} E\big ((t,\infty );\lambda \big ):=\sum _{m=0}^{\infty }E\big (m,(t,\infty )\big )(1-\lambda )^m, \end{aligned}$$
(1.17)

which, as a consequence of Theorem 1.8 can also be evaluated in terms of the solution of RHP 1.3:

Corollary 1.11

For every \((t,\lambda )\in \mathbb {R}\times [0,1]\),

$$\begin{aligned} E\big ((t,\infty );\lambda \big )&=\exp \left[ -\frac{1}{8}\int _t^{\infty }(x-t)\left| y\Big (\frac{x}{2};\bar{\lambda }\Big )\right| ^2{\mathrm d}x\right] \nonumber \\&\quad \sqrt{\frac{\lambda -1-\cosh \mu (t;\bar{\lambda })+\sqrt{\bar{\lambda }}\sinh \mu (t;\bar{\lambda })}{\lambda -2}}, \end{aligned}$$
(1.18)

with \(\bar{\lambda }:=2\lambda -\lambda ^2\), the above function \(y(x;\lambda )=2\mathrm {i}Y_1^{12}(x,\lambda )\) and the antiderivative (1.15).

Formula (1.18) is a simple consequence of the inclusion–exclusion principle; see Sect. 6. The generating function is of interest from the random matrix theory viewpoint as it allows one to compute the limiting distribution function \(F_m(t)\) of the mth largest edge scaled real eigenvalue (\(m=1\) is the largest) in the GinOE in recursive form,

$$\begin{aligned} F_{m+1}(t)-F_m(t)=\frac{(-1)^m}{m!}\frac{{\mathrm d}^m}{{\mathrm d}\lambda ^m}E\big ((t,\infty );\lambda \big )\bigg |_{\lambda =1},\ \ \ m\in \mathbb {Z}_{\ge 0}: \ \ F_0(t)\equiv 0, \end{aligned}$$

see [5, Section 6.3.2] for the standard probabilistic argument used in the derivation of such recursions in random matrix theory.

Remark 1.12

The analogue of (1.18) for the GOE was first derived in [25, Theorem 2.1] and then used for the computation of the limiting distribution function of the largest eigenvalue in the thinned GOE; see for example [13, Proposition 1.1]. For the GinOE, we will proceed in the reverse direction and first prove (1.14).

1.3 Tail Expansions

One major advantage of the explicit formula (1.14)—besides the fact that it places the thinned GinOE on firm integrable systems ground—originates from its usefulness in the derivation of tail expansions. Indeed, once the Riemann–Hilbert problem connection is in place, it is somewhat straightforward to obtain asymptotic information for the distribution function \(P(t;\gamma )\) in (1.12) as \(t\rightarrow \pm \infty \). We summarize the relevant estimates in our fifth result below.

Theorem 1.13

Let \(\gamma \in [0,1]\). We have, as \(t\rightarrow +\infty \),

$$\begin{aligned} P(t;\gamma )=1-\frac{\gamma }{4}\text {erfc}(t)+\mathcal {O}\big (t^{-1}\mathrm {e}^{-2t^2}\big ), \end{aligned}$$
(1.19)

with the complementary error function \(w=\text {erfc}(z):=\frac{2}{\sqrt{\pi }}\int _z^{\infty }\mathrm {e}^{-t^2}{\mathrm d}t\), see [54, 7.2.2]. On the other hand, as \(t\rightarrow -\infty \),

$$\begin{aligned} P(t;\gamma )=\mathrm {e}^{c_1(\gamma )t+c_0(\gamma )}\big (1+o(1)\big ), \end{aligned}$$
(1.20)

with

$$\begin{aligned} c_1(\gamma )=\frac{1}{2\sqrt{2\pi }}\text {Li}_{\frac{3}{2}}(\bar{\gamma }), \ \ \ \ c_0(\gamma )=\frac{1}{2}\ln \left( \frac{2}{2-\gamma }\right) +\frac{1}{4\pi }\int _0^{\bar{\gamma }}\left( \big (\text {Li}_{\frac{1}{2}}(x)\big )^2-\frac{x\pi }{1-x}\right) \frac{{\mathrm d}x}{x}, \end{aligned}$$
(1.21)

in terms of the polylogarithm \(w=\text {Li}_s(z):=\sum _{n=1}^{\infty }z^n/n^s\), see [54, 25.12.10].

Expansion (1.19) was first derived in [41] for \(\gamma =1\). The leading order exponential decay of the left tail (1.20) appeared in [55, (1.11)] for \(\gamma =1\) and for \(\gamma \in [0,1]\) in [38, (2.30)], albeit in somewhat implicit form. The notoriously difficult constant factor \(c_0(\gamma )\) in (1.20), difficult because it cannot be obtained via trace norm estimates (compare our discussion in (1.23)), was recently computed in [36, (3)] for \(\gamma =1\) using probabilistic arguments. In this paper, we derive (1.20) for all \(\gamma \in [0,1)\) by nonlinear steepest descent techniques. The evaluation of \(c_0(1)\) would require further analysis and we choose not to rederive \(c_0(1)\) in this paper. Nonetheless, we note that our result (1.20), (1.21) matches formally onto [55, (1.11)], [36, (3)], i.e., onto the \(t\rightarrow -\infty \) expansion

$$\begin{aligned} P(t;1)&=\exp \left[ \!\frac{t}{2\sqrt{2\pi }}\zeta \left( \frac{3}{2}\right) +\frac{1}{2}\ln 2+\frac{1}{4\pi }\sum _{n=1}^{\infty }\frac{1}{n}\left( \!-\pi +\sum _{m=1}^{n-1}\frac{1}{\sqrt{m(n-m)}}\!\right) \!\right] \nonumber \\&\quad \times \big (1+o(1)\big ), \end{aligned}$$
(1.22)

since \(c_1(1)=\frac{1}{2\sqrt{2\pi }}\zeta \left( \frac{3}{2}\right) \) and since \(c_0(\gamma )\) in (1.21) satisfies the following property

Lemma 1.14

The function \(c_0(\gamma )\) is continuous in \(\gamma \in [0,1]\) and equals

$$\begin{aligned} c_0(\gamma )=\frac{1}{2}\ln \left( \frac{2}{2-\gamma }\right) +\frac{1}{4\pi }\sum _{n=1}^{\infty }\frac{1}{n}\left( -\pi +\sum _{m=1}^{n-1}\frac{1}{\sqrt{m(n-m)}}\right) \bar{\gamma }^n. \end{aligned}$$
(1.23)

As it is standard (for instance in invariant random matrix theory ensembles), the right tail (1.19) of the extreme value distribution \(P(t;\gamma )\) follows from elementary considerations and does not need RHP 1.3. The left tail, however, is much more subtle since

$$\begin{aligned} \mathop {\mathrm {tr}}\limits _{L^2(\mathbb {R})}(\chi _tT\chi _t)&=\int _t^{\infty }T(x,x){\mathrm d}x=\frac{1}{2\pi }\sqrt{\frac{\pi }{2}}\int _t^{\infty }\text {erfc}(\sqrt{2}x){\mathrm d}x\\&=-\frac{t}{\pi }\sqrt{\frac{\pi }{2}}+\mathcal {O}(1),\ \ \ \ \ t\rightarrow -\infty , \end{aligned}$$

becomes unbounded, yet the distribution function \(P(t;\gamma )\) converges to zero. It is this well-known issue which requires the full use of RHP 1.3 and associated nonlinear steepest descent techniques for its asymptotic analysis; see Sect. 7.

Remark 1.15

The explicit computation of constant factors such as \(c_0(\gamma )\) in (1.21) is a well-known challenge in the asymptotic analysis of correlation and distribution functions in nonlinear mathematical physics. Without aiming for completeness, we mention the following contributions to the field: In the theory of exactly solvable lattice models, the works [6, 7, 11, 12, 61]. In classical invariant random matrix theory, the works [3, 23, 24, 33, 34, 49], and most recently on \(\tau \)-function connection problems for Painlevé transcendents the works [46, 47]. Finally, related to thinned ensembles in random matrix theory, the works [13,14,15,16,17,18,19, 22].

1.4 Numerics

The Fredholm determinant formula (1.13) provides us with an efficient way to evaluate \(P(t;\gamma )\) numerically, cf. [9]. Indeed, in order to showcase the applicability of (1.13) we now provide the following numerical evaluations for the limiting distribution of \(\max _j z_j^{\gamma }(\mathbf{X})\): First, Table 1 shows a few centralized moments for varying \(\gamma \).

Table 1 Some moments of the thinned real GinOE process

Second, probability density and distribution function plots for varying \(\gamma \in [0,1]\) are shown in Fig. 1.

Fig. 1
figure 1

The distribution functions \(P(t;\gamma )\) of the largest real eigenvalue in the thinned real GinOE process for varying values of \(\gamma \). The plots were generated in MATLAB with \(m=50\) quadrature points using the Nyström method with Gauss–Legendre quadrature. On the left cdfs, on the right pdfs

Third, we compare our asymptotic expansions (1.19) and (1.20) to the numerical results obtained from (1.13) in Figs. 2 and 3.

Fig. 2
figure 2

The distribution functions \(P(t;\gamma )\) in red for \(\gamma =1\) (left) and \(\gamma =0.75\) (right). We compare the numerical computed values from (1.13) to the right tail expansion (1.19). Again we used the Nyström method with Gauss–Legendre quadrature and \(m=50\) quadrature points

Fig. 3
figure 3

The distribution functions \(P(t;\gamma )\) in red for \(\gamma =1\) (left) and \(\gamma =0.75\) (right). We compare the numerical computed values from (1.13) to the left tail expansion (1.20) in a semilogarithmic plot. Again we used the Nyström method with Gauss–Legendre quadrature and \(m=50\) quadrature points

1.5 Methodology and Outline of Paper

The remainder of the paper is organized as follows. We prove Lemma 1.7 in Sect. 2 using a simple probabilistic argument. Afterward, we use (1.10) and carefully simplify the regularized Fredholm determinant in order to arrive at a finite n formula which is amenable to asymptotics. Our approach is somewhat similar to the ones carried out in [25, 56]; however, two issues arise along the way: One, the absence of Christoffel–Darboux structures throughout forces us to rely on the Fourier tricks used in [2, Section 2 and 3] in the derivation of (1.14). The other, unlike in the invariant ensembles, our computations depend heavily on the parity of n. We first work out the necessary details for even n in Sect. 3 and afterward develop a comparison argument to treat all odd n, see Sect. 3.3. The content of Subsect. 3.3 seemingly marks the first time that the extreme value statistics in the GinOE for odd n have been computed rigorously. Even for \(\gamma =1\), typos in [56, Section 4.2] have been pointed out in [55, Appendix B], but these had not been fixed until now. After several initial steps in Sect. 3 we complete the proof of Theorem 1.8 in Sect. 4. Once Theorem 1.8 has been derived, our proof of Theorem 1.9 in Sect. 5 is rather short, making essential use of the inverse scattering theory connection worked out in our previous paper [2]. This is followed by our short proof of (1.11) for the eigenvalue generating function in Sect. 6. Afterward, we prove Theorem 1.13 in Sect. 7. In fact, the asymptotic analysis is split into two parts, one part which deals with a total integral of \(y=y(x;\gamma )\) and a second part which computes the constant factor in the asymptotic expansion of the determinant

$$\begin{aligned} \det (1-\gamma \chi _t T\chi _t\upharpoonright _{L^2(\mathbb {R})}). \end{aligned}$$

Unlike for invariant matrix ensembles (compare the discussion in [16, page 492,493]), we are here able to efficiently employ the \(\gamma \)-derivative method in the computation of the constant factor without having a differential equation in the spectral variable. Indeed, since our nonlinear steepest descent analysis in Appendix 7.3 does not use any local model functions, the cumbersome double integration in the \(\gamma \)-derivative method becomes manageable. This feature is comparable with Deift’s proof of the strong Szegő limit theorem in [26, Example 3] and the details of our analysis can be found in Sect. 7. The final two sections of the paper in Appendices 7.3 and 7.3 prove two curious integral identities used in the proof of Theorem 1.8 and present a streamlined version of the nonlinear steepest descent analysis of [2, Section 5] which is crucial in our proof of Theorem 1.13.

Remark 1.16

Our final remark in this Section concerns two possible routes for future investigations of the thinned real GinOE process. On the one hand, the large t asymptotic behavior of the cumulants of the counting function \(N(t,\gamma ):=\#\{k\in \mathbb {N}: z_k^{\gamma }(\mathbf{X})\le t\}\) is seemingly within reach by (1.18) and (1.20) and can afterward be contrasted to recent results for the sine, Airy, Bessel or Pearcey processes, cf. [17, 18, 22]. On the other hand, with access to the eigenvalue generating function through (1.14) and (1.20), one can wonder whether the refined rigidity analysis of the determinantal Airy process in [20] can be extended to the Pfaffian point process formed by the real eigenvalues \(z_j(\mathbf{X})\) of some \(\mathbf{X}\in \text {GinOE}\).

2 Proof of Lemma 1.7

It is known from [52] that the eigenvalues \(\{z_j(\mathbf{X})\}_{j=1}^n\subset \mathbb {C}\) of \(\mathbf{X}\in \mathbb {R}^{n\times n}\) drawn from the \(\text {GinOE}\) are distributed according to a random point process whose correlation functions are computable as Pfaffians, cf. [10, 41, 59]. In particular, the real eigenvalues form a Pfaffian process whose correlations are given by

(2.1)

with the skew-symmetric \(2\times 2\)-matrix kernel

$$\begin{aligned} \mathbf{K}_n^{\mathbb {R},\mathbb {R}}(x,y):=\begin{bmatrix}\epsilon (x,y)-(IS_n)(x,y) &{} S_n(y,x)\\ -S_n(x,y) &{} -(DS_n^{*})(x,y)\end{bmatrix}. \end{aligned}$$

Note that for any distinct points \(w_j\in \mathbb {R}\),

$$\begin{aligned}&\rho _{\ell }(w_1,\ldots ,w_{\ell })\nonumber \\&\quad =\lim _{\Delta w_i\rightarrow 0}\frac{\mathbb {P}(\text {one real GinOE eigenvalue in each}(w_i,w_i+\Delta w_i))}{\Delta w_1\cdot \ldots \cdot \Delta w_{\ell }}. \end{aligned}$$

Thus, if \(\rho _{\ell }^{\gamma }\) denotes the \(\ell \)-th correlation function in the thinned real GinOE process, we find with \(1\le \ell \le m_{\gamma ,n}\),

since each eigenvalue is removed independently with likelihood \(1-\gamma \). In short, \(\rho _{\ell }^{\gamma }=\gamma ^{\ell }\rho _{\ell }\) which shows that the thinned Pfaffian point process is also a Pfaffian process and its kernel is simply given by \(\gamma \mathbf{K}_n^{\mathbb {R},\mathbb {R}}(x,y)\). Equipped with this insight, one now repeats the computations in [56, page 1630] and arrives at (1.10).

3 Proof of Theorem 1.8—first steps

Abbreviate

$$\begin{aligned} F_{n}\equiv F_{n}(t,\gamma ):=\det _2\big (1-\gamma \chi _t\mathbf{K}_{n}\chi _t{\upharpoonright }_{L^2(\mathbb {R})\oplus L^2(\mathbb {R})}\big ),\ \ \ n\in \mathbb {Z}_{\ge 2}. \end{aligned}$$

We will first simplify \(F_n\) for n even and afterward take the limit as \(n\rightarrow \infty \) with n even. Once done, we then compare the odd n case with the even n case and prove existence of the limit (1.12) all together.

3.1 Finite Even n Calculations

We consider \(F_{2n}\). Our overall approach follows closely [56, page 1640], keeping throughout track of the \(\gamma \)-modifications due to (1.10). First, the kernel \(\chi _t\mathbf{K}_{2n}\chi _t\) can be factorized as

$$\begin{aligned} \begin{bmatrix}\rho ^{-1}\chi _tD\rho &{} 0\\ 0 &{} \rho \chi _t\rho ^{-1}\end{bmatrix}\begin{bmatrix}-\rho ^{-1}\epsilon S_{2n}\chi _t\rho &{} \rho ^{-1}S_{2n}^{*}\chi _t\rho ^{-1}\\ -\rho (\epsilon S_{2n}-\epsilon )\chi _t\rho &{} \rho S_{2n}^{*}\chi _t\rho ^{-1}\end{bmatrix}, \end{aligned}$$
(3.1)

and by using (1.5) we can move the factor on the left in (3.1) to the right, so \(F_{2n}(t,\gamma )\) equals the regularized 2-determinant of the operator with kernel

$$\begin{aligned} \gamma \begin{bmatrix}-\rho ^{-1}\epsilon S_{2n}\chi _tD\rho &{} \rho ^{-1}S_{2n}^{*}\chi _t\rho ^{-1}\\ -\rho (\epsilon S_{2n}-\epsilon )\chi _tD\rho &{} \rho S_{2n}^{*}\chi _t\rho ^{-1}\end{bmatrix}. \end{aligned}$$

Next, we observe that the traces of the last operator’s powers of \(2,3,\ldots \) match the corresponding traces of the operator with kernel

$$\begin{aligned} \gamma \begin{bmatrix}-\rho ^{-1}(\epsilon S_{2n}\chi _tD-S_{2n}^{*}\chi _t)\rho &{} \rho ^{-1}S_{2n}^{*}\chi _t\rho ^{-1}\\ \rho \epsilon \chi _tD\rho &{} 0\end{bmatrix}. \end{aligned}$$

Hence, by the Plemelj–Smithies formula for \(\det _2\), see [57, Theorem 9.3],

$$\begin{aligned} F_{2n}(t,\gamma )=\det _2\begin{bmatrix}1+\gamma \rho ^{-1}(\epsilon S_{2n}\chi _tD-S_{2n}^{*}\chi _t)\rho &{} -\gamma \rho ^{-1}S_{2n}^{*}\chi _t\rho ^{-1}\\ -\gamma \rho \epsilon \chi _tD\rho &{} 1\end{bmatrix}. \end{aligned}$$

Factorizing the underlying kernel, we then obtain

$$\begin{aligned} F_{2n}&=\det _2\left( \begin{bmatrix}1 &{} -\gamma \rho ^{-1}S_{2n}^{*}\chi _t\rho ^{-1}\\ 0 &{} 1\end{bmatrix}\right. \\&\left. \quad \begin{bmatrix}1+\gamma \rho ^{-1}(\epsilon S_{2n}\chi _tD-S_{2n}^{*}\chi _t-\gamma S_{2n}^{*}\chi _t\epsilon \chi _tD)\rho &{} 0\\ 0 &{} 1\end{bmatrix} \begin{bmatrix}1 &{} 0\\ -\gamma \rho \epsilon \chi _tD\rho &{} 1\end{bmatrix}\right) , \end{aligned}$$

and since both triangular factors are of the form identity plus block operator as in Remark 1.2, we are allowed to use (1.4). In fact the regularized 2-determinant of those triangular factors equals one, so we have just shown that the original determinant in (1.1) for even n simplifies to

$$\begin{aligned} F_{2n}(t,\gamma )=\det _2\left( 1+\begin{bmatrix}\gamma \rho ^{-1}(\epsilon S_{2n}\chi _tD-S_{2n}^{*}\chi _t-\gamma S_{2n}^{*}\chi _t\epsilon \chi _tD)\rho &{} 0\\ 0 &{} 0\end{bmatrix}{\upharpoonright }_{L^2(\mathbb {R})\oplus L^2(\mathbb {R})}\right) . \end{aligned}$$
(3.2)

