# Large gap probabilities of complex and symplectic
spherical ensembles with point charges

Sung-Soo Byun  Department of Mathematical Sciences and Research Institute of
Mathematics, Seoul National University, Seoul 151-747, Republic of Korea
<EMAIL_ADDRESS>and  Seongjae Park  Department of Mathematical Sciences,
Seoul National University, Seoul 151-747, Republic of Korea
<EMAIL_ADDRESS>
###### Abstract.

We consider $n$ eigenvalues of complex and symplectic induced spherical
ensembles, which can be realised as two-dimensional determinantal and Pfaffian
Coulomb gases on the Riemann sphere under the insertion of point charges. For
both cases, we show that the probability that there are no eigenvalues in a
spherical cap around the poles has an asymptotic behaviour as $n\to\infty$ of
the form

$\exp\Big{(}c_{1}n^{2}+c_{2}n\log n+c_{3}n+c_{4}\sqrt{n}+c_{5}\log
n+c_{6}+\mathcal{O}(n^{-\frac{1}{12}})\Big{)}$

and determine the coefficients explicitly. Our results provide the second
example of precise (up to and including the constant term) large gap
asymptotic behaviours for two-dimensional point processes, following a recent
breakthrough by Charlier.

###### Key words and phrases:

Spherical ensemble, gap probability, large deviation, orthogonal polynomial,
asymptotic analysis

###### 2020 Mathematics Subject Classification:

Primary 60B20; Secondary 33B20

Sung-Soo Byun was partially supported by the POSCO TJ Park Foundation (POSCO
Science Fellowship), by the New Faculty Startup Fund at Seoul National
University and by the National Research Foundation of Korea funded by the
Korea government (NRF-2016K2A9A2A13003815, RS-2023-00301976).

##  1\. Introduction and main results

###  1.1. Spherical ensembles

We consider two-dimensional Coulomb gases [63] (also known as one-component
plasmas) $\boldsymbol{x}=\\{x_{j}\\}_{j=1}^{n}$ on the Riemann sphere
$\mathbb{S}=\\{x\in\mathbb{R}^{3}:\|x\|=1\\}$, under the insertion of point
charges at the south and north poles, ${\rm{S}}=(0,0,-1)$ and
${\rm{N}}=(0,0,1)$ respectively. Here, $\|\cdot\|$ is the Euclidean metric in
$\mathbb{R}^{3}$. In addition to the point charge insertions, we consider both
determinantal and Pfaffian Coulomb gases. Consequently, for given fixed non-
negative integers $\alpha$ and $c$ determining point charges at the poles, the
joint probability distributions of the models are given by

(1.1) $\displaystyle
d\boldsymbol{\mathcal{P}}_{n}^{\mathbb{C}}(\boldsymbol{x})$
$\displaystyle=\frac{1}{\mathcal{Z}_{n}^{\mathbb{C}}}\prod_{j>k=1}^{n}\|x_{j}-x_{k}\|^{2}\prod_{j=1}^{n}\|x_{j}-{\rm{N}}\|^{2\alpha}\,\|x_{j}-{\rm{S}}\|^{2c}\,dA_{\mathbb{S}}(x_{j}),$
(1.2) $\displaystyle
d\boldsymbol{\mathcal{P}}_{n}^{\mathbb{H}}(\boldsymbol{x})$
$\displaystyle=\frac{1}{\mathcal{Z}_{n}^{\mathbb{H}}}\prod_{j>k=1}^{n}\|x_{j}-x_{k}\|^{2}\|x_{j}-\overline{x}_{k}\|^{2}\prod_{j=1}^{n}\|x_{j}-\overline{x}_{j}\|^{2}\|x_{j}-{\rm{N}}\|^{4\alpha}\,\|x_{j}-{\rm{S}}\|^{4c}\,dA_{\mathbb{S}}(x_{j}),$

where $dA_{\mathbb{S}}$ is the area measure on $\mathbb{S}$, and we use the
convention $\overline{x}=(x_{1},-x_{2},x_{3})$ for $x=(x_{1},x_{2},x_{3})$.
Here, $\mathcal{Z}_{n}^{\mathbb{C}}$ and $\mathcal{Z}_{n}^{\mathbb{H}}$ are
partition functions that make $\boldsymbol{\mathcal{P}}_{n}^{\mathbb{C}}$ and
$\boldsymbol{\mathcal{P}}_{n}^{\mathbb{H}}$ probability measures. The
superscripts $\mathbb{C}$ and $\mathbb{H}$ denoting determinantal and Pfaffian
Coulomb gases, respectively, will become clear below.