Clearly, the determinant in (3.2) on \(L^2(\mathbb {R})\oplus L^2(\mathbb {R})\) is really a determinant on \(L^2(\mathbb {R})\) alone,

$$\begin{aligned} F_{2n}(t,\gamma )=\det _2\big (1+\gamma \rho ^{-1}(\epsilon S_{2n}\chi _tD-S_{2n}^{*}\chi _t-\gamma S_{2n}^{*}\chi _t\epsilon \chi _tD)\rho {\upharpoonright }_{L^2(\mathbb {R})}\big ), \end{aligned}$$
(3.3)

and as our upcoming computations will show (see in particular (3.8)) the operator \(\epsilon S_{2n}\chi _tD-S_{2n}^{*}\chi _t-\gamma S_{2n}^{*}\chi _t\epsilon \chi _tD\) is of finite rank, i.e., the regularized 2-determinant in (3.3) is an ordinary Fredholm determinant by Remark 1.2 and the conjugation with \(\rho \) now redundant. We have thus arrived at the following replacement of the equation right above [56, (4.6)],

$$\begin{aligned} F_{2n}(t,\gamma )=\det \big (1-\gamma S_{2n}^{*}\chi _t+\gamma \epsilon S_{2n}\chi _tD-\gamma ^2S_{2n}^{*}\chi _t\epsilon \chi _tD{\upharpoonright }_{L^2(\mathbb {R})}\big ). \end{aligned}$$
(3.4)

In order to simplify (3.4) further, we now record

Lemma 3.1

([56, page 1640]) For any \(n\in \mathbb {Z}_{\ge 1}\),

$$\begin{aligned} \epsilon S_{2n}=S_{2n}^{*}\epsilon . \end{aligned}$$
(3.5)

Proof

The stated identity follows easily by induction on \(n\in \mathbb {Z}_{\ge 1}\) using only that

$$\begin{aligned} S_n(x,y)=\frac{1}{\sqrt{2\pi }}\mathrm {e}^{-\frac{1}{2}(x^2+y^2)}\mathfrak {e}_{n-2}(xy)+\frac{x^{n-1}\mathrm {e}^{-\frac{1}{2}x^2}}{\sqrt{2\pi }(n-2)!}\int _0^yu^{n-2}\mathrm {e}^{-\frac{1}{2}u^2}{\mathrm d}u,\ \ \ n\in \mathbb {Z}_{\ge 2}. \end{aligned}$$

\(\square \)

Inserting (3.5) into (3.4), we find

$$\begin{aligned} F_{2n}(t,\gamma )=\det \big (1-\gamma S_{2n}^{*}\chi _t+\gamma S_{2n}^{*}(1-\gamma \chi _t)\epsilon \chi _tD\upharpoonright _{L^2(\mathbb {R})}\big ). \end{aligned}$$
(3.6)

We write \(\alpha \otimes \beta \) for a general rank one integral operator on \(L^2(\mathbb {R})\) with kernel \((\alpha \otimes \beta )(x,y)=\alpha (x)\beta (y)\). Noting \(\epsilon D\chi _t=-\chi _t\) and applying the commutator identity, cf. [62, (16)],

$$\begin{aligned} \epsilon \big [\chi _t,D\big ]=-\epsilon _t\otimes \delta _t+\epsilon _{\infty }\otimes \delta _{\infty }\ \ \ \text {with} \ \ \ \int _{-\infty }^{\infty }f(x)\delta _a(x){\mathrm d}x:=f(a),\\ {\left\{ \begin{array}{ll}\epsilon _t(x):=\frac{1}{2}\text {sgn}(t-x),\ \ t\in \mathbb {R}&{}\\ \epsilon _{\infty }(x):=\frac{1}{2}&{}\end{array}\right. }, \end{aligned}$$

one part in (3.6) simplifies to

$$\begin{aligned} (1-\gamma \chi _t)\epsilon \chi _t D=(1-\gamma \chi _t)(-\epsilon _t\otimes \delta _t+\epsilon _{\infty }\otimes \delta _{\infty })-(1-\gamma )\chi _t, \end{aligned}$$

which (since \(\epsilon _t=\frac{1}{2}-\chi _{[t,\infty )},\epsilon _{\infty }=\frac{1}{2}\)) yields

$$\begin{aligned} (1-\gamma \chi _t)\epsilon \chi _t D=-\big ((1-\gamma \chi _t)\epsilon _{\infty }\big )\otimes (\delta _t-\delta _{\infty })+(1-\gamma )\big (\chi _{[t,\infty )}\otimes \delta _t\big )-(1-\gamma )\chi _t. \end{aligned}$$
(3.7)

Substituting (3.7) back into (3.6), we have thus (recall \(\bar{\gamma }=2\gamma -\gamma ^2\))

$$\begin{aligned} F_{2n}(t,\gamma )=\det \left( 1-\bar{\gamma }S_{2n}^{*}\chi _t-\gamma S_{2n}^{*}\Big (\big ((1-\gamma \chi _t)\epsilon _{\infty }\big )\otimes (\delta _t-\delta _{\infty })\Big )\right. \nonumber \\ \left. +\gamma (1-\gamma )S_{2n}^{*}(\chi _{[t,\infty )}\otimes \delta _t)\upharpoonright _{L^2(\mathbb {R})}\right) . \end{aligned}$$
(3.8)

Next, from the definition of \(S_n\) in Proposition 1.1, we may write, see [56, (4.7)],

$$\begin{aligned} S_n^{*}(x,y)= & {} \frac{1}{\sqrt{2\pi }}\mathrm {e}^{-\frac{1}{2}(x^2+y^2)}\mathfrak {e}_{n-2}(xy)+\frac{y^{n-1}\mathrm {e}^{-\frac{1}{2}y^2}}{\sqrt{2\pi }(n-2)!}\int _0^xu^{n-2}\mathrm {e}^{-\frac{1}{2}u^2}{\mathrm d}u\\=: & {} T_n(x,y) +\big (\phi _n\otimes \psi _n\big )(x,y), \end{aligned}$$

where \(T_n(x,y)\) is a symmetric kernel and

$$\begin{aligned} \phi _n(x):=\sqrt{\frac{\sqrt{n}}{\sqrt{2\pi }(n-2)!}}\int _0^xu^{n-2}\mathrm {e}^{-\frac{1}{2}u^2}{\mathrm d}u, \ \ \ \ \psi _n(y):=\sqrt{\frac{1}{\sqrt{2\pi n}(n-2)!}}y^{n-1}\mathrm {e}^{-\frac{1}{2}y^2}. \end{aligned}$$
(3.9)

Lemma 3.2

Given \(t\ge 0\) and \(n\in \mathbb {Z}_{\ge 2}\), the trace class operator \(T_n:L^2(t,\infty )\rightarrow L^2(t,\infty )\) with kernel \(T_n(x,y)\) satisfies \(0\le T_n\le 1\) and \(1-\gamma T_n\) is invertible on \(L^2(t,\infty )\) for all \(\gamma \in [0,1]\).

Proof

For every \(f\in L^2(t,\infty )\),

$$\begin{aligned} \langle f,T_n f\rangle _{L^2(t,\infty )}=\frac{1}{\sqrt{2\pi }}\sum _{k=0}^{n-2}\frac{1}{k!}\left| \int _t^{\infty }f(x)\mathrm {e}^{-\frac{1}{2}x^2}x^k{\mathrm d}x\right| ^2, \end{aligned}$$

which implies nonnegativity of \(T_n\). For the upper bound, we apply Schur’s test,

$$\begin{aligned} \Vert T_n\Vert&\le \sup _{y>t}\int _t^{\infty }\big |T_n(x,y)\big |{\mathrm d}x\le \sup _{y>t}\frac{1}{\sqrt{2\pi }}\int _t^{\infty }\mathrm {e}^{-\frac{1}{2}(|x|-|y|)^2}{\mathrm d}x\\&\le \frac{1}{\sqrt{\pi }}\int _{-\infty }^{\infty }\mathrm {e}^{-v^2}{\mathrm d}v=1\ \ \forall t\ge 0, \end{aligned}$$

and conclude by self-adjointness of \(T_n\) that

$$\begin{aligned} \sup _{\Vert f\Vert _{L^2(t,\infty )}=1}\big |\langle f,T_nf\rangle _{L^2(t,\infty )}\big |=\Vert T_n\Vert \le 1, \end{aligned}$$

i.e., \(T_n\le 1\) for any \(n\in \mathbb {Z}_{\ge 2}\). Next, using that \(\Vert T_n\Vert \le 1\), the invertibility of \(1-\gamma T_n\) on \(L^2(t,\infty )\) follows readily from the underlying Neumann series provided \(\gamma \in [0,1)\). The case \(\gamma =1\) has been addressed in [56, Lemma 4.2]. This concludes our proof. \(\square \)

In the following we will use the result of Lemma 3.2 for the operator \(\bar{\gamma }\chi _tT_{2n}\chi _t\) which acts on \(L^2(\mathbb {R})\). Inserting the operator decomposition \(S_n^{*}=T_n+\phi _n\otimes \psi _n\) into (3.8) and using the general identities \((\alpha \otimes \beta )(\gamma \otimes \delta )=\langle \beta ,\gamma \rangle (\alpha \otimes \delta )\) and \(A(\beta \otimes \gamma )D=(A\beta )\otimes (D^{*}\gamma )\) (for arbitrary operators AD),

$$\begin{aligned} F_{2n}(t,\gamma )&=\det \Big (1-\bar{\gamma }T_{2n}\chi _t-\bar{\gamma }\phi _{2n}\otimes (\chi _t\psi _{2n})-\gamma \big (\langle \psi _{2n},(1-\gamma \chi _t)\epsilon _{\infty }\rangle \phi _{2n}\nonumber \\&\quad +T_{2n}(1-\gamma \chi _t)\epsilon _{\infty }\big )\otimes (\delta _t-\delta _{\infty })\nonumber \\&\quad +\gamma (1-\gamma )\big (\langle \psi _{2n},\chi _{[t,\infty )}\rangle \phi _{2n}+T_{2n}\chi _{[t,\infty )}\big )\otimes \delta _t\upharpoonright _{L^2(\mathbb {R})}\Big ). \end{aligned}$$
(3.10)

Here, \(\langle \cdot ,\cdot \rangle \) is the standard \(L^2(\mathbb {R})\) inner product. Since \(\chi _t^2=\chi _t,\chi _t^{*}=\chi _t\) and

$$\begin{aligned} \alpha \otimes (\delta _t-\delta _{\infty })=\alpha \otimes \big (\chi _t(\delta _t-\delta _{\infty })\big ),\ \ \ \ \alpha \otimes \delta _t=\alpha \otimes (\chi _t\delta _t), \end{aligned}$$

we can then rewrite (3.10) by Sylvester’s identity [43, Chapter IV, (5.9)] as

$$\begin{aligned} F_{2n}(t,\gamma )&=\det \Big (1-\bar{\gamma }\chi _tT_{2n}\chi _t-\bar{\gamma }(\chi _t\phi _{2n})\otimes (\chi _t\psi _{2n})-\gamma \big (\langle \psi _{2n},(1-\gamma \chi _t)\epsilon _{\infty }\rangle \chi _t\phi _{2n}\nonumber \\&+\chi _tT_{2n}(1-\gamma \chi _t)\epsilon _{\infty }\big )\otimes (\delta _t-\delta _{\infty })+\gamma (1-\gamma )\big (\langle \psi _{2n},\chi _{[t,\infty )}\rangle \chi _t\phi _{2n} \nonumber \\&\quad +\chi _tT_{2n}\chi _{[t,\infty )}\big )\otimes \delta _t\upharpoonright _{L^2(\mathbb {R})}\Big ). \end{aligned}$$
(3.11)

Next, using that \(1-\bar{\gamma }\chi _tT_n\chi _t\) is invertible on \(L^2(\mathbb {R})\) for \(t\ge 0\) by Lemma 3.2, we factorize \(F_{2n}(t,\gamma )\) as follows:

$$\begin{aligned} F_{2n}(t,\gamma )=\det \big (1-\bar{\gamma }\chi _tT_{2n}\chi _t\upharpoonright _{L^2(\mathbb {R})}\big )\det \left( 1-\sum _{k=1}^3\alpha _k\otimes \beta _k\upharpoonright _{L^2(\mathbb {R})}\right) . \end{aligned}$$
(3.12)

Here, \(\alpha _j,\beta _k\) denote the six functions

$$\begin{aligned} \alpha _1:=&\bar{\gamma }(1-\bar{\gamma }\chi _tT_{2n}\chi _t)^{-1}\chi _t\phi _{2n},\ \beta _1:=\chi _t\psi _{2n},\ \ \ \ \beta _2:=\delta _t-\delta _{\infty },\ \ \ \ \beta _3:=\delta _t,\\ \alpha _2:=&\gamma \langle \psi _{2n},(1-\gamma \chi _t)\epsilon _{\infty }\rangle (1-\bar{\gamma }\chi _tT_{2n}\chi _t)^{-1}\chi _t\phi _{2n}\\&+\gamma (1-\bar{\gamma }\chi _tT_{2n}\chi _t)^{-1}\chi _tT_{2n}(1-\gamma \chi _t)\epsilon _{\infty },\\ \alpha _3:=&-\gamma \big \langle \psi _{2n},(1-\gamma \chi _t)\chi _{[t,\infty )}\big \rangle (1-\bar{\gamma }\chi _tT_{2n}\chi _t)^{-1}\chi _t\phi _{2n}\\&-\gamma (1-\bar{\gamma }\chi _tT_{2n}\chi _t)^{-1}\chi _tT_{2n}(1-\gamma \chi _t)\chi _{[t,\infty )}. \end{aligned}$$

By general theory, cf. [43, Chapter I, (3.3)]

$$\begin{aligned} \det \left( 1-\sum _{k=1}^3\alpha _k\otimes \beta _k\upharpoonright _{L^2(\mathbb {R})}\right) =\det \big [\delta _{jk}-\langle \alpha _j,\beta _k\rangle \big ]_{j,k=1}^3,\ \ \ \ \delta _{jk}:={\left\{ \begin{array}{ll}1,&{}j=k\\ 0,&{}j\ne k\end{array}\right. } \end{aligned}$$
(3.13)

with the \(L^2(\mathbb {R})\) inner product \(\langle \cdot ,\cdot \rangle \). We conclude our finite n calculation for even n with the following further algebraic simplifications.

Lemma 3.3

We have

$$\begin{aligned} \langle \alpha _1,\beta _1\rangle&=\big \langle \chi _t\phi _{2n},\bar{\gamma }(1-\bar{\gamma }\chi _tT_{2n}\chi _t)^{-1}\chi _t\psi _{2n}\big \rangle ,\ \langle \alpha _1,\beta _2\rangle \\&=\bar{\gamma }\big ((1-\bar{\gamma }\chi _tT_{2n}\chi _t)^{-1}\chi _t\phi _{2n}\big )(t)-\bar{\gamma }\phi _{2n}(\infty ), \end{aligned}$$

followed by

$$\begin{aligned} \langle \alpha _1,\beta _3\rangle =\bar{\gamma }\big ((1-\bar{\gamma }\chi _tT_{2n}\chi _t)^{-1}\chi _t\phi _{2n}\big )(t). \end{aligned}$$

Next, with \(R_n:=\bar{\gamma }T_n\chi _t(1-\bar{\gamma }\chi _tT_n\chi _t)^{-1},n\in \mathbb {Z}_{\ge 1}\) which is well defined as operator on \(L^2(\mathbb {R})\) by Lemma 3.2 for any \(t\ge 0\),

$$\begin{aligned} \langle \alpha _2,\beta _1\rangle =&\frac{\gamma }{2\bar{\gamma }}\left[ c_{2n}\langle \alpha _1,\beta _1\rangle -c_{2n}+\int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )\big ((1+R_{2n})\psi _{2n}\big )(x){\mathrm d}x\right] ,\\ \langle \alpha _2,\beta _2\rangle =&\frac{\gamma }{2\bar{\gamma }}\left[ c_{2n}\langle \alpha _1,\beta _2\rangle +\int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )R_{2n}(x,t){\mathrm d}x\right] ,\\ \langle \alpha _2,\beta _3\rangle =&\frac{\gamma }{2\bar{\gamma }}\left[ c_{2n}\langle \alpha _1,\beta _3\rangle +\int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )R_{2n}(x,t){\mathrm d}x\right] , \end{aligned}$$

where \(c_n:=\langle \psi _n,1-\gamma \chi _{[t,\infty )}\rangle \). Moreover

$$\begin{aligned} \langle \alpha _3,\beta _1\rangle =&-\frac{\gamma }{\bar{\gamma }}\left[ d_{2n}\langle \alpha _1,\beta _1\rangle -d_{2n}\right. \\&\quad \left. +\int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )\chi _{[t,\infty )}(x)\big ((1+R_{2n})\psi _{2n}\big )(x){\mathrm d}x\right] ,\\ \langle \alpha _3,\beta _2\rangle =&-\frac{\gamma }{\bar{\gamma }}\left[ d_{2n}\langle \alpha _1,\beta _2\rangle +\int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )\chi _{[t,\infty )}(x)R_{2n}(x,t){\mathrm d}x\right] ,\\ \langle \alpha _3,\beta _3\rangle =&\ -\frac{\gamma }{\bar{\gamma }}\left[ d_{2n}\langle \alpha _1,\beta _3\rangle +\int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )\chi _{[t,\infty )}(x)R_{2n}(x,t){\mathrm d}x\right] , \end{aligned}$$

with \(d_n:=\langle \psi _n,(1-\gamma \chi _t)\chi _{[t,\infty )}\rangle \).

Proof

We use self-adjointness of the operator \(T_n\) and write \(R_n(x,t)\) for (cf. [62, page 732])

$$\begin{aligned} \lim _{\begin{array}{c} y\rightarrow t \\ y>t \end{array}}R_n(x,y). \end{aligned}$$

\(\square \)

With Lemma 3.3 in place, we finally evaluate the Fredholm determinant in (3.13). Noting that the terms \(c_n,d_n\) cancel out due to multilinearity of the finite-dimensional determinant, we obtain

$$\begin{aligned} \det \left( 1-\sum _{k=1}^3\alpha _k\otimes \beta _k\upharpoonright _{L^2(\mathbb {R})}\right) =\det \begin{bmatrix}1-\langle \alpha _1,\beta _1\rangle &{} -\langle \alpha _1,\beta _2\rangle &{} -\langle \alpha _1,\beta _3\rangle \\ -\frac{\gamma }{2\bar{\gamma }}I_3 &{} 1-\frac{\gamma }{2\bar{\gamma }}I_1 &{} -\frac{\gamma }{2\bar{\gamma }}I_1\\ \frac{\gamma }{\bar{\gamma }}I_4 &{} \frac{\gamma }{\bar{\gamma }}I_2 &{} 1+\frac{\gamma }{\bar{\gamma }}I_2\end{bmatrix}, \end{aligned}$$
(3.14)

in terms of the three inner products \(\langle \alpha _1,\beta _k\rangle ,k=1,2,3\) and the four integrals \(I_j=I_j(t,\gamma ,2n)\) with

$$\begin{aligned} I_1:=&\int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )R_{2n}(x,t){\mathrm d}x,\nonumber \\ I_2:=&\int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )\chi _{[t,\infty )}(x)R_{2n}(x,t){\mathrm d}x,\nonumber \\ I_3:=&\int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )\big ((1+R_{2n})\psi _{2n}\big )(x){\mathrm d}x,\nonumber \\ I_4:=&\int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )\chi _{[t,\infty )}(x)\big ((1+R_{2n})\psi _{2n}\big )(x){\mathrm d}x. \end{aligned}$$
(3.15)

Identities (3.12) and (3.14) conclude our calculations for finite n, provided n is even.

Remark 3.4

The \(3\times 3\) determinant (3.14) is the analogue of the GOE computation [25, (3.63)].

3.2 The Limit \(n\rightarrow \infty \), n Even

In order to pass to the large n limit, we first shift the independent variable t according to \(t\mapsto t+\sqrt{2n}\); compare the left-hand side of (1.7). Under this scaling, we have

$$\begin{aligned} \det (1-\bar{\gamma }\chi _{t+\sqrt{2n}}T_{2n}\chi _{t+\sqrt{2n}}\upharpoonright _{L^2(\mathbb {R})})=\det \big (1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t\upharpoonright _{L^2(\mathbb {R})}\big ), \end{aligned}$$

where \(\widetilde{T}_n:L^2(\mathbb {R})\rightarrow L^2(\mathbb {R})\) has kernel

$$\begin{aligned} \widetilde{T}_n(x,y):=T_n(x+\sqrt{n},y+\sqrt{n}). \end{aligned}$$
(3.16)

Moreover, the entries in the \(3\times 3\) determinant (3.14) transform in a similar fashion, for instance

$$\begin{aligned} \langle \alpha _1,\beta _1\rangle \mapsto \langle \widetilde{\alpha }_1,\widetilde{\beta }_1\rangle =\langle \chi _t\widetilde{\phi }_{2n},\bar{\gamma }(1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\chi _t\widetilde{\psi }_{2n}\rangle , \end{aligned}$$

and likewise

$$\begin{aligned} \langle \alpha _1,\beta _2\rangle \mapsto \langle \widetilde{\alpha }_1,\widetilde{\beta }_2\rangle =&\bar{\gamma }\big ((1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\chi _t\widetilde{\phi }_{2n}\big )(t)-\bar{\gamma }\widetilde{\phi }_{2n}(\infty ),\\ \langle \alpha _1,\beta _3\rangle \mapsto \langle \widetilde{\alpha }_1,\widetilde{\beta }_3\rangle =&\bar{\gamma }\big ((1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\chi _t\widetilde{\phi }_{2n}\big )(t), \end{aligned}$$

which involve \(\widetilde{\phi }_n(x):=\phi _n(x+\sqrt{n})\) and \(\widetilde{\psi }_n(x):=\psi _n(x+\sqrt{n})\). The remaining four integrals \(I_k\), see (3.15), are treated the same way and every occurrence of \(R_n\) in them gets replaced by \(\widetilde{R}_n\) with

$$\begin{aligned} \widetilde{R}_n=\bar{\gamma }\widetilde{T}_n\chi _t(1-\bar{\gamma }\chi _t\widetilde{T}_n\chi _t)^{-1} \end{aligned}$$

defined in terms of \(\widetilde{T}_n:L^2(\mathbb {R})\rightarrow L^2(\mathbb {R})\) with kernel (3.16). At this point, we collect a sequence of technical limits.

Lemma 3.5

Uniformly in \(x\in \mathbb {R}\) chosen from compact subsets,

$$\begin{aligned} \lim _{n\rightarrow \infty }\widetilde{\phi }_n(x)&=\frac{1}{\sqrt{2\pi }}\int _{-\infty }^x\mathrm {e}^{-y^2}{\mathrm d}y{\mathop {=}\limits ^{1.9}}\frac{1}{\sqrt{2}}G(x),\nonumber \\ \lim _{n\rightarrow \infty }\widetilde{\psi }_n(x)&=\frac{1}{\sqrt{2\pi }}\mathrm {e}^{-x^2} {\mathop {=}\limits ^{1.9}}\frac{1}{\sqrt{2}}g(x); \end{aligned}$$
(3.17)

and for any fixed \(s\in \mathbb {R}\) with \(p\in \{1,2\}\),

$$\begin{aligned} \lim _{n\rightarrow \infty }\left\| \widetilde{\psi }_n-\frac{g}{\sqrt{2}}\right\| _{L^p(s,\infty )}=0,\ \ \ \ \ \ \ \ \lim _{n\rightarrow \infty }\left\| \widetilde{\phi }_n-\widetilde{\phi }_n(\infty )+\frac{1}{2\sqrt{2}}\text {erfc}\right\| _{L^p(s,\infty )}=0. \end{aligned}$$
(3.18)

Here, \(w=\text {erfc}(z)\) denotes the complementary error function, cf. [54, 7.2.2].

Proof

The limits (3.17), (3.18) are mentioned en route in [56, page 1640] and we thus only give a few details: as \(n\rightarrow \infty \), uniformly in \(x\in \mathbb {R}\),

$$\begin{aligned} \widetilde{\psi }_n(x)=\frac{1}{\sqrt{2\pi }}\left( 1+\frac{x}{\sqrt{n}}\right) ^{n-1}\mathrm {e}^{-x\sqrt{n}-\frac{1}{2}x^2}\Big (1+\mathcal {O}\big (n^{-1}\big )\Big ). \end{aligned}$$

But on compact subsets of \(\mathbb {R}\ni x\),

$$\begin{aligned} \left( 1+\frac{x}{\sqrt{n}}\right) ^{n-1}\mathrm {e}^{-x\sqrt{n}}{\mathop {\longrightarrow }\limits ^{n\rightarrow \infty }}\mathrm {e}^{-\frac{1}{2}x^2}, \end{aligned}$$

which yields the second limit in (3.17). Since also for any \(x>0\) and \(n\in \mathbb {Z}_{\ge 2}\),

$$\begin{aligned} \left( 1+\frac{x}{\sqrt{n}}\right) ^{n-1}\mathrm {e}^{-x\sqrt{n}}\le \mathrm {e}^{-x/\sqrt{n}}\le 1, \end{aligned}$$

the dominated convergence theorem yields the first \(L^p(s,\infty )\) convergence in (3.18). For the limits involving \(\widetilde{\phi }_n\), we note that as \(n\rightarrow \infty \), uniformly in \(x\in \mathbb {R}\),

$$\begin{aligned} \widetilde{\phi }_n(x)=\frac{1}{\sqrt{2}}P\left( \frac{1}{2}(n-1),\frac{1}{2}\big (x+\sqrt{n}\big )^2\right) \Big (1+\mathcal {O}\big (n^{-1}\big )\Big ), \end{aligned}$$

with the normalized incomplete gamma function \(w=P(z)\), cf. [54, 8.2.4]. But on compact subsets of \(\mathbb {R}\ni x\), see [54, 8.11.10],

$$\begin{aligned} P\left( \frac{1}{2}(n-1),\frac{1}{2}\big (x+\sqrt{n}\big )^2\right) {\mathop {\longrightarrow }\limits ^{n\rightarrow \infty }}\frac{1}{2}\text {erfc}(-x)=\frac{1}{\sqrt{\pi }}\int _{-\infty }^x\mathrm {e}^{-y^2}{\mathrm d}y \end{aligned}$$

which yields the first limit in (3.17). For the outstanding limit in (3.18), we use that as \(n\rightarrow \infty \), uniformly in \(x\in \mathbb {R}\),

$$\begin{aligned} \widetilde{\phi }_n(x)=\widetilde{\phi }_n(\infty )-\frac{1}{\sqrt{2}}\frac{\Gamma (\frac{1}{2}(n-1),\frac{1}{2}(x+\sqrt{n})^2)}{\Gamma (\frac{1}{2}(n-1))}\Big (1+\mathcal {O}\big (n^{-1}\big )\Big ) \end{aligned}$$
(3.19)

with \(w=\Gamma (a,z)\) the incomplete Gamma function, see [54, 8.2.2]. But since for \(x>0\) and \(a\ge 1\) such that \(x+1-a>0\),

$$\begin{aligned} \Gamma (a,x)=x^a\mathrm {e}^{-x}\int _0^{\infty }(1+y)^{a-1}\mathrm {e}^{-yx}{\mathrm d}y\le x^a\mathrm {e}^{-x}\int _0^{\infty }\mathrm {e}^{y(a-1)}\mathrm {e}^{-yx}{\mathrm d}y=\frac{x^a\mathrm {e}^{-x}}{x+1-a}, \end{aligned}$$

we find for any \(x>0\) and \(n\in \mathbb {Z}_{\ge 2}\),

$$\begin{aligned} \frac{\Gamma (\frac{1}{2}(n-1),\frac{1}{2}(x+\sqrt{n})^2)}{\Gamma (\frac{1}{2}(n-1))}\le c\mathrm {e}^{-\frac{1}{2}x^2},\ \ \ \ c>0. \end{aligned}$$
(3.20)

Using also that on compact subsets of \(\mathbb {R}\ni x\),

$$\begin{aligned} \frac{\Gamma (\frac{1}{2}(n-1),\frac{1}{2}(x+\sqrt{n})^2)}{\Gamma (\frac{1}{2}(n-1))}&=1-P\left( \!\frac{1}{2}(n-1),\frac{1}{2}\big (x+\sqrt{n}\big )^2\!\right) {\mathop {\longrightarrow }\limits ^{n\rightarrow \infty }}1-\frac{1}{2}\text {erfc}(-x)\nonumber \\&=\frac{1}{2}\text {erfc}(x), \end{aligned}$$
(3.21)

the second limit in (3.18) follows from (3.19), (3.20), (3.21) and the dominated convergence theorem. This completes our proof. \(\square \)

The next limits concern the large n-behavior of the kernel function \(\widetilde{T}_n(x,y)\) and its total integrals. Recall the kernel T(xy) defined in (1.8).

Lemma 3.6

([56, page 1642-644]) Uniformly in \(x,y\in \mathbb {R}\) chosen from compact subsets,

$$\begin{aligned} \lim _{n\rightarrow \infty }\widetilde{T}_n(x,y)= & {} T(x,y){\mathop {=}\limits ^{1.9}}\frac{1}{\pi }\!\int _0^{\infty }\! \mathrm {e}^{-(x+u)^2}\mathrm {e}^{-(y+u)^2}{\mathrm d}u,\ \ \ \lim _{n\rightarrow \infty }\int _{-\infty }^{\infty }\widetilde{T}_n(x,y){\mathrm d}y\nonumber \\= & {} \int _{-\infty }^{\infty }T(x,y){\mathrm d}y. \end{aligned}$$
(3.22)

Moreover, for any fixed \(s,t\in \mathbb {R}\) and with \(p\in \{1,2\}\),

$$\begin{aligned}&\lim _{n\rightarrow \infty }\Big \Vert \widetilde{T}_n(\cdot ,t)-T(\cdot ,t)\Big \Vert _{L^p(s,\infty )}=0,\nonumber \\&\lim _{n\rightarrow \infty }\Big \Vert \int _{-\infty }^{\infty }\widetilde{T}_n(\cdot ,y){\mathrm d}y-\int _{-\infty }^{\infty }T(\cdot ,y){\mathrm d}y\Big \Vert _{L^p(s,\infty )}=0. \end{aligned}$$
(3.23)

Proof

The limits (3.22) and (3.23) follow from the detailed discussion on page 1642 and 1643 in [56], see also [56, (4.16)]. We omit details. \(\square \)

Finally we state the central convergence result for the operator \(\chi _t\widetilde{T}_n\chi _t\) on \(L^2(\mathbb {R})\).

Lemma 3.7

([56, Lemma 4.2]) Given \(t\in \mathbb {R}\), the operator \(\chi _t\widetilde{T}_n\chi _t\) converges in trace norm on \(L^2(\mathbb {R})\) and in \(L^p(\mathbb {R})\) operator norm with \(p\in \{1,2,\infty \}\) to the operator \(\chi _tT\chi _t\). Additionally, for any \(\gamma \in [0,1]\),

$$\begin{aligned} (1-\bar{\gamma }\chi _t\widetilde{T}_n\chi _t)^{-1}{\mathop {\longrightarrow }\limits ^{n\rightarrow \infty }}(1-\bar{\gamma }\chi _tT\chi _t)^{-1} \end{aligned}$$
(3.24)

in \(L^p(\mathbb {R})\) operator norm with \(p\in \{1,2,\infty \}\).

Proof

The convergences have been proven for \(\gamma =1\) in [56, Lemma 4.2]. The extension to \(\gamma \in [0,1)\) follows from the Neumann series expansion of the resolvents in (3.24); compare Lemma 3.2 and [2, Lemma 2.1]. \(\square \)

We now apply Lemmas 3.5, 3.6 and Lemma 3.7 in the large n analysis of the Fredholm determinants back in (3.12), after the rescaling \(t\mapsto t+\sqrt{2n}\). First the leading factor:

Lemma 3.8

For any \(\gamma \in [0,1]\) and \(t\in \mathbb {R}\),

$$\begin{aligned} \lim _{n\rightarrow \infty }\det \big (1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t{\upharpoonright }_{L^2(\mathbb {R})}\big )=\det \big (1-\bar{\gamma }\chi _tT\chi _t{\upharpoonright }_{L^2(\mathbb {R})}\big ). \end{aligned}$$

Proof

We know from Lemma 3.2 that \(\widetilde{T}_n\) is trace class on \(L^2(t,\infty )\) and the same applies to T (since it is a product of Hilbert–Schmidt operators). Thus, with [43, Chapter IV, (5.14)],

$$\begin{aligned}&\Big |\det \big (1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t{\upharpoonright }_{L^2(\mathbb {R})}\big )-\det \big (1-\bar{\gamma }\chi _tT\chi _t{\upharpoonright }_{L^2(\mathbb {R})}\big )\Big |\\&\quad \le \bar{\gamma }\Vert \chi _t\widetilde{T}_{2n}\chi _t-\chi _tT\chi _t\Vert _{1}\exp \big (1+\bar{\gamma }\Vert \chi _t\widetilde{T}_{2n}\chi _t\Vert _1+\bar{\gamma }\Vert \chi _tT\chi _t\Vert _1\big ). \end{aligned}$$

But the operator difference in trace norm converges to zero by Lemma 3.7 and \(\Vert \chi _t\widetilde{T}_{2n}\chi _t\Vert _1\) remains bounded by the same result. This completes our proof. \(\square \)

Next, we move on to the \(L^2(\mathbb {R})\) inner products which appear in (3.14).

Lemma 3.9

For any \(\gamma \in [0,1]\) and \(t\in \mathbb {R}\),

$$\begin{aligned} \lim _{n\rightarrow \infty }\langle \widetilde{\alpha }_1,\widetilde{\beta }_1\rangle= & {} \frac{\bar{\gamma }}{2}\int _t^{\infty }G(x)\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x,\\ \lim _{n\rightarrow \infty }\langle \widetilde{\alpha }_1,\widetilde{\beta }_2\rangle= & {} \frac{\bar{\gamma }}{\sqrt{2}}\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}G\big )(t)-\frac{\bar{\gamma }}{\sqrt{2}},\\ \lim _{n\rightarrow \infty }\langle \widetilde{\alpha }_1,\widetilde{\beta }_3\rangle= & {} \frac{\bar{\gamma }}{\sqrt{2}}\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}G\big )(t). \end{aligned}$$

Proof

In the first inner product, we write

$$\begin{aligned} \langle \widetilde{\alpha }_1,\widetilde{\beta }_1\rangle =&\left\langle \chi _t\big (\widetilde{\phi }_{2n}(\infty )-\frac{1}{2\sqrt{2}}\text {erfc}\big ),\bar{\gamma }(1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\chi _t\widetilde{\psi }_{2n}\right\rangle \nonumber \\&\ \ \ \ \ \ +\left\langle \chi _t\big (\widetilde{\phi }_{2n}-\widetilde{\phi }_{2n}(\infty )+\frac{1}{2\sqrt{2}}\text {erfc}\big ),\bar{\gamma }(1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\chi _t\widetilde{\psi }_{2n}\right\rangle \end{aligned}$$
(3.25)

and now use that, uniformly in \(x\in \mathbb {R}\),

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( \widetilde{\phi }_{2n}(\infty )-\frac{1}{2\sqrt{2}}\text {erfc}(x)\right) =\frac{1}{\sqrt{2}}-\frac{1}{2\sqrt{2}}\text {erfc}(x) {\mathop {=}\limits ^{1.9}}\frac{1}{\sqrt{2}}G(x). \end{aligned}$$

But from Lemma 3.5 and Lemma 3.7, we also know that for \(p\in \{1,2\}\),

$$\begin{aligned} \lim _{n\rightarrow \infty }\left\| (1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\chi _t\widetilde{\psi }_{2n}-\frac{1}{\sqrt{2}}(1-\bar{\gamma }\chi _tT\chi _t)^{-1}\chi _tg\right\| _{L^p(\mathbb {R})}=0, \end{aligned}$$

so with (3.18) and Hölder’s inequality therefore back in (3.25)

$$\begin{aligned} \lim _{n\rightarrow \infty }\langle \widetilde{\alpha }_1,\widetilde{\beta }_1\rangle =\left\langle \chi _t\frac{1}{\sqrt{2}}G,\bar{\gamma }(1-\bar{\gamma }\chi _tT\chi _t)^{-1}\frac{1}{\sqrt{2}}g\right\rangle \\ =\frac{\bar{\gamma }}{2}\int _t^{\infty }G(x)\big ((1-\bar{\gamma }T\chi _t\upharpoonright _{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x, \end{aligned}$$

as claimed. For the second inner product, we write instead (with \(\chi _{[t,\infty )}(t)=1\))

$$\begin{aligned} \langle \widetilde{\alpha }_1,\widetilde{\beta }_2\rangle =&\bar{\gamma }\big (\widetilde{\phi }_{2n}(t)-\widetilde{\phi }_{2n}(\infty )\big )+\bar{\gamma }^2\big (\widetilde{T}_{2n}\chi _t(1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\chi _t\widetilde{\phi }_{2n}\big )(t)\\ =&\bar{\gamma }\big (\widetilde{\phi }_{2n}(t){-}\widetilde{\phi }_{2n}(\infty )\big ){+}\bar{\gamma }^2\!\int _t^{\infty }\!\!\widetilde{T}_{2n}(t,x)\big ((1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\chi _t\widetilde{\phi }_{2n}\big )(x){\mathrm d}x \end{aligned}$$

and recall the previous decomposition of \(\widetilde{\phi }_n\) used in (3.25). Hence, with Lemma 3.7 and (3.17), (3.23) we find from Hölder’s inequality,

$$\begin{aligned} \lim _{n\rightarrow \infty }\langle \widetilde{\alpha }_1, \widetilde{\beta }_2\rangle&=\frac{\bar{\gamma }}{\sqrt{2}}G(t)-\frac{\bar{\gamma }}{\sqrt{2}}+\frac{\bar{\gamma }^2}{\sqrt{2}}\int _t^{\infty }T(t,x)\big ((1-\bar{\gamma }\chi _tT\chi _t)^{-1}\chi _tG\big )(x){\mathrm d}x\\&=\frac{\bar{\gamma }}{\sqrt{2}}G(t)-\frac{\bar{\gamma }}{\sqrt{2}}+\frac{\bar{\gamma }}{\sqrt{2}}\big (\bar{\gamma }\chi _tT\chi _t(1-\bar{\gamma }\chi _tT\chi _t)^{-1}\chi _tG\big )(t)\\&=\frac{\bar{\gamma }}{\sqrt{2}}\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}G\big )(t)-\frac{\bar{\gamma }}{\sqrt{2}}. \end{aligned}$$

The derivation of the third inner product is completely analogous. \(\square \)

At this point, we are left with the computation of the large n limits of the rescaled integrals \(I_k\). Let R(xy) denote the kernel of the resolvent \(R=\bar{\gamma }T\chi _t(1-\bar{\gamma }\chi _tT\chi _t)^{-1}\) on \(L^2(\mathbb {R})\).

Lemma 3.10

For every \(\gamma \in [0,1]\) and \(t\in \mathbb {R}\),

$$\begin{aligned} \lim _{n\rightarrow \infty }I_1(t+\sqrt{2n},\gamma ,2n)= & {} \int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )R(x,t){\mathrm d}x,\\ \lim _{n\rightarrow \infty }I_2(t+\sqrt{2n},\gamma ,2n)= & {} \int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )\chi _{[t,\infty )}(x)R(x,t){\mathrm d}x,\\ \lim _{n\rightarrow \infty }I_3(t+\sqrt{2n},\gamma ,2n)= & {} \frac{1}{\sqrt{2}}\int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x-\frac{1}{\sqrt{2}},\\ \lim _{n\rightarrow \infty }I_4(t+\sqrt{2n},\gamma ,2n)= & {} \frac{1}{\sqrt{2}}\int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )\chi _{[t,\infty )}(x)\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x. \end{aligned}$$

Proof

We begin with the kernel function identity (cf. [62, page 748]),

$$\begin{aligned} \widetilde{R}_n(x,t)=&\bar{\gamma }\big (\widetilde{T}_n\chi _t(1-\bar{\gamma }\chi _t\widetilde{T}_n\chi _t)^{-1}\big )(x,t)=\big ((1-\bar{\gamma }\widetilde{T}_n\chi _t)^{-1}\bar{\gamma }\widetilde{T}_n\chi _t\big )(x,t)\\ =&\bar{\gamma }\widetilde{T}_n(x,t)+\bar{\gamma }\big (\widetilde{T}_n\chi _t(1-\bar{\gamma }\chi _t\widetilde{T}_n\chi _t)^{-1}\bar{\gamma }\widetilde{T}_n\big )(x,t),\ \ \ \ x\in \mathbb {R}, \end{aligned}$$

which, upon insertion into the integrand of \(I_1(t+\sqrt{2n},\gamma ,2n)\), leads to four integrals,

$$\begin{aligned} \int _t^{\infty }\widetilde{T}_{2n}(x,t){\mathrm d}x, \ \ \ \int _{-\infty }^t\widetilde{T}_{2n}(x,t){\mathrm d}x, \ \ \ \int _t^{\infty }\big (\widetilde{T}_{2n}\chi _t(1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\widetilde{T}_{2n}\big )(x,t){\mathrm d}x, \end{aligned}$$
(3.26)

and

$$\begin{aligned} \int _{-\infty }^t\big (\widetilde{T}_{2n}\chi _t(1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\widetilde{T}_{2n}\big )(x,t){\mathrm d}x. \end{aligned}$$

Apply Lemma 3.6 and conclude for the first two integrals

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _t^{\infty }\widetilde{T}_{2n}(x,t)&{\mathrm d}x{\mathop {=}\limits ^{3.23}}\int _t^{\infty }T(x,t){\mathrm d}x,\\ \lim _{n\rightarrow \infty }\int _{-\infty }^t\widetilde{T}_{2n}(x,t)&{\mathrm d}x=\lim _{n\rightarrow \infty }\left[ \int _{-\infty }^{\infty }\widetilde{T}_{2n}(x,t){\mathrm d}x-\int _t^{\infty }\widetilde{T}_{2n}(x,t){\mathrm d}x\right] \\&\underset{3.23}{\overset{3.22}{=}}\int _{-\infty }^tT(x,t){\mathrm d}x. \end{aligned}$$

For the third integral in (3.26), we write

$$\begin{aligned}&\int _t^{\infty }\big (\widetilde{T}_{2n}\chi _t(1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\widetilde{T}_{2n}\big )(x,t)dx\\&\quad =\left\langle \chi _{[t,\infty )}\int _t^{\infty }\widetilde{T}_{2n}(x,\cdot ){\mathrm d}x,\big ((1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\chi _t\widetilde{T}_{2n}\big )(\cdot ,t)\right\rangle , \end{aligned}$$

and note that each entry of the last \(L^2(\mathbb {R})\) inner product converges to its formal limits in \(L^2(\mathbb {R})\) sense, cf. Lemma 3.7, equation (3.23) and the workings in [56, page 1643]. The outstanding fourth integral is treated similarly, the difference being that the first entry in the corresponding \(L^2(\mathbb {R})\) inner product equals

$$\begin{aligned} \chi _{[t,\infty )}\int _{-\infty }^t\widetilde{T}_{2n}(x,\cdot ){\mathrm d}x=\chi _{[t,\infty )}\int _{-\infty }^{\infty }\widetilde{T}_{2n}(x,\cdot ){\mathrm d}x-\chi _{[t,\infty )}\int _t^{\infty }\widetilde{T}_{2n}(x,\cdot ){\mathrm d}x. \end{aligned}$$

Since both terms converge to their formal limits in \(L^2(\mathbb {R})\) sense (compare our reasoning above and (3.23)), we find all together,

$$\begin{aligned} \lim _{n\rightarrow \infty }I_1(t+\sqrt{2n},\gamma ,2n)=\int _{-\infty }^{\infty }\big (1-\gamma \chi _{[t,\infty )}(x)\big )\big (\bar{\gamma }T\chi _t(1-\bar{\gamma }\chi _tT\chi _t)^{-1}\big )(x,t){\mathrm d}x, \end{aligned}$$

which is the desired formula for \(I_1\), given that \(R=\bar{\gamma }T\chi _t(1-\bar{\gamma }\chi _tT\chi _t)^{-1}\). The derivation of the limit for \(I_2\) is completely analogous and in fact simpler since no integrals over \((-\infty ,t)\) occur. Moving ahead, the limit evaluation of \(I_3(t+\sqrt{2n},\gamma ,2n)\) also requires four integrals,

$$\begin{aligned} \int _t^{\infty }\widetilde{\psi }_{2n}(x){\mathrm d}x,\ \ \ \ \int _{-\infty }^t\widetilde{\psi }_{2n}(x){\mathrm d}x,\ \ \ \ \int _t^{\infty }\big (\widetilde{T}_{2n}\chi _t(1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\widetilde{\psi }_{2n}\big )(x){\mathrm d}x, \end{aligned}$$
(3.27)
$$\begin{aligned} \int _{-\infty }^t\big (\widetilde{T}_{2n}\chi _t(1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\widetilde{\psi }_{2n}\big )(x){\mathrm d}x. \end{aligned}$$

Note that

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _t^{\infty }\widetilde{\psi }_{2n}(x){\mathrm d}x&{\mathop {=}\limits ^{(3.18)}}\frac{1}{\sqrt{2}}\int _t^{\infty }g(x){\mathrm d}x,\\ \lim _{n\rightarrow \infty }\int _{-\infty }^t\widetilde{\psi }_{2n}(x){\mathrm d}x&=\lim _{n\rightarrow \infty }\left[ \int _{-\infty }^{\infty }\widetilde{\psi }_{2n}(x){\mathrm d}x-\int _t^{\infty }\widetilde{\psi }_{2n}(x){\mathrm d}x\right] \\&=-\lim _{n\rightarrow \infty }\int _t^{\infty }\widetilde{\psi }_{2n}(x){\mathrm d}x, \end{aligned}$$

since \(\psi _{2n}\) is an odd function. Also

$$\begin{aligned}&\int _t^{\infty }\big (\widetilde{T}_{2n}\chi _t(1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\widetilde{\psi }_{2n}\big )(x){\mathrm d}x \\&\quad =\left\langle \chi _{[t,\infty )}\int _t^{\infty }\widetilde{T}_{2n}(x,\cdot ){\mathrm d}x,\big ((1-\bar{\gamma }\chi _t\widetilde{T}_{2n}\chi _t)^{-1}\chi _t\widetilde{\psi }_{2n}\big )(\cdot )\right\rangle , \end{aligned}$$

which converges to its formal limit as \(n\rightarrow \infty \); compare our reasoning for \(I_1\) and (3.18). The same is true for the remaining fourth integral and we obtain all together, as \(n\rightarrow \infty \),

$$\begin{aligned}&I_3(t+\sqrt{2n},\gamma ,2n)\rightarrow (1-\gamma )\frac{1}{\sqrt{2}}\int _t^{\infty }\big ((1+R)g\big )(x){\mathrm d}x\\&\quad +\frac{1}{\sqrt{2}}\int _{-\infty }^t\big ((1+R)g\big )(x){\mathrm d}x-\frac{1}{\sqrt{2}}\int _{-\infty }^{\infty }g(x){\mathrm d}x, \end{aligned}$$

which is the claimed identity. The derivation for \(I_4\) is again similar and does not use any integrals along \((-\infty ,t)\). This completes our proof. \(\square \)

With Lemma 3.8, 3.9 and 3.10 in place, we now obtain the following result.

Proposition 3.11

As \(n\rightarrow \infty \), uniformly for \(t\in \mathbb {R}\) chosen from compact subsets and any \(\gamma \in [0,1]\),

$$\begin{aligned}&F_{2n}(t+\sqrt{2n},\gamma )\rightarrow \det \big (1-\bar{\gamma }\chi _tT\chi _t{\upharpoonright }_{L^2(\mathbb {R})}\big )\nonumber \\&\quad \det \begin{bmatrix}1-u &{} -v+\frac{\bar{\gamma }}{\sqrt{2}} &{} -v\\ -\frac{\gamma }{2\bar{\gamma }}(p+q-\frac{1}{\sqrt{2}}) &{} 1-\frac{\gamma }{2\bar{\gamma }}(r+w) &{} -\frac{\gamma }{2\bar{\gamma }}(r+w)\\ \frac{\gamma }{\bar{\gamma }}p &{} \frac{\gamma }{\bar{\gamma }}r &{} 1+\frac{\gamma }{\bar{\gamma }}r\end{bmatrix}, \end{aligned}$$
(3.28)

where uvpqrw are the following six functions of \((t,\gamma )\in \mathbb {R}\times [0,1]\),

$$\begin{aligned} u:=&\frac{\bar{\gamma }}{2}\int _t^{\infty }G(x)\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x,\nonumber \\ v:=&\frac{\bar{\gamma }}{\sqrt{2}}\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}G\big )(t),\\ p:=&\frac{1-\gamma }{\sqrt{2}}\int _t^{\infty }\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x,\nonumber \\ q:=&\frac{1}{\sqrt{2}}\int _{-\infty }^t\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x,\nonumber \\ r:=&(1-\gamma )\int _t^{\infty }R(x,t){\mathrm d}x,\qquad w:=\int _{-\infty }^tR(x,t){\mathrm d}x.\nonumber \end{aligned}$$
(3.29)

3.3 The Limit (1.12) for Odd n

In this subsection, we will compute the limit \(F_{2n+1}(t+\sqrt{2n+1},\gamma )\) using a comparison argument. Precisely, we show how the computations in Subsect. 3.1 have to be modified in order to account for odd \(n\in \mathbb {Z}_{\ge 3}\). These additional manipulations are necessary given the different structure of the operator \(IS_n\) in Proposition 1.1 for odd n. The details are as follows. We first relate \(\mathbf{K}_{2n+1}\) to \(\mathbf{K}_{2n}\):

Proposition 3.12

For any \(n\in \mathbb {Z}_{\ge 3}\),

$$\begin{aligned} S_n=S_{n-1}+\psi _n\otimes \phi _n-f_n\psi _{n-1}\otimes \phi _{n+1},\ \ \ \ f_n^2:=\frac{n-1}{n-2}\sqrt{\frac{n-1}{n+1}}{\mathop {\longrightarrow }\limits ^{n\rightarrow \infty }} 1,\ \ \ f_n>0, \end{aligned}$$
(3.30)

and thus in turn,

$$\begin{aligned} IS_{2n+1}=\epsilon S_{2n}+\epsilon \psi _{2n+1}\otimes \phi _{2n+1}-\phi _{2n+1}\otimes \epsilon \psi _{2n+1},\ \ n\in \mathbb {Z}_{\ge 1}. \end{aligned}$$
(3.31)

Proof

Identity (3.30) follows from the equality

$$\begin{aligned} S_n(x,y)=S_{n-1}(x,y)+\frac{1}{\sqrt{2\pi }(n-2)!}\int _0^y(x-u)(xu)^{n-2}\mathrm {e}^{-\frac{1}{2}(x^2+u^2)}{\mathrm d}u,\ \ n\in \mathbb {Z}_{\ge 3}, \end{aligned}$$
(3.32)

which appears in [56, page 1628] and which can be proven by induction on \(n\in \mathbb {Z}_{\ge 3}\) using the original definition of \(S_n(x,y)\) given in Proposition 1.1. Once (3.32) is known we find immediately (3.30) by comparison with (3.9). On the other hand,

$$\begin{aligned} (\epsilon S_{2n+1})(x,y)&=(\epsilon S_{2n})(x,y)-\frac{1}{\sqrt{2\pi }(2n-1)!}\int _0^xz^{2n}\mathrm {e}^{-\frac{1}{2}z^2}{\mathrm d}z\int _0^yu^{2n-1}\mathrm {e}^{-\frac{1}{2}u^2}{\mathrm d}u\nonumber \\&\quad -\frac{1}{\sqrt{2\pi }(2n-1)!}\int _x^{\infty }z^{2n-1}\mathrm {e}^{-\frac{1}{2}z^2}{\mathrm d}z\int _0^yu^{2n}\mathrm {e}^{-\frac{1}{2}u^2}{\mathrm d}u\nonumber \\&=(\epsilon S_{2n})(x,y)+(\epsilon \psi _{2n+1})(x)\phi _{2n+1}(y)\nonumber \\&\quad -\phi _{2n+1}(x)(\epsilon \psi _{2n+1})(y)-\frac{2^{n-1}}{\sqrt{2\pi }}\frac{\Gamma (n)}{\Gamma (2n)}\int _0^yu^{2n}\mathrm {e}^{-\frac{1}{2}u^2}{\mathrm d}u,\ \ \ \ n\in \mathbb {Z}_{\ge 1}, \end{aligned}$$
(3.33)

which used (3.9) and

$$\begin{aligned} (\epsilon \psi _{2n+1})(x)=-\int _0^x\psi _{2n+1}(y){\mathrm d}y, \end{aligned}$$

in the last equality. However, by the Legendre duplication formula [54, 5.5.5], for any \(n\in \mathbb {Z}_{\ge 1}\),

$$\begin{aligned} \frac{2^{n-1}}{\sqrt{2\pi }}\frac{\Gamma (n)}{\Gamma (2n)}=\frac{1}{2^{n+\frac{1}{2}}\Gamma (n+\frac{1}{2})}, \end{aligned}$$

so (3.33) yields

$$\begin{aligned}&(\epsilon S_{2n+1})(x,y)+\frac{1}{2^{n+\frac{1}{2}}\Gamma (n+\frac{1}{2})}\int _0^yu^{2n}\mathrm {e}^{-\frac{1}{2}u^2}{\mathrm d}u\nonumber \\&\quad =\big (\epsilon S_{2n}+\epsilon \psi _{2n+1}\otimes \phi _{2n+1}-\phi _{2n+1}\otimes \epsilon \psi _{2n+1}\big )(x,y), \end{aligned}$$

and this is (3.31) after comparison with the kernel of \(IS_{2n+1}\) written in Proposition 1.1. \(\square \)

Inserting (3.30) and (3.31) into formula (1.2) for \(\mathbf{K}_{2n+1}\), we find that \(\mathbf{K}_{2n+1}=\mathbf{K}_{2n}+\mathbf{E}_{2n}\) where the operator \(\mathbf{E}_n\) has kernel

$$\begin{aligned} \mathbf{E}_n:=\begin{bmatrix}\rho ^{-1}(\psi _{n+1}\otimes \phi _{n+1}-f_{n+1}\psi _n\otimes \phi _{n+2})\rho &{} \rho ^{-1}(D\phi _{n+1}\otimes \psi _{n+1}-f_{n+1}D\phi _{n+2}\otimes \psi _n)\rho ^{-1}\\ \rho (\epsilon \psi _{n+1}\otimes \phi _{n+1}-\phi _{n+1}\otimes \epsilon \psi _{n+1})\rho &{} \rho (\phi _{n+1}\otimes \psi _{n+1}-f_{n+1}\phi _{n+2}\psi _n)\rho ^{-1}\end{bmatrix}. \end{aligned}$$

Note that \(\chi _t\mathbf{E}_n\chi _t\) is finite rank on \(L^2(\mathbb {R})\oplus L^2(\mathbb {R})\), so in particular trace class. Also, since for any \(x\in \mathbb {R}\),

$$\begin{aligned} (\epsilon \psi _{n+1})(x)&=-\underbrace{\sqrt{\frac{n}{\sqrt{(n+1)(n+2)}}}}_{\rightarrow 1,\ n\rightarrow \infty }\phi _{n+2}(x),\\ (D\phi _{n+1})(x)&=\underbrace{\sqrt{\frac{\sqrt{n(n+1)}}{n-1}}}_{\rightarrow 1\ n\rightarrow \infty }\psi _n(x),\ \ \ \ \ \ n\in \mathbb {Z}_{\ge 2}, \end{aligned}$$

Lemma 3.5 and triangle inequality yield that, in trace norm,

$$\begin{aligned} \Vert \chi _{t+\sqrt{n}}\mathbf{E}_n\chi _{t+\sqrt{n}}\Vert _1\rightarrow 0\ \ \ \text {as}\ \ n\rightarrow \infty . \end{aligned}$$
(3.34)

But \(1-\gamma \chi _{t+\sqrt{2n}}\mathbf{K}_{2n}\chi _{t+\sqrt{2n}}\) is invertible for sufficiently large n and any \((t,\gamma )\in \mathbb {R}\times [0,1]\) by the working of Sect. 3 and Remark 1.2. Hence, we use (1.4) and obtain for \(n\ge n_0\),

$$\begin{aligned}&F_{2n+1}(t+\sqrt{2n},\gamma )\\&\quad =\det _2(1-\gamma \chi _{t+\sqrt{2n}}\mathbf{K}_{2n}\chi _{t+\sqrt{2n}}-\gamma \chi _{t+\sqrt{2n}}\mathbf{E}_{2n}\chi _{t+\sqrt{2n}}{\upharpoonright }_{L^2(\mathbb {R})\oplus L^2(\mathbb {R})})\\&\quad = F_{2n}(t+\sqrt{2n},\gamma )\det \big (1-\gamma (1-\gamma \chi _{t+\sqrt{2n}}\mathbf{K}_{2n}\chi _{t+\sqrt{2n}})^{-1}\\&\qquad \quad \chi _{t+\sqrt{2n}}\mathbf{E}_{2n}\chi _{t+\sqrt{2n}}{\upharpoonright }_{L^2(\mathbb {R})\oplus L^2(\mathbb {R})}\big ), \end{aligned}$$

where the second (finite rank) determinant converges to one as \(n\rightarrow \infty \) because of (3.34). This shows that

$$\begin{aligned} \lim _{n\rightarrow \infty }F_{2n+1}(t+\sqrt{2n},\gamma )=&\lim _{n\rightarrow \infty }F_{2n}(t+\sqrt{2n},\gamma ) \end{aligned}$$
(3.35)

for any \((t,\gamma )\in \mathbb {R}\times [0,1]\). In fact, the above convergence is uniform in \((t,\gamma )\in \mathbb {R}\times [0,1]\) chosen from compact subsets and since \(F_{2n+1}(t+\sqrt{2n},\gamma )\) is at least differentiable in \(t\in \mathbb {R}\) (this can be seen directly from (1.1) by scaling t into the kernel and then using the logic behind [1, Lemma 2.20]), we find

$$\begin{aligned} F_{2n+1}(t+\sqrt{2n},\gamma )=F_{2n+1}(t+\sqrt{2n+1},\gamma )+o(1),\ \ \ n\rightarrow \infty \end{aligned}$$
(3.36)

on compact subsets of \((t,\gamma )\in \mathbb {R}\times [0,1]\). Hence, combining (3.35) with (3.36) we arrive at the analogue of (3.28) for odd n, i.e.,

Proposition 3.13

Proposition 3.11 holds with \(F_{2n}(t+\sqrt{2n},\gamma )\) in the left-hand side of (3.28) replaced by \(F_{2n+1}(t+\sqrt{2n+1},\gamma )\).

Finally, merging Propositions 3.11 and 3.13 we have now established the existence of the limit (1.12). This completes the current section.

4 Proof of Theorem 1.8—final steps

In order to prove the outstanding representation (1.13) we now find a new representation for the \(3\times 3\) determinant in (3.28). To begin with, we list four algebraic relations between the functions uvpqr and w in Corollary 4.3. These follow from the next lemma. Recall \(R=(1-\bar{\gamma }T\chi _t\upharpoonright _{L^2(\mathbb {R})})^{-1}-1\) and the definitions of g and G in (1.9).

Lemma 4.1

For every \((t,\gamma )\in \mathbb {R}\times [0,1]\),

$$\begin{aligned} \big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}G\big )(t)=&\int _{-\infty }^t\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x,\\ \int _t^{\infty }R(x,t){\mathrm d}x=&\bar{\gamma }\int _t^{\infty }\big (1-G(x)\big )\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x,\\ \int _{-\infty }^tR(x,t){\mathrm d}x=&\bar{\gamma }\int _t^{\infty }G(x)\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x,\\ 1+\int _t^{\infty }R(x,t){\mathrm d}x=&\int _{-\infty }^{\infty }\big ((1-\bar{\gamma }T\chi _t\upharpoonright _{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x. \end{aligned}$$

Proof

The first equality follows from [2, (4.9)] with the formal replacements \(\gamma \mapsto \bar{\gamma },G^{\gamma }\mapsto G\) and \(g^{\gamma }\mapsto g\), see [2, (4.3)]. The second and third are a consequence of (A.3). Indeed, we have

$$\begin{aligned} \int _t^{\infty }R(x,t){\mathrm d}x=&\int _0^{\infty }R(x+t,t){\mathrm d}x\\ =&\int _0^{\infty }\big (\bar{\gamma }T\chi _t(1-\bar{\gamma }\chi _tT\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}\big )(x+t,t){\mathrm d}x\\ =&\sum _{k=1}^{\infty }\bar{\gamma }^k\int _0^{\infty }(T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^k(x+t,t){\mathrm d}x, \end{aligned}$$

and, similarly,

$$\begin{aligned} \int _{-\infty }^tR(x,t){\mathrm d}x=&\int _{-\infty }^0R(x+t,t){\mathrm d}x\\ =&\int _{-\infty }^0\big (\bar{\gamma }T\chi _t(1-\bar{\gamma }\chi _tT\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}\big )(x+t,t){\mathrm d}x\\ =&\sum _{k=1}^{\infty }\bar{\gamma }^k\int _{-\infty }^0(T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^k(x+t,t){\mathrm d}x. \end{aligned}$$

Now choose \(K=T\) (which is self-adjoint since \(\phi (x)=\psi (x)=g(x)\) in (A.1)) and \(I=(0,\infty )\) in (A.3), so that

$$\begin{aligned} \int _t^{\infty }R(x,t){\mathrm d}x=&\sum _{k=1}^{\infty }\bar{\gamma }^k\int _t^{\infty }\left[ \int _x^{\infty }g(v){\mathrm d}v\right] \big ((T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{k-1}g\big )(x){\mathrm d}x\\ =&\bar{\gamma }\int _t^{\infty }\big (1-G(x)\big )\big ((1-\bar{\gamma }T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x, \end{aligned}$$

which is the second integral identity. For the third, we simply choose \(I=(-\infty ,0)\) in (A.3), and for the fourth we use (A.5), self-adjointness of T and \(\int _{-\infty }^{\infty }g(x){\mathrm d}x=1\) to find that

$$\begin{aligned} \int _t^{\infty }R(x,t){\mathrm d}x=&\int _t^{\infty }R(t,x){\mathrm d}x=\int _0^{\infty }R(t,x+t){\mathrm d}x\\&=\sum _{k=1}^{\infty }\bar{\gamma }^k\int _0^{\infty }(T\chi _t\upharpoonright _{L^2(\mathbb {R})})^k(t,x+t){\mathrm d}x\\ {\mathop {=}\limits ^{(A.5)}}&\sum _{k=1}^{\infty }\bar{\gamma }^k\int _{-\infty }^{\infty }\big ((T\chi _t\upharpoonright _{L^2(\mathbb {R})})^kg\big )(u){\mathrm d}u\\&=-1+\int _{-\infty }^{\infty }\big ((1-\bar{\gamma }T\chi _t\upharpoonright _{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x. \end{aligned}$$

This completes our proof. \(\square \)

Remark 4.2

The first and third integral identities in Lemma 4.1 are the \(\bar{\gamma }\)-generalizations of the equalities [55, (2.6),(2.8),(2.10)] and [55, (2.3),(2.9)]. The second and fourth identities are seemingly new.

Corollary 4.3

For any \((t,\gamma )\in \mathbb {R}\times [0,1]\),

$$\begin{aligned} v=\bar{\gamma }q, \ \ \ \ \ r=-2(1-\gamma )u+\bar{\gamma }\sqrt{2}p, \ \ \ \ \ w=2u,\ \ \ \ \ \ \ 2u+\sqrt{2}q+\sqrt{2}\left( \frac{1-\bar{\gamma }}{1-\gamma }\right) p=1. \end{aligned}$$
(4.1)

Proof

These follow from inserting the integral identities of Lemma 4.1 into the definitions of uvpqr and w. \(\square \)

Once we substitute (4.1) into the \(3\times 3\) determinant (3.28) we are left with two unknown, p and q, say, and the determinant simplifies to

$$\begin{aligned} \det&\begin{bmatrix}1-u &{} -v+\frac{\bar{\gamma }}{\sqrt{2}} &{} -v\\ -\frac{\gamma }{2\bar{\gamma }}(p+q-\frac{1}{\sqrt{2}}) &{} 1-\frac{\gamma }{2\bar{\gamma }}(r+w) &{} -\frac{\gamma }{2\bar{\gamma }}(r+w)\\ \frac{\gamma }{\bar{\gamma }}p &{} \frac{\gamma }{\bar{\gamma }}r &{} 1+\frac{\gamma }{\bar{\gamma }}r\end{bmatrix}\nonumber \\&=\frac{1}{2(2-\gamma )}\left[ (2-\gamma )\left( p+q+\frac{1}{\sqrt{2}}\right) ^2-\gamma \left( p-q+\frac{1}{\sqrt{2}}\right) ^2\right] . \end{aligned}$$
(4.2)

Next, we define the two functions

$$\begin{aligned} \tau _k=\tau _k(t,\bar{\gamma })&:=\int _{-\infty }^{\infty }\big ((1-\bar{\gamma }T\chi _t\upharpoonright _{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x+(-1)^{k-1}\sqrt{\bar{\gamma }}\\&\quad \int _t^{\infty }\big ((1-\bar{\gamma }T\chi _t\upharpoonright _{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x \end{aligned}$$

for \((t,\gamma )\in \mathbb {R}\times [0,1],k=1,2\) and note that by (3.29)

$$\begin{aligned} p=\frac{1-\gamma }{2\sqrt{2\bar{\gamma }}}\big (\tau _1-\tau _2\big ),\ \ \ \ \ q=\frac{1}{2\sqrt{2\bar{\gamma }}}\Big [\sqrt{\bar{\gamma }}\big (\tau _1+\tau _2\big )-\big (\tau _1-\tau _2\big )\Big ]. \end{aligned}$$
(4.3)

Inserting (4.3) in (4.2), we find in turn

$$\begin{aligned} \text {RHS in}\ (4.2)=\frac{(1-\gamma )(1+\tau _1\tau _2)+\tau _1+\tau _2-\sqrt{\bar{\gamma }}(\tau _1-\tau _2)}{2(2-\gamma )}, \end{aligned}$$
(4.4)

and now set out to simplify \(\tau _k\). First, by the second and fourth identity in Lemma 4.1,

$$\begin{aligned} \tau _k&=1+(-1)^{k-1}\sqrt{\bar{\gamma }}\int _t^{\infty }\left[ 1+(-1)^{k-1}\sqrt{\bar{\gamma }}\int _x^{\infty }g(y){\mathrm d}y\right] \nonumber \\&\quad \big ((1-\bar{\gamma }T\chi _t\upharpoonright _{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x. \end{aligned}$$
(4.5)

Second, making essential use of the regularization scheme for Fredholm determinant and inner product manipulations in [62, Section VIII], we have the following two analogues of [37, (4.18),(4.21)] which we will use with \(a=\pm \sqrt{\bar{\gamma }}\).

Lemma 4.4

For any \((t,a)\in \mathbb {R}\times [-1,1]\),

$$\begin{aligned} 1-a\int _t^{\infty }\left[ 1-a\int _x^{\infty }g(y){\mathrm d}y\right] \big ((1-a^2 T\chi _t\upharpoonright _{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x\\ =\langle \chi _0,(1+aS_t\upharpoonright _{L^2(0,\infty )})^{-1}\delta _0\rangle _{L^2(0,\infty )}, \end{aligned}$$

where, for any test function f,

$$\begin{aligned} \int _0^{\infty }f(x)\delta _0(x){\mathrm d}x:=f(0),\chi _0(x):={\left\{ \begin{array}{ll}1,&{}x\ge 0\\ 0,&{}x<0\end{array}\right. }, \end{aligned}$$

and \(S_t:L^2(0,\infty )\rightarrow L^2(0,\infty )\) denotes the trace class integral operator on \(L^2(0,\infty )\) with kernel

$$\begin{aligned} S_t(x,y):=\frac{1}{\sqrt{\pi }}\mathrm {e}^{-(x+y+t)^2},\ \ \ x,y\ge 0. \end{aligned}$$

Proof

Note that

$$\begin{aligned} 1-a\int _{x+t}^{\infty }g(y){\mathrm d}y=1-a\int _0^{\infty }S_t(x,y){\mathrm d}y=\big ((1-aS_t\upharpoonright _{L^2(0,\infty )})\chi _0\big )(x). \end{aligned}$$

On the other hand, if \(T_t:L^2(0,\infty )\rightarrow L^2(0,\infty )\) has kernel \(T_t(x,y):=T(x+t,y+t)\), then

$$\begin{aligned} \big ((1-a^2T\chi _t\upharpoonright _{L^2(\mathbb {R})})^{-1}g\big )(x+t)=\big ((1-a^2 T_t\upharpoonright _{L^2(0,\infty )})^{-1}S_t\delta _0\big )(x) \end{aligned}$$

and we have \(T_t=S_tS_t\). Thus, without explicitly writing the underlying Hilbert spaces,

$$\begin{aligned} \int _t^{\infty }&\bigg [1-a\int _x^{\infty }g(y){\mathrm d}y\bigg ]\big ((1-a^2T\chi _t\upharpoonright _{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x\\&=\big \langle (1-aS_t)\chi _0,(1-a^2T_t)^{-1}S_t\delta _0,\big \rangle _{L^2(0,\infty )}=\big \langle \chi _0,(1+aS_t)^{-1}S_t\delta _0\big \rangle _{L^2(0,\infty )} \end{aligned}$$

by self-adjointness of \(S_t\) and [2, Lemma 6.1]. Now using that \((1+aS_t)^{-1}=1-a(1+aS_t)^{-1}S_t\), we obtain at once the claimed identity from \(\langle \chi _0,\delta _0\rangle _{L^2(0,\infty )} =1\). \(\square \)

Lemma 4.5

For every \((t,a)\in \mathbb {R}\times [-1,1]\),

$$\begin{aligned} \det (1-aS_t\upharpoonright _{L^2(0,\infty )})=\det (1+aS_t\upharpoonright _{L^2(0,\infty )})\big \langle \chi _0,(1+aS_t\upharpoonright _{L^2(0,\infty )})^{-1}\delta _0\big \rangle _{L^2(0,\infty )}. \end{aligned}$$
(4.6)

Proof

As outlined in [2, (6.9)], identity (4.6) is equivalent to

$$\begin{aligned} \mathop {\mathrm {tr}}\limits _{L^2(0,\infty )}\Big ((1-a^2 S_t^2)^{-1}a\frac{{\mathrm d}S_t}{{\mathrm d}t}\Big )=-\frac{1}{2}\frac{{\mathrm d}}{{\mathrm d}t}\ln \big \langle \delta _0,(1+aS_t)^{-1}\chi _0\big \rangle _{L^2(0,\infty )}, \end{aligned}$$

and thus to

$$\begin{aligned} \big \langle \delta _0,(1-a^2S_t^2)^{-1}aS_t\delta _0\rangle _{L^2(0,\infty )}=\frac{{\mathrm d}}{{\mathrm d}t}\ln \big \langle \delta _0,(1+aS_t)^{-1}\chi _0\big \rangle _{L^2(0,\infty )}, \end{aligned}$$
(4.7)

where we do not indicate the underlying Hilbert spaces for compact notation. In proving (4.7), we use the following straightforward a-generalization of [2, (6.11)],

$$\begin{aligned} \frac{{\mathrm d}}{{\mathrm d}t}(1+aS_t)^{-1}=(1-a^2S_t^2)^{-1}aS_tD+(1-a^2S_t^2)^{-1}aS_t\Delta _0(1+aS_t)^{-1}, \end{aligned}$$

where \(\Delta _0\) denotes multiplication by \(\delta _0(x)\) and \(D(=\frac{{\mathrm d}}{{\mathrm d}x})\) differentiation. We have thus

$$\begin{aligned} \frac{{\mathrm d}}{{\mathrm d}t}\ln \big \langle \delta _0,&(1+aS_t)^{-1}\chi _0\big \rangle _{L^2(0,\infty )}=\frac{\langle \delta _0,\frac{{\mathrm d}}{{\mathrm d}t}(1+aS_t)^{-1}\chi _0\rangle _{L^2(0,\infty )}}{\langle \delta _0,(1+aS_t)^{-1}\chi _0\rangle _{L^2(0,\infty )}}\\&{\mathop {=}\limits ^{D\chi _0=0}}\frac{\langle \delta _0,(1-a^2S_t^2)^{-1}aS_t\Delta _0(1+aS_t)^{-1}\chi _0\rangle }{\langle \delta _0,(1+aS_t)^{-1}\chi _0\rangle }\\&=\frac{\langle \delta _0,(1-a^2 S_t^2)^{-1}aS_t\delta _0\rangle \langle \delta _0,(1+aS_t)^{-1}\chi _0\rangle }{\langle \delta _0,(1+aS_t)^{-1}\chi _0\rangle }\\&=\big \langle \delta _0,(1-a^2S_t^2)^{-1}aS_t\delta _0\big \rangle _{L^2(0,\infty )}, \end{aligned}$$

i.e., identity (4.7). This completes our proof. \(\square \)

Hence, given that \(\det (1\mp \sqrt{\bar{\gamma }} S_t\upharpoonright _{L^2(0,\infty )})=\det (1\mp \sqrt{\bar{\gamma }}\chi _tS\chi _t\upharpoonright _{L^2(\mathbb {R})})>0\) with \(S:L^2(\mathbb {R})\rightarrow L^2(\mathbb {R})\) as in the formulation of Theorem 1.8, we obtain the following result from Lemma 4.4 and (4.6) with \(a=\pm \sqrt{\bar{\gamma }}\).

Proposition 4.6

For any \((t,\gamma )\in \mathbb {R}\times [0,1]\),

$$\begin{aligned} \tau _1(t,\bar{\gamma })=\frac{\det (1+\sqrt{\bar{\gamma }}\chi _tS\chi _t\upharpoonright _{L^2(\mathbb {R})})}{\det (1-\sqrt{\bar{\gamma }}\chi _tS\chi _t\upharpoonright _{L^2(\mathbb {R})})},\ \ \ \ \ \ \tau _2(t,\bar{\gamma })=\frac{1}{\tau _1(t,\bar{\gamma })} \end{aligned}$$
(4.8)

We now return to (4.4) and first use that \(\tau _1\tau _2=1\), so after simplification

$$\begin{aligned} \sqrt{\text {RHS in}\ 4.2}=\sqrt{\frac{1-\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\sqrt{\tau _1}+\sqrt{\frac{1+\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\sqrt{\tau _2}. \end{aligned}$$
(4.9)

But since

$$\begin{aligned} \det \big (1-\bar{\gamma }\chi _tT\chi _t\upharpoonright _{L^2(\mathbb {R})}\big )=\det \big (1-\sqrt{\bar{\gamma }}\chi _tS\chi _t\upharpoonright _{L^2(\mathbb {R})}\big )\det \big (1+\sqrt{\bar{\gamma }}\chi _tS\chi _t\upharpoonright _{L^2(\mathbb {R})}\big ), \end{aligned}$$
(4.10)

we then find (1.13) from (1.10), Propositions 3.11, 3.13 and equations (4.8), (4.9), (4.10). This completes our proof of Theorem 1.8.

5 Proof of Theorem 1.9

Our proof begins with the following analogue of [37, (4.12)].

Lemma 5.1

For any \((t,\gamma )\in \mathbb {R}\times [0,1]\), we have with \(\mu (t;\gamma )\) as in (1.15),

$$\begin{aligned} \mathrm {e}^{-\mu (t;\bar{\gamma })}=\tau _2(t,\bar{\gamma }),\ \ \ \ \ \ \ \ \ \ \mathrm {e}^{\mu (t;\bar{\gamma })}=\tau _1(t,\bar{\gamma }). \end{aligned}$$
(5.1)

Proof

By definition of p and u in Proposition 3.11,

$$\begin{aligned} \tau _2(t,\bar{\gamma })=&1-\sqrt{\bar{\gamma }}\big (1-\sqrt{\bar{\gamma }}\big )\int _t^{\infty }\big ((1-\bar{\gamma }T\chi _t\upharpoonright _{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x\\&-\bar{\gamma }\int _t^{\infty }G(x)\big ((1-\bar{\gamma }T\chi _t\upharpoonright _{L^2(\mathbb {R})})^{-1}g\big )(x){\mathrm d}x\\ =&1-\sqrt{2\bar{\gamma }}\left( \frac{1-\sqrt{\bar{\gamma }}}{1-\gamma }\right) p-2u. \end{aligned}$$

But with the formal replacement \(\gamma \mapsto \bar{\gamma }\) in [2, Section 4], we have

$$\begin{aligned} u=\frac{1}{2}\big (1+\sqrt{\bar{\gamma }}\sinh \mu (t;\bar{\gamma })-\cosh \mu (t;\bar{\gamma })\big ),\ \ \ \ (t,\gamma )\in \mathbb {R}\times [0,1], \end{aligned}$$
(5.2)

see [2, (4.19)], where (compare (1.15) and [2, Proposition 3.10]Footnote 1)

$$\begin{aligned} \mu (t;\gamma )=\int _t^{\infty }Y_1^{12}\left( \frac{x}{2},\gamma \right) {\mathrm d}x=-\frac{\mathrm {i}}{2}\int _t^{\infty }y\left( \frac{x}{2};\gamma \right) {\mathrm d}x. \end{aligned}$$

On the other hand, from [2, (4.11), (4.18)] after dividing out \(\sqrt{2\gamma }\) and replacing \(\gamma \mapsto \bar{\gamma }\),

$$\begin{aligned} q=\frac{1}{\sqrt{2}}\left( \cosh \mu (t;\bar{\gamma })-\frac{1}{\sqrt{\bar{\gamma }}}\sinh \mu (t;\bar{\gamma })\right) ,\ \ \ \ (t,\gamma )\in \mathbb {R}\times [0,1], \end{aligned}$$
(5.3)

so that with (5.2) and (5.3) back in the fourth equation in (4.1),

$$\begin{aligned} p=\frac{1-\gamma }{\sqrt{2\bar{\gamma }}}\sinh \mu (t;\bar{\gamma }),\ \ \ \ (t,\gamma )\in \mathbb {R}\times [0,1]. \end{aligned}$$
(5.4)

Thus, all together,

$$\begin{aligned} \tau _2(t,\bar{\gamma })=1-\sqrt{2\bar{\gamma }}\left( \frac{1-\sqrt{\bar{\gamma }}}{1-\gamma }\right) p-2u {\mathop {=}\limits ^{[5.2]{5.4}}}\mathrm {e}^{-\mu (t;\bar{\gamma })}, \end{aligned}$$

which is the analogue of [37, (4.12)]). The outstanding formula for \(\tau _1(t,\bar{\gamma })\) follows from (4.8). \(\square \)

In order to arrive at (1.14), we now apply (3.28), Proposition 3.13 and (4.9),

$$\begin{aligned} P(t;\gamma )=\sqrt{\det \big (1-\bar{\gamma }\chi _tT\chi _t\upharpoonright _{L^2(\mathbb {R})}\big )}\left( \sqrt{\frac{1-\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\sqrt{\tau _1}+\sqrt{\frac{1+\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\sqrt{\tau _2}\right) ,\\ \ \ (t,\gamma )\in \mathbb {R}\times [0,1]. \end{aligned}$$

But for any \(a\in [0,1]\), see [2, (3.33)],

$$\begin{aligned} \det (1-a\chi _tT\chi _t\upharpoonright _{L^2(\mathbb {R})})=\exp \left[ -\frac{1}{4}\int _t^{\infty }(x-t)\Big |y\Big (\frac{x}{2};a\Big )\Big |^2{\mathrm d}x\right] , \end{aligned}$$
(5.5)

so with (5.1),

$$\begin{aligned} P(t;\gamma )= & {} \exp \left[ -\frac{1}{8}\int _t^{\infty }(x-t)\Big |y\left( \frac{x}{2};\bar{\gamma }\right) \Big |^2{\mathrm d}x\right] \\&\left( \sqrt{\frac{1-\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\mathrm {e}^{\frac{1}{2}\mu (t;\bar{\gamma })}+\sqrt{\frac{1+\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\mathrm {e}^{-\frac{1}{2}\mu (t;\bar{\gamma })}\right) . \end{aligned}$$

This is exactly (1.14).

6 Proof of Corollary 1.11

Note that with the abbreviation (1.12), for any \(\gamma \in [0,1]\),

$$\begin{aligned} P(t;\gamma )= & {} \lim _{n\rightarrow \infty }\mathbb {P}\big (\text {no edge scaled eigenvalues}\ \mu _j^{\gamma }(\mathbf{X})\ \text {in}\ (t,\infty )\big )\nonumber \\= & {} \sum _{m=0}^{\infty }E(m,(t,\infty ))(1-\gamma )^m, \end{aligned}$$
(6.1)

since each eigenvalue is removed independently with likelihood \(1-\gamma \). But comparing the latter with (1.17) we find immediately (1.18). Note also that since \(\mu (x;\gamma )\) is in fact real analytic in \(x\in \mathbb {R}\) for any fixed \(\gamma \in [0,1]\) (see [2, Corollary 3.6] for continuity in x, real analyticity follows by a similar argument using the analytic Fredholm alternative in [64]) we obtain from Taylor’s theorem, (6.1) and (1.17),

$$\begin{aligned} E\big (m,(t,\infty )\big )=\frac{(-1)^m}{m!}\frac{\partial ^m}{\partial \xi ^m}E\big ((t,\infty );\xi \big )\bigg |_{\xi =1},\ \ \ m\in \mathbb {Z}_{\ge 0},\ \ t\in \mathbb {R}. \end{aligned}$$

This is the standard relation between the generating function and eigenvalue occupation probability known for any continuous one-dimensional statistical mechanical system, cf. [40, (8.1)].

7 Proof of Theorem 1.13 and Lemma 1.14

We prove (1.19), (1.20) and Lemma 1.14 in the upcoming three subsections.

7.1 Right Tail Asymptotics—Proof of (1.19)

From (5.5), i.e., [2, (3.33)],

$$\begin{aligned} \exp \left[ -\frac{1}{8}\int _t^{\infty }(x-t)\Big |y\Big (\frac{x}{2};\bar{\gamma }\Big )\Big |^2{\mathrm d}x\right] =\sqrt{\det (1-\bar{\gamma }\chi _tT\chi _t{\upharpoonright }_{L^2(\mathbb {R})})}, \end{aligned}$$
(7.1)

and thus from [2, Lemma 3.11], as \(t\rightarrow +\infty \),

$$\begin{aligned} \exp \left[ -\frac{1}{8}\int _t^{\infty }(x-t)\Big |y\Big (\frac{x}{2};\bar{\gamma }\Big )\Big |^2{\mathrm d}x\right] =1-\frac{\bar{\gamma }}{\sqrt{2\pi }}\frac{\text {erfc}(\sqrt{2}t)}{16t}\big (1+\mathcal {O}(t^{-2})\big ), \end{aligned}$$
(7.2)

uniformly in \(\gamma \in [0,1]\). Moreover, from [2, (3.31), Proposition 3.10], as \(x\rightarrow +\infty \),

$$\begin{aligned} y\left( \frac{x}{2};a\right) =2\mathrm {i}\sqrt{\frac{a}{\pi }}\mathrm {e}^{-x^2}\Big (1+\mathcal {O}\big (\mathrm {e}^{-x^2}\big )\Big ),\ \ \ a\in [0,1], \end{aligned}$$

so that in (1.15), as \(t\rightarrow +\infty \),

$$\begin{aligned} \mu (t;\bar{\gamma })=\frac{\sqrt{\bar{\gamma }}}{2}\text {erfc}(t)+\mathcal {O}\big (t^{-1}\mathrm {e}^{-2t^2}\big ), \end{aligned}$$
(7.3)

uniformly in \(\gamma \in [0,1]\). Inserting (7.3) into (1.16) and combining the so-obtained result with (7.2) yields immediately (1.19).

7.2 Left Tail Asymptotics—Proof of (1.20)

From [2, Proposition 5.7], as \(t\rightarrow -\infty \) for any fixed \(a\in [0,1)\),

$$\begin{aligned} -\frac{1}{8}\int _t^{\infty }(x-t)\left| y\left( \frac{x}{2};a\right) \right| ^2{\mathrm d}x=\frac{t}{2\sqrt{2\pi }}\text {Li}_{\frac{3}{2}}(a)+D_1(a)+o(1), \end{aligned}$$
(7.4)

with the polylogarithm \(\text {Li}_s(z)\), [54, 25.12.10] and an unknown, t-independent, term \(D_1(a)\). Moreover, from [2, page 492], as \(t\rightarrow -\infty \) and fixed \(a\in [0,1)\),

$$\begin{aligned} \mu (t;a)=D_2(a)+o(1), \end{aligned}$$
(7.5)

with another unknown, t-independent, term \(D_2(a)\). Thus combining (7.4) and (7.5) in the right-hand side of (1.14) we find that for \(\gamma \in [0,1)\), as \(t\rightarrow -\infty \),

$$\begin{aligned} \ln P(t;\gamma )=\frac{t}{2\sqrt{2\pi }}\text {Li}_{\frac{3}{2}}(\bar{\gamma })+\eta (\gamma )+o(1),\ \ \ \gamma \in [0,1), \end{aligned}$$
(7.6)

where \(\eta (\gamma )\) is, as of now, unknown. Since \(\eta (\gamma )\) comes from \(D_1(\bar{\gamma })\) and \(D_2(\bar{\gamma })\), we split its computation into two parts.

7.2.1 Total Integral Computation

We first address the computation of \(D_2(\bar{\gamma })\). Since

$$\begin{aligned} \mu (t;\gamma )&=\int _t^{\infty }Y_1^{12}\left( \frac{x}{2},\gamma \right) {\mathrm d}x{\mathop {=}\limits ^{7.5}}\int _{-\infty }^{\infty }Y_1^{12}\left( \frac{x}{2},\gamma \right) {\mathrm d}x+o(1),\\&\quad t\rightarrow -\infty ,\ \gamma \in [0,1), \end{aligned}$$

we need to evaluate a total integral. In order to achieve this, we follow the approach developed in [4], our net result being an analogue of [4, (28)]. Recall that \(\mathbf{Y}(z)=\mathbf{Y}(z;x,\gamma )\) solves RHP 1.3.

Lemma 7.1

The well-defined and invertible limit

$$\begin{aligned} \mathbf{V}(x,\gamma ):=\lim _{\begin{array}{c} z\rightarrow 0\\ \Im z<0 \end{array}}\mathbf{Y}(z;x,\gamma ),\ \ \ \ \ \ \ (x,\gamma )\in \mathbb {R}\times [0,1], \end{aligned}$$

satisfies

$$\begin{aligned} \mathbf{V}(x,\gamma )=\begin{bmatrix}\cosh \nu &{} \mathrm {i}\sinh \nu \\ -\mathrm {i}\sinh \nu &{} \cosh \nu \end{bmatrix}{} \mathbf{V}(x_0,\gamma ), \ \ \ \ \nu =\nu (x,x_0,\gamma ):=2\int _{x_0}^xY_1^{12}(u,\gamma ){\mathrm d}u, \end{aligned}$$
(7.7)

for arbitrary \(x,x_0\in \mathbb {R}\) and \(\gamma \in [0,1]\).

Proof

Define \(\mathbf{W}(z;x,\gamma ):=\mathbf{Y}(z;x,\gamma )\mathrm {e}^{-\mathrm {i}zx\sigma _3}\) for \(z\in \mathbb {C}\setminus \mathbb {R}\) with \((x,\gamma )\in \mathbb {R}\times [0,1]\) and where \(\sigma _3:=\bigl [{\begin{matrix}1 &{} 0\\ 0 &{} -1\end{matrix}}\bigr ]\). It is well-known, cf. [2, page 479], that \(\mathbf{W}=\mathbf{W}(z;x,\gamma )\) solves the Zakharov–Shabat system

$$\begin{aligned} \frac{\partial \mathbf{W}}{\partial x}=\left\{ -\mathrm {i}z\sigma _3+2\mathrm {i}\begin{bmatrix}0 &{} Y_1^{12}\\ -Y_1^{12} &{} 0\end{bmatrix}\right\} \mathbf{W},\ \ \ \ Y_1^{12}=Y_1^{12}(x,\gamma ). \end{aligned}$$

Taking the limit \(z\rightarrow 0\) with \(\Im z<0\), we find that

$$\begin{aligned} \frac{\partial \mathbf{V}}{\partial x}=2\mathrm {i}\begin{bmatrix}0 &{} Y_1^{12}\\ -Y_1^{12} &{} 0\end{bmatrix}{} \mathbf{V} \end{aligned}$$

with general solution (7.7). This completes our proof. \(\square \)

We now compute the limits of \(\cosh \nu \) and \(\sinh \nu \) as \(x\rightarrow +\infty \) and \(x_0\rightarrow -\infty \). By (7.7) these limits follow from the x-asymptotic behavior of \(\mathbf{V}(x,\gamma )\) and thus from \(\mathbf{Y}(z;x,\gamma )\). Some aspects of the asymptotic analysis of \(\mathbf{Y}(z;x,\gamma )\) were carried out in [2, Section 3.4], others can be found in Appendix 7.3.

Proposition 7.2

Let \(Y_1^{12}(x,\gamma )\) denote the (12)-entry of the matrix coefficient \(\mathbf{Y}_1(x,\gamma )\) in RHP 1.3, condition (3). Then for any fixed \(\gamma \in [0,1)\),

$$\begin{aligned} \int _{-\infty }^{\infty }Y_1^{12}(u,\gamma ){\mathrm d}u=\frac{1}{4}\ln \left( \frac{1+\sqrt{\gamma }}{1-\sqrt{\gamma }}\right) . \end{aligned}$$

Proof

Since \(Y_1^{12}(\cdot ,\gamma )\in L^1(\mathbb {R})\) for any \(\gamma \in [0,1]\), see [2, Corollary 3.6] and [2, page 481,492], we will take \(x\rightarrow +\infty \) and \(x_0\rightarrow -\infty \) in (7.7) in order to compute the desired total integral. First, consider the limit

$$\begin{aligned} \lim _{x\rightarrow +\infty }{} \mathbf{V}(x,\gamma ),\ \ \ \gamma \in [0,1]. \end{aligned}$$

By [2, (3.28)], for any \(x>0\) and \(\gamma \in [0,1]\),

$$\begin{aligned} \mathbf{V}(x,\gamma )=\mathbf{T}(0;2x,\gamma )\begin{bmatrix}1 &{} \mathrm {i}\sqrt{\gamma }\\ 0 &{} 1\end{bmatrix}, \end{aligned}$$

in terms of the solution \(\mathbf{T}(z;x,\gamma )\) of [2, RHP 3.12] evaluated at \(z=0\). But [2, (3.29),(3.30)] imply that \(\mathbf{T}(0;2x,\gamma )\rightarrow \mathbb {I}\) as \(x\rightarrow +\infty \), hence

$$\begin{aligned} \lim _{x\rightarrow +\infty }{} \mathbf{V}(x,\gamma )=\begin{bmatrix}1 &{} \mathrm {i}\sqrt{\gamma }\\ 0 &{} 1\end{bmatrix},\ \ \gamma \in [0,1]. \end{aligned}$$
(7.8)

Second, we compute

$$\begin{aligned} \lim _{x\rightarrow -\infty }{} \mathbf{V}(x,\gamma ),\ \ \ \gamma \in [0,1) \end{aligned}$$

using the results of the nonlinear steepest descent analysis in Appendix 7.3. From (B.3) and (B.5), for any \(x<0\) and \(\gamma \in [0,1)\),

$$\begin{aligned} \mathbf{V}(x,\gamma )=\mathbf{M}(0;2x,\gamma )\exp \left[ -\frac{\sigma _3}{2\pi \mathrm {i}}\text {pv}\int _{-\infty }^{\infty }h(s;2x,\gamma )\frac{{\mathrm d}s}{s}\right] \begin{bmatrix}1 &{} 0\\ \mathrm {i}\sqrt{\gamma } &{} 1\end{bmatrix}(1-\gamma )^{-\frac{1}{2}\sigma _3}, \end{aligned}$$
(7.9)

in terms of the solution \(\mathbf{M}(z;x,\gamma )\) of RHP B.4 evaluated at \(z=0\), where

$$\begin{aligned} h(s;t,\gamma )=-\ln \big (1-\gamma \mathrm {e}^{-\frac{1}{2}t^2s^2}\big ),\ \ \ s,t\in \mathbb {R} \end{aligned}$$

But since the integrand in (7.9) is an odd function of s, the principal value integral in (7.9) equals zero. Furthermore, (B.8) implies that \(\mathbf{M}(0;2x,\gamma )\rightarrow \mathbb {I}\) as \(x\rightarrow -\infty \). Thus,

$$\begin{aligned} \lim _{x\rightarrow -\infty }{} \mathbf{V}(x,\gamma )=\begin{bmatrix}1 &{} 0\\ \mathrm {i}\sqrt{\gamma } &{} 1\end{bmatrix}(1-\gamma )^{-\frac{1}{2}\sigma _3},\ \ \gamma \in [0,1). \end{aligned}$$
(7.10)

Combining (7.8) and (7.10),

$$\begin{aligned} \mathbf{V}(+\infty ,\gamma )\big (\mathbf{V}(-\infty ,\gamma )\big )^{-1}=\frac{1}{\sqrt{1-\gamma }}\begin{bmatrix} 1 &{} \mathrm {i}\sqrt{\gamma }\\ -\mathrm {i}\sqrt{\gamma } &{} 1\end{bmatrix},\ \ \gamma \in [0,1), \end{aligned}$$

which inserted into (7.7) yields

$$\begin{aligned} \cosh \big (\nu (+\infty ,-\infty ,\gamma )\big )=\frac{1}{\sqrt{1-\gamma }},\ \ \ \ \ \ \ \ \ \ \ \sinh \big (\nu (+\infty ,-\infty ,\gamma )\big )=\sqrt{\frac{\gamma }{1-\gamma }}, \end{aligned}$$

and thus after simplification (with \(Y_1^{12}\in \mathbb {R}\)) the claimed integral identity. This completes our proof. \(\square \)

With Proposition 7.2 at hand, we obtain in turn

Corollary 7.3

For every fixed \(\gamma \in [0,1)\), as \(t\rightarrow -\infty \),

$$\begin{aligned} \mu (t;\gamma )=\frac{1}{2}\ln \left( \frac{1+\sqrt{\gamma }}{1-\sqrt{\gamma }}\right) +o(1), \end{aligned}$$

and thus

$$\begin{aligned} \sqrt{\frac{1-\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\mathrm {e}^{\frac{1}{2}\mu (t;\bar{\gamma })}+\sqrt{\frac{1+\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\mathrm {e}^{-\frac{1}{2}\mu (t;\bar{\gamma })}=\sqrt{2}\sqrt{\frac{1-\gamma }{2-\gamma }}+o(1). \end{aligned}$$
(7.11)

The last corollary concludes our computation of \(D_2(\gamma )\) in (7.5).

7.2.2 Resolvent Integration

We now compute \(D_1(\bar{\gamma })\) in (7.4) using a different set of techniques. To be precise, we first recall from [2, Proposition 3.3],

$$\begin{aligned} \det (1-\gamma \chi _tT\chi _t{\upharpoonright }_{L^2(\mathbb {R})})=\det (1-\gamma T\chi _t{\upharpoonright }_{L^2(\mathbb {R})})=\det (1-G{\upharpoonright }_{L^2(\Omega )}), \end{aligned}$$
(7.12)

with the oriented contour (see Fig. 4 in Appendix 7.3)

$$\begin{aligned} \Omega =\mathbb {R}\sqcup (\mathbb {R}+\mathrm {i}\omega ), \end{aligned}$$

where \(\omega >0\) will be determined in Lemma 7.5 and \(G:L^2(\Omega ,|{\mathrm d}\lambda |)\rightarrow L^2(\Omega ,|{\mathrm d}\lambda |)\) has kernel

(7.13)

The algebraic form (7.13) of its kernel identifies the operator G as an integrable operator, cf. [45], whose resolvent \(R=-1+(1-G)^{-1}\) on \(L^2(\Omega )\), if existent, has the form (B.1). Choosing right-sided limits for definiteness, we have from (B.1) and (B.2), for \(\lambda ,\mu \in \mathbb {R}\),

$$\begin{aligned} \mathbf{F}(\lambda )=\mathbf{S}_-(\lambda )\mathbf{f}(\lambda ),\ \ \ \ \mathbf{G}(\mu )=\big (\mathbf{S}_-^{\intercal }(\mu )\big )^{-1}{} \mathbf{g}(\mu ), \end{aligned}$$
(7.14)

and for \(\lambda ,\mu \in \mathbb {R}+\mathrm {i}\omega \),

$$\begin{aligned} \mathbf{F}(\lambda )&=\mathbf{S}(\lambda )\begin{bmatrix}1 &{} 0\\ \mathrm {i}\sqrt{\gamma }\mathrm {e}^{-\frac{1}{4}\lambda ^2+\mathrm {i}t\lambda } &{} 1\end{bmatrix}{} \mathbf{f}(\lambda ),\nonumber \\ \mathbf{G}(\mu )&=\big (\mathbf{S}^{\intercal }(\mu )\big )^{-1}\begin{bmatrix}1 &{} -\mathrm {i}\sqrt{\gamma }\mathrm {e}^{-\frac{1}{4}\mu ^2+\mathrm {i}t\mu }\\ 0 &{} 1\end{bmatrix}{} \mathbf{g}(\mu ), \end{aligned}$$
(7.15)

where \(\mathbf{S}(z)\) connects to RHP 1.3 via \(\mathbf{S}(z;t,\gamma )=\mathbf{Y}(z;\frac{t}{2},\gamma ),z\in \mathbb {C}\setminus \mathbb {R}\); compare [2, (3.20)]. Next, we record the following standard differential identity.

Proposition 7.4

For any \((t,\gamma )\in \mathbb {R}\times [0,1]\),

$$\begin{aligned} \frac{\partial }{\partial \gamma }\ln \det (1-\gamma \chi _tT\chi _t{\upharpoonright }_{L^2(\mathbb {R})})=-\frac{1}{2\gamma }\int _{\Omega }R(\lambda ,\lambda ){\mathrm d}\lambda , \end{aligned}$$
(7.16)

with the kernel \(R(\lambda ,\mu )\) of the resolvent \(R=-1+(1-G)^{-1}\) on \(L^2(\Omega )\).

Proof

We know from [2, page 475] that the resolvent operator exists for any \((t,\gamma )\in \mathbb {R}\times [0,1]\), thus by straightforward differentiation of (7.12) and (7.13),

$$\begin{aligned}&\frac{\partial }{\partial \gamma }\ln \det (1-\gamma \chi _tT\chi _t{\upharpoonright }_{L^2(\mathbb {R})})\\&\quad = \frac{\partial }{\partial \gamma }\ln \det (1-G{\upharpoonright }_{L^2(\Omega )})=-\mathop {\mathrm {tr}}\limits _{L^2(\Omega )}\left( (1-G)^{-1}\frac{\partial G}{\partial \gamma }\right) \\&\quad {\mathop {=}\limits ^{7.3}} -\frac{1}{2\gamma }\mathop {\mathrm {tr}}\limits _{L^2(\Omega )}\big ((1-G)^{-1}G\big )=-\frac{1}{2\gamma }\int _{\Omega }R(\lambda ,\lambda ){\mathrm d}\lambda . \end{aligned}$$

This concludes our proof. \(\square \)

In order to apply (7.16), we use the explicit formula (B.1) for the kernel of \(R(\lambda ,\mu )\) (see [45] for regularity properties of \(R(\lambda ,\mu )\)),

$$\begin{aligned} R(\lambda ,\lambda )=\big (\mathbf{F}^{\intercal }(\lambda )\big )'\mathbf{G}(\lambda ),\ \ \ (')=\frac{\partial }{\partial \lambda },\ \ \ \ \ \lambda \in \Omega , \end{aligned}$$

and combine it with the asymptotic results of Appendix 7.3, afterward we integrate in (7.16). In more detail, once the \(t\rightarrow -\infty \) asymptotic expansion of the kernel \(R(\lambda ,\lambda )\) is known uniformly with respect to fixed \(\gamma \in [0,1)\) and any \(\lambda \in \Omega \) we simply integrate

$$\begin{aligned} \ln \det (1-\gamma \chi _tT\chi _t{\upharpoonright }_{L^2(\mathbb {R})})= & {} \int _0^{\gamma }\frac{\partial }{\partial \gamma '}\ln \det (1-\gamma '\chi _tT\chi _t{\upharpoonright }_{L^2(\mathbb {R})}){\mathrm d}\gamma '\nonumber \\= & {} -\frac{1}{2}\int _0^{\gamma }\left[ \int _{\Omega }R(\lambda ,\lambda ){\mathrm d}\lambda \right] \frac{{\mathrm d}\gamma '}{\gamma '} \end{aligned}$$
(7.17)

and arrive at (1.20) and (1.21). The detailed steps of this approach are as follows: From (B.3), (B.5) for any \((t,\gamma )\in (-\infty ,0)\times [0,1)\), provided we choose \(\omega >0\) so that \(\Sigma _\mathbf{M}\cap (\mathbb {R}+\mathrm {i}\frac{\omega }{|t|})=\emptyset \), see Lemma 7.5,

(7.18)

for \((\lambda ,\mu )\in \Omega \times \Omega \) with the unimodular factors

$$\begin{aligned} \mathbf{A}_1(\lambda ):= & {} \exp \left[ -\frac{\sigma _3}{2\pi \mathrm {i}}\text {pv}\int _{-\infty }^{\infty }\frac{h(s;1,\gamma )}{s-\lambda }{\mathrm d}s\right] \begin{bmatrix}1 &{} 0\\ \mathrm {i}\sqrt{\gamma }\mathrm {e}^{-\frac{1}{4}\lambda ^2+\mathrm {i}t\lambda } &{} 1\end{bmatrix}\\&\big (1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}\big )^{-\frac{1}{2}\sigma _3},\ \ \ \lambda \in \mathbb {R}, \end{aligned}$$

and

$$\begin{aligned} \mathbf{A}_2(\lambda )&:=\exp \left[ -\frac{\sigma _3}{2\pi \mathrm {i}}\int _{-\infty }^{\infty }\frac{h(s;1,\gamma )}{s-\lambda }{\mathrm d}s\right] \\&\quad \begin{bmatrix}(1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2})^{-1}&{}-\mathrm {i}\sqrt{\gamma }(1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2})^{-1}\mathrm {e}^{-\frac{1}{4}\lambda ^2-\mathrm {i}t\lambda }\\ \mathrm {i}\sqrt{\gamma }\mathrm {e}^{-\frac{1}{4}\lambda ^2+\mathrm {i}t\lambda } &{} 1\end{bmatrix}, \ \lambda \in \mathbb {R}+\mathrm {i}\omega . \end{aligned}$$

Inserting (7.18) into the right-hand side of (7.16), we obtain after a short computation

$$\begin{aligned} \int _{\Omega }R(\lambda ,\lambda ){\mathrm d}\lambda =\int _{\Omega }\big (\mathbf{f}^{\intercal }(\lambda )\big )'&\mathbf{g}(\lambda ){\mathrm d}\lambda +\int _{\mathbb {R}}\mathbf{f}^{\intercal }(\lambda )\big (\mathbf{A}_1^{\intercal }(\lambda )\big )'\big (\mathbf{A}_1^{\intercal }(\lambda )\big )^{-1}\mathbf{g}(\lambda ){\mathrm d}\lambda \\ \nonumber&+\int _{\mathbb {R}+\mathrm {i}\omega }\mathbf{f}^{\intercal }(\lambda )\big (\mathbf{A}_2^{\intercal }(\lambda )\big )'\big (\mathbf{A}_2^{\intercal }(\lambda )\big )^{-1}\mathbf{g}(\lambda ){\mathrm d}\lambda \\ \nonumber&+\int _{\Omega }{} \mathbf{f}^{\intercal }(\lambda )\mathbf{E}(\lambda )\mathbf{g}(\lambda ){\mathrm d}\lambda , \end{aligned}$$
(7.19)

with

Given the particular shape of \(\mathbf{f}(\lambda )\) and \(\mathbf{g}(\lambda )\) in (7.13), the first integral in (7.19) evaluates to zero. For the fourth integral, we record the following estimate.

Lemma 7.5

There exists \(c>0\) such that for every fixed \(\gamma \in [0,1)\) we can find \(t_0=t_0(\gamma )>0\) so that

$$\begin{aligned} \left| -\frac{1}{2\gamma }\int _{\Omega }\mathbf{f}^{\intercal }(\lambda )\mathbf{E}(\lambda )\mathbf{g}(\lambda ){\mathrm d}\lambda \right| \le c\left( \frac{s(\gamma )}{\sqrt{\gamma }|\ln \gamma |}\right) \mathrm {e}^{\frac{t}{2}\sqrt{-\ln \gamma }},\ \ \ s(\gamma ):=\frac{\gamma ^{\frac{1}{4}}}{1-\sqrt{\gamma }}, \end{aligned}$$
(7.20)

for all \((-t)\ge t_0\).

Proof

If \(\gamma =0\), then \(\mathbf{A}_k(\lambda )\equiv \mathbb {I}\) for \(k=1,2\) and likewise \(\mathbf{M}(z)\equiv \mathbb {I}\); compare RHP B.4. Thus, the integral in question is identically zero and the claim trivially true. If \(\gamma \in (0,1)\) is fixed, pick \(\omega :=\frac{1}{4}\sqrt{-\ln \gamma }>0\) so that \(0<\frac{\omega }{|t|}<\delta _{t\gamma }\) for \((-t)\ge t_0\), and first note from (B.8),

$$\begin{aligned} \left| \int _{\mathbb {R}}{} \mathbf{f}^{\intercal }(\lambda )\mathbf{E}(\lambda )\mathbf{g}(\lambda ){\mathrm d}\lambda \right| \le c\sqrt{\gamma }s(\gamma )\frac{t^{-2}\mathrm {e}^{t\sqrt{-\ln \gamma }}}{\text {dist}^2(\Sigma _\mathbf{M},\mathbb {R})}\ \ \ \forall (-t)\ge t_0, \end{aligned}$$

thus, since \(\text {dist}(\Sigma _\mathbf{M},\mathbb {R})\ge \delta _{t\gamma }>0\), we indeed obtain the right-hand side in (7.20) as upper bound. On the other hand, by explicit computation using again (B.8),

$$\begin{aligned} \left| \int _{\mathbb {R}+\mathrm {i}\omega }{} \mathbf{f}^{\intercal }(\lambda )\mathbf{E}(\lambda )\mathbf{g}(\lambda ){\mathrm d}\lambda \right| \le c\sqrt{\gamma }s(\gamma )\frac{t^{-2}\mathrm {e}^{t\sqrt{-\ln \gamma }}}{\text {dist}^2(\Sigma _\mathbf{M},\mathbb {R}+\mathrm {i}\frac{\omega }{|t|})}\mathrm {e}^{-2t\omega }, \end{aligned}$$

where \(\text {dist}(\Sigma _\mathbf{M},\mathbb {R}+\mathrm {i}\frac{\omega }{|t|})\ge \frac{3}{4}\delta _{t\gamma }>0\) by choice of \(\omega \). We thus also obtain the right-hand side of (7.20) as upper bound and have therefore completed our proof. \(\square \)

The remaining two integrals in (7.19) yield non-trivial contributions. We first state a lemma which is used in their evaluation.

Lemma 7.6

For any \(a,b,\omega >0\),

$$\begin{aligned} \int _{\mathbb {R}+\mathrm {i}\omega }\int _{\mathbb {R}}\frac{\mathrm {e}^{-\frac{a}{2}\lambda ^2-\frac{b}{2}s^2}}{(s-\lambda )^2}{\mathrm d}s{\mathrm d}\lambda =-\frac{2\pi \sqrt{ab}}{a+b}. \end{aligned}$$
(7.21)

Proof

Integration by parts in the variable s, as well as in the variable \(\lambda \), yields

$$\begin{aligned} \text {LHS in}\ (7.21)&=-b\int _{\mathbb {R}+\mathrm {i}\omega }\int _{\mathbb {R}}\frac{s}{s-\lambda }\mathrm {e}^{-\frac{a}{2}\lambda ^2-\frac{b}{2}s^2}{\mathrm d}s{\mathrm d}\lambda \\&=a\int _{\mathbb {R}+\mathrm {i}\omega }\int _{\mathbb {R}}\frac{\lambda }{s-\lambda }\mathrm {e}^{-\frac{a}{2}\lambda ^2-\frac{b}{2}s^2}{\mathrm d}s{\mathrm d}\lambda , \end{aligned}$$

and therefore

$$\begin{aligned} -\left( \frac{1}{a}+\frac{1}{b}\right) \times \text {LHS in}\ (7.22)=\int _{\mathbb {R}+\mathrm {i}\omega }\int _{\mathbb {R}}\mathrm {e}^{-\frac{a}{2}\lambda ^2-\frac{b}{2}s^2}{\mathrm d}s{\mathrm d}\lambda =\frac{2\pi }{\sqrt{ab}}, \end{aligned}$$

since both remaining integrals are standard Gaussians. This proves (7.21). \(\square \)

We now compute the two outstanding integrals in (7.19)

Lemma 7.7

For every \(\gamma \in [0,1)\),

$$\begin{aligned}&-\frac{1}{2\gamma }\int _{\mathbb {R}+\mathrm {i}\omega }\mathbf{f}^{\intercal }(\lambda )\big (\mathbf{A}_2^{\intercal }(\lambda )\big )'\big (\mathbf{A}_2^{\intercal }(\lambda )\big )^{-1}\mathbf{g}(\lambda ){\mathrm d}\lambda \nonumber \\&\quad =\frac{\partial }{\partial \gamma }\left[ \frac{t}{2\sqrt{2\pi }}\text {Li}_{\frac{3}{2}}(\gamma )\right] +\frac{1}{4\pi \gamma }\big (\text {Li}_{\frac{1}{2}}(\gamma )\big )^2. \end{aligned}$$
(7.22)

Proof

Inserting the formulae for \(\mathbf{f}(\lambda ),\mathbf{g}(\lambda )\) and \(\mathbf{A}_2(\lambda )\), we find

$$\begin{aligned} \text {LHS in}&\ 7.22=-\frac{1}{4\pi \sqrt{\gamma }}\int _{\mathbb {R}+\mathrm {i}\omega }\mathrm {e}^{-\frac{1}{4}\lambda ^2+\mathrm {i}t\lambda }\begin{bmatrix}0\\ 1\end{bmatrix}^{\intercal }\big (\mathbf{A}_2^{\intercal }(\lambda )\big )'\big (\mathbf{A}_2^{\intercal }(\lambda )\big )^{-1}\begin{bmatrix}1\\ 0\end{bmatrix}{\mathrm d}\lambda \\ =&\frac{\mathrm {i}}{4\pi }\int _{\mathbb {R}+\mathrm {i}\omega }\mathrm {e}^{-\frac{1}{4}\lambda ^2+\mathrm {i}t\lambda }\frac{{\mathrm d}}{{\mathrm d}\lambda }\left[ \frac{\mathrm {e}^{-\frac{1}{4}\lambda ^2-\mathrm {i}t\lambda }}{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}\right] {\mathrm d}\lambda \\&-\frac{1}{4\pi ^2}\int _{\mathbb {R}+\mathrm {i}\omega }\frac{\mathrm {e}^{-\frac{1}{2}\lambda ^2}}{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}\frac{{\mathrm d}}{{\mathrm d}\lambda }\left[ \int _{-\infty }^{\infty }\frac{h(s;1,\gamma )}{s-\lambda }{\mathrm d}s\right] {\mathrm d}\lambda . \end{aligned}$$

Integrating by parts, collapsing \(\mathbb {R}+\mathrm {i}\omega \) to \(\mathbb {R}\) and using the oddness of a part of the integrand, we see that the first remaining integral yields

$$\begin{aligned} \frac{\mathrm {i}}{4\pi }\int _{\mathbb {R}+\mathrm {i}\omega }\mathrm {e}^{-\frac{1}{4}\lambda ^2+\mathrm {i}t\lambda } \frac{{\mathrm d}}{{\mathrm d}\lambda }\left[ \frac{\mathrm {e}^{-\frac{1}{4}\lambda ^2-\mathrm {i}t\lambda }}{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}\right] {\mathrm d}\lambda&=\frac{t}{4\pi }\int _{\mathbb {R}}\frac{\mathrm {e}^{-\frac{1}{2}\lambda ^2}}{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}{\mathrm d}\lambda \\&=\frac{t}{4\pi \gamma }\int _{-\infty }^{\infty }\left( \sum _{n=1}^{\infty }\gamma ^n\mathrm {e}^{-\frac{n}{2}\lambda ^2}\right) {\mathrm d}\lambda \\&=\frac{t}{2\sqrt{2\pi }}\sum _{n=1}^{\infty }\frac{\gamma ^{n-1}}{\sqrt{n}}=\frac{\partial }{\partial \gamma }\left[ \frac{t}{2\sqrt{2\pi }}\text {Li}_{\frac{3}{2}}(\gamma )\right] . \end{aligned}$$

In the second (double) integral, we use geometric progression and the power series expansion \(\ln (1-z)=-\sum _{n=1}^{\infty }\frac{1}{n}z^n,|z|<1\) for \(h(s;1,\gamma )\),

$$\begin{aligned} -\frac{1}{4\pi ^2}\int _{\mathbb {R}+\mathrm {i}\omega }\frac{\mathrm {e}^{-\frac{1}{2}\lambda ^2}}{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}&\frac{{\mathrm d}}{{\mathrm d}\lambda }\left[ \int _{-\infty }^{\infty }\frac{h(s;1,\gamma )}{s-\lambda }{\mathrm d}s\right] {\mathrm d}\lambda \\&=-\frac{1}{4\pi ^2\gamma }\sum _{n,m=1}^{\infty }\frac{\gamma ^{n+m}}{m}\int _{\mathbb {R}+\mathrm {i}\omega }\int _{\mathbb {R}}\frac{\mathrm {e}^{-\frac{n}{2}\lambda ^2-\frac{m}{2}s^2}}{(s-\lambda )^2}{\mathrm d}s{\mathrm d}\lambda \\ {\mathop {=}\limits ^{7.21}}&\frac{1}{2\pi \gamma }\sum _{n,m=1}^{\infty }\frac{\gamma ^{n+m}}{n+m}\sqrt{\frac{n}{m}}\\&=\frac{1}{2\pi \gamma }\int _0^{\gamma }\text {Li}_{-\frac{1}{2}}(x)\text {Li}_{\frac{1}{2}}(x)\frac{{\mathrm d}x}{x}=\frac{1}{4\pi \gamma }\big (\text {Li}_{\frac{1}{2}}(\gamma )\big )^2, \end{aligned}$$

since \(\frac{{\mathrm d}}{{\mathrm d}x}\text {Li}_{\frac{1}{2}}(x)=\frac{1}{x}\text {Li}_{-\frac{1}{2}}(x)\) and \(\text {Li}_{-\frac{1}{2}}(0)=\text {Li}_{\frac{1}{2}}(0)=0\). This completes our proof. \(\square \)

Lemma 7.8

For every \(\gamma \in [0,1)\),

$$\begin{aligned} -\frac{1}{2\gamma }\int _{\mathbb {R}}\mathbf{f}^{\intercal }(\lambda )\big (\mathbf{A}_1^{\intercal }(\lambda )\big )'\big (\mathbf{A}_1^{\intercal }(\lambda )\big )^{-1}\mathbf{g}(\lambda ){\mathrm d}\lambda =\frac{\partial }{\partial \gamma }\left[ \frac{t}{2\sqrt{2\pi }}\text {Li}_{\frac{3}{2}}(\gamma )\right] +\frac{1}{4\pi \gamma }\big (\text {Li}_{\frac{1}{2}}(\gamma )\big )^2. \end{aligned}$$
(7.23)

Proof

Using the above formula for \(\mathbf{A}_1(\lambda )\) and (7.13) we find at once

$$\begin{aligned}&-\frac{1}{2\gamma }\int _{\mathbb {R}}\mathbf{f}^{\intercal }(\lambda )\big (\mathbf{A}_1^{\intercal }(\lambda )\big )'\big (\mathbf{A}_1^{\intercal }(\lambda )\big )^{-1}\mathbf{g}(\lambda ){\mathrm d}\lambda =\frac{t}{4\pi }\int _{-\infty }^{\infty }\frac{\mathrm {e}^{-\frac{1}{2}\lambda ^2}{\mathrm d}\lambda }{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}\\&\qquad -\frac{1}{4\pi ^2}\int _{-\infty }^{\infty }\frac{\mathrm {e}^{-\frac{1}{2}\lambda ^2}}{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}\frac{{\mathrm d}}{{\mathrm d}\lambda }\left[ \text {pv}\int _{-\infty }^{\infty }\frac{h(s;1,\gamma )}{s-\lambda }{\mathrm d}s\right] {\mathrm d}\lambda . \end{aligned}$$

Here, the first remaining integral was already computed in the proof of Lemma 7.7,

$$\begin{aligned} \frac{t}{4\pi }\int _{-\infty }^{\infty }\frac{\mathrm {e}^{-\frac{1}{2}\lambda ^2}{\mathrm d}\lambda }{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}=\frac{\partial }{\partial \gamma }\left[ \frac{t}{2\sqrt{2\pi }}\text {Li}_{\frac{3}{2}}(\gamma )\right] ,\ \ \ \gamma \in [0,1). \end{aligned}$$

For the second one, we use the Plemelj–Sokhotski formula,

$$\begin{aligned} \lim _{\begin{array}{c} z\rightarrow \lambda \in \mathbb {R}\\ \Im z>0 \end{array}}\int _{-\infty }^{\infty }\frac{h(s;1,\gamma )}{s-z}{\mathrm d}s=\mathrm {i}\pi h(\lambda ;1,\gamma )+\text {pv}\int _{-\infty }^{\infty }\frac{h(s;1,\gamma )}{s-\lambda }{\mathrm d}s, \end{aligned}$$
(7.24)

and note that by oddness of the integrand,

$$\begin{aligned} \int _{-\infty }^{\infty }h(\lambda ;1,\gamma )\frac{{\mathrm d}}{{\mathrm d}\lambda }\left[ \frac{\mathrm {e}^{-\frac{1}{2}\lambda ^2}}{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}\right] {\mathrm d}\lambda =0. \end{aligned}$$
(7.25)

Thus, integrating by parts and adding (7.25), we find

$$\begin{aligned}&-\frac{1}{4\pi ^2}\int _{-\infty }^{\infty }\frac{\mathrm {e}^{-\frac{1}{2}\lambda ^2}}{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}\frac{{\mathrm d}}{{\mathrm d}\lambda }\left[ \text {pv}\int _{-\infty }^{\infty }\frac{h(s;1,\gamma )}{s-\lambda }{\mathrm d}s\right] {\mathrm d}\lambda \\&\quad = \frac{1}{4\pi ^2}\int _{-\infty }^{\infty }\left( \frac{{\mathrm d}}{{\mathrm d}\lambda }\left[ \frac{\mathrm {e}^{-\frac{1}{2}\lambda ^2}}{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}\right] \right) \left( \text {pv}\int _{-\infty }^{\infty }\frac{h(s;1,\gamma )}{s-\lambda }{\mathrm d}s+\mathrm {i}\pi h(\lambda ;1,\gamma )\right) {\mathrm d}\lambda . \end{aligned}$$

Now change the contour \(\mathbb {R}\ni \lambda \) to \(\mathbb {R}+\mathrm {i}\omega \) by Cauchy’s theorem while using the analytic continuation (7.24) for the second round bracket. The result equals

$$\begin{aligned}&-\frac{1}{4\pi ^2}\int _{-\infty }^{\infty }\frac{\mathrm {e}^{-\frac{1}{2}\lambda ^2}}{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}\frac{{\mathrm d}}{{\mathrm d}\lambda }\left[ \text {pv}\int _{-\infty }^{\infty }\frac{h(s;1,\gamma )}{s-\lambda }{\mathrm d}s\right] {\mathrm d}\lambda \\&\quad = \frac{1}{4\pi ^2}\int _{\mathbb {R}+\mathrm {i}\omega }\left( \frac{{\mathrm d}}{{\mathrm d}\lambda }\left[ \frac{\mathrm {e}^{-\frac{1}{2}\lambda ^2}}{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}\right] \right) \int _{-\infty }^{\infty }\frac{h(s;1,\gamma )}{s-\lambda }{\mathrm d}s{\mathrm d}\lambda \\&\quad = -\frac{1}{4\pi ^2}\int _{\mathbb {R}+\mathrm {i}\omega }\frac{\mathrm {e}^{-\frac{1}{2}\lambda ^2}}{1-\gamma \mathrm {e}^{-\frac{1}{2}\lambda ^2}}\frac{{\mathrm d}}{{\mathrm d}\lambda }\left[ \int _{-\infty }^{\infty }\frac{h(s;1,\gamma )}{s-\lambda }{\mathrm d}s\right] {\mathrm d}\lambda \end{aligned}$$

after another integration by parts in the last equality. The obtained result is identical to the second (double) integral in the proof of Lemma 7.7, and we therefore find (7.23) all together. \(\square \)

We now combine (7.20), (7.23), (7.22) and (7.17) to obtain the following result.

Proposition 7.9

There exists \(c>0\) such that for every fixed \(\gamma \in [0,1)\) we can find \(t_0=t_0(\gamma )>0\) so that

$$\begin{aligned} \ln \det (1-\gamma \chi _tT\chi _t\upharpoonright _{L^2(\mathbb {R})})=\frac{t}{\sqrt{2\pi }}\text {Li}_{\frac{3}{2}}(\gamma )+\frac{1}{2\pi }\int _0^{\gamma }\big (\text {Li}_{\frac{1}{2}}(x)\big )^2\frac{{\mathrm d}x}{x} + r(t,\gamma ) \end{aligned}$$
(7.26)

for \((-t)\ge t_0\) where the error term \(r(t,\gamma )\) is differentiable with respect to \(\gamma \) and satisfies

$$\begin{aligned} \big |r(t,\gamma )\big |\le c\left( \frac{\gamma ^{\frac{3}{4}}}{(1-\sqrt{\gamma })|\ln \gamma |}\right) \mathrm {e}^{\frac{t}{2}\sqrt{-\ln \gamma }}\ \ \ \ \ \ \forall (-t)\ge t_0. \end{aligned}$$

Proof

We have, as \(t\rightarrow -\infty \),

$$\begin{aligned} -\frac{1}{2\gamma }\int _{\Omega }R(\lambda ,\lambda ){\mathrm d}\lambda =\frac{\partial }{\partial \gamma }\left[ \frac{t}{\sqrt{2\pi }}\text {Li}_{\frac{3}{2}}(\gamma )\right] +\frac{1}{2\pi \gamma }\big (\text {Li}_{\frac{1}{2}}(\gamma )\big )^2+\mathcal {O}\left( \frac{s(\gamma )}{\sqrt{\gamma }|\ln \gamma |}\mathrm {e}^{\frac{t}{2}\sqrt{-\ln \gamma }}\right) , \end{aligned}$$

uniformly in \(\gamma \in [0,1)\). Integrating this expansion in (7.17) from 0 to \(\gamma <1\) yields immediately the two leading terms in (7.26), and for the error term we estimate as follows:

$$\begin{aligned} \left| \int _0^{\gamma }\frac{x^{-\frac{1}{4}}}{(1-\sqrt{x})\ln x}\mathrm {e}^{\frac{t}{2}\sqrt{-\ln x}}{\mathrm d}x\right|&\le -\frac{\mathrm {e}^{\frac{t}{2}\sqrt{-\ln \gamma }}}{1-\sqrt{\gamma }}\int _0^{\gamma }\frac{x^{-\frac{1}{4}}}{\ln x}{\mathrm d}x\\&\le c\left( \frac{\gamma ^{\frac{3}{4}}}{(1-\sqrt{\gamma })|\ln \gamma |}\right) \mathrm {e}^{\frac{t}{2}\sqrt{-\ln \gamma }}. \end{aligned}$$

This completes our proof. \(\square \)

.

Combining our results, we finally arrive at (1.20).

Corollary 7.10

As \(t\rightarrow -\infty \), for any \(\gamma \in [0,1)\),

$$\begin{aligned} P(t;\gamma )=\exp \left[ \frac{t}{2\sqrt{2\pi }}\text {Li}_{\frac{3}{2}}(\bar{\gamma })+\frac{1}{2}\ln \left( \frac{2}{2-\gamma }\right) +\frac{1}{4\pi }\int _0^{\bar{\gamma }}\left( \big (\text {Li}_{\frac{1}{2}}(x)\big )^2-\frac{\pi x}{1-x}\right) \frac{{\mathrm d}x}{x}\right] \big (1+o(1)\big ) \end{aligned}$$

Proof

From (1.14), (7.1), (7.4), (7.11) and (7.26) (substituting \(\gamma \mapsto \bar{\gamma }\) in the last equation),

$$\begin{aligned} P(t;\gamma )=&\sqrt{\det (1-\bar{\gamma }\chi _tT\chi _t\upharpoonright _{L^2(\mathbb {R})})}\left( \sqrt{\frac{1-\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\mathrm {e}^{\frac{1}{2}\mu (t;\bar{\gamma })}+\sqrt{\frac{1+\sqrt{\bar{\gamma }}}{2(2-\gamma )}}\mathrm {e}^{-\frac{1}{2}\mu (t;\bar{\gamma })}\right) \nonumber \\ =&\exp \left[ \frac{t}{2\sqrt{2\pi }}\text {Li}_{\frac{3}{2}}(\bar{\gamma })+\frac{1}{4\pi }\int _0^{\bar{\gamma }}\big (\text {Li}_{\frac{1}{2}}(x)\big )^2\frac{{\mathrm d}x}{x}\right] \sqrt{2}\sqrt{\frac{1-\gamma }{2-\gamma }}\big (1+o(1)\big ) \end{aligned}$$
(7.27)

as \(t\rightarrow -\infty \). The claim follows now after writing

$$\begin{aligned} \ln (1-\gamma )=\frac{1}{2}\ln \big (1-\bar{\gamma }\big )=-\frac{1}{2}\int _0^{\bar{\gamma }}\frac{{\mathrm d}x}{1-x}. \end{aligned}$$

\(\square \)

7.3 Proof of Lemma 1.14

Since \(\text {Li}_{\frac{1}{2}}(x)=x+\mathcal {O}(x^2)\) as \(x\rightarrow 0\) and, cf. [54, 25.12.12],

$$\begin{aligned} \text {Li}_{\frac{1}{2}}(x)=\sqrt{\frac{\pi }{1-x}}\big (1+\mathcal {O}(x-1)\big )+\mathcal {O}(1),\ \ \ x\uparrow 1, \end{aligned}$$

we see that

$$\begin{aligned} \int _0^{\bar{\gamma }}\left( \big (\text {Li}_{\frac{1}{2}}(x)\big )^2-\frac{\pi x}{1-x}\right) \frac{{\mathrm d}x}{x} \end{aligned}$$

converges as \(\bar{\gamma }\uparrow 1\), so \(c_0(\gamma )\) is indeed continuous in \(\gamma \in [0,1]\). On the other hand, from the power series representation of the polylogarithm,

$$\begin{aligned} \big (\text {Li}_{\frac{1}{2}}(x)\big )^2-\frac{\pi x}{1-x}=\sum _{n,m=1}^{\infty }\frac{x^{n+m}}{\sqrt{nm}}-\pi \sum _{n=1}^{\infty }x^n=\sum _{n=1}^{\infty }a_nx^n,\ \ \ |x|<1, \end{aligned}$$

with

$$\begin{aligned} a_n:=-\pi +\sum _{m=1}^{n-1}\frac{1}{\sqrt{m(n-m)}},\ \ n\in \mathbb {Z}_{\ge 1}\ \ \ \ \ \text {and}\ \ \ \\ \nonumber \sum _{m=1}^{n-1}\frac{1}{\sqrt{m(n-m)}}=\pi -\frac{c}{\sqrt{n}}+\mathcal {O}\big (n^{-1}\big ),\ n\rightarrow \infty , \end{aligned}$$
(7.28)

for some \(c>0\). Thus, for any \(0\le t<1\),

$$\begin{aligned} \int _0^t\left( \big (\text {Li}_{\frac{1}{2}}(x)\big )^2-\frac{\pi x}{1-x}\right) \frac{{\mathrm d}x}{x}=\sum _{n=1}^{\infty }\frac{1}{n}\left( -\pi +\sum _{m=1}^{n-1}\frac{1}{\sqrt{m(n-m)}}\right) t^n, \end{aligned}$$

which verifies (1.23) for \(0\le \gamma <1\) through (1.21). But using again [54, 25.12.12], we also have that

$$\begin{aligned} \int _t^1\left( \big (\text {Li}_{\frac{1}{2}}(x)\big )^2-\frac{\pi x}{1-x}\right) \frac{{\mathrm d}x}{x}=o(1),\ \ \ t\uparrow 1, \end{aligned}$$

so by Abel’s convergence theorem,

$$\begin{aligned}&\int _0^1\left( \big (\text {Li}_{\frac{1}{2}}(x)\big )^2-\frac{\pi x}{1-x}\right) \frac{{\mathrm d}x}{x}\\&\quad =\lim _{t\uparrow 1}\left[ \int _0^t\left( \big (\text {Li}_{\frac{1}{2}}(x)\big )^2-\frac{\pi x}{1-x}\right) \frac{{\mathrm d}x}{x}+\int _t^1\left( \big (\text {Li}_{\frac{1}{2}}(x)\big )^2-\frac{\pi x}{1-x}\right) \frac{{\mathrm d}x}{x}\right] \\&\quad = \lim _{t\uparrow 1}\left[ \sum _{n=1}^{\infty }\frac{1}{n}a_nt^n+o(1)\right] =\sum _{n=1}^{\infty }\frac{1}{n}a_n, \end{aligned}$$

since \(\frac{1}{n}a_n\) is summable, see (7.28). The proof of Lemma 1.14 is now complete.