For readers who are inclined towards the statistical physics perspective, it
suffices to regard (1.1) and (1.2) as the primary subjects of our
investigations. On the other hand, for readers from the random matrix theory
community, let us stress that these models have realisations of eigenvalues of
random matrix models called complex [61, 64, 77] and symplectic [28, 65, 87,
88] spherical induced ensembles. We refer to [26, Section 2.5] and [27,
Section 6.3] for comprehensive reviews, see also a recent work [91] and
references therein. A simple way to define these random matrix models is as
$n\times n$ complex or quaternion matrices, denoted by G, whose matrix
probability distribution function is proportional to

(1.3)
$\frac{\det(\textbf{G}\textbf{G}^{*})^{c\beta/2}}{\det(\mathbbm{1}_{n}+\textbf{G}\textbf{G}^{*})^{(2n+c+\alpha)\beta/2}},\qquad\beta=\begin{cases}2&\textup{for
the complex case},\vskip 3.0pt plus 1.0pt minus 1.0pt\\\ 4&\textup{for the
quaternionic case}.\end{cases}$

For the special case when $c=\alpha=0$ (i.e. without point charges), these
models are called the complex and symplectic spherical ensembles. In this
case, an element-wise realisation of random matrices distributed as (1.3) is
due to Krishnapur [77]. To be more concrete, the matrix distribution (1.3)
follows from $\textbf{G}_{1}\textbf{G}_{2}^{-1}$, where $\textbf{G}_{1}$ and
$\textbf{G}_{2}$ are independent complex or symplectic Ginibre matrices whose
elements are given by i.i.d. complex or quaternionic Gaussian random
variables. It is evident from the construction that the spherical ensembles
are closely related to generalised eigenvalue problems [65, 54], which find
applications in random plane geometry. Beyond the spherical cases, the
presence of point charges can also be realised at the level of matrix models
via a so-called inducing procedure, see e.g. [61, 88, 60]. This construction
requires the use of rectangular Ginibre matrices as well as Haar-distributed
matrices within the symmetry classes under consideration. In this inducing
procedure, one needs to assume that $\alpha$ and $c$ are non-negative integer
values since they are related to the rectangular parameters.

It follows from (1.3) that the joint distributions of eigenvalues
$\boldsymbol{z}=\\{z_{j}\\}_{j=1}^{n}$ are given by

(1.4) $\displaystyle d\mathbf{P}_{n}^{\mathbb{C}}(\boldsymbol{z})$
$\displaystyle=\frac{1}{Z_{n}^{\mathbb{C}}}\prod_{j>k=1}^{n}|z_{j}-z_{k}|^{2}\prod_{j=1}^{n}e^{-nQ_{n}(z_{j})}\,dA(z_{j}),$
(1.5) $\displaystyle d\mathbf{P}_{n}^{\mathbb{H}}(\boldsymbol{z})$
$\displaystyle=\frac{1}{Z_{n}^{\mathbb{H}}}\prod_{j>k=1}^{n}|z_{j}-z_{k}|^{2}|z_{j}-\overline{z}_{k}|^{2}\prod_{j=1}^{n}|z_{j}-\overline{z}_{j}|^{2}e^{-2nQ_{n}(z_{j})}\,dA(z_{j}),$

where $dA(z)=d^{2}z/\pi$ and the external potential $Q_{n}$ is given by

(1.6) $Q_{n}(z):=\frac{n+\alpha+c+1}{n}\log(1+|z|^{2})-\frac{2c}{n}\log|z|.$

The models (1.4) and (1.5) are again two-dimensional Coulomb gas ensembles in
the complex plane, with an $n$-dependent potential. Due to their special
integrable structures, they are determinantal and Pfaffian Coulomb gas
ensembles, respectively. We mention that the partition functions
$Z_{n}^{\mathbb{C}}$ and $Z_{n}^{\mathbb{H}}$ can be expressed in terms of the
gamma functions as

(1.7)
$Z_{n}^{\mathbb{C}}=n!\prod_{k=0}^{n-1}\frac{\Gamma(k+c+1)\Gamma(n+\alpha-k)}{\Gamma(n+\alpha+c+1)},\qquad
Z_{n}^{\mathbb{H}}=n!\,2^{n}\,\prod_{k=0}^{n-1}\frac{\Gamma(2k+2c+2)\Gamma(2n+2\alpha-2k)}{\Gamma(2n+2\alpha+2c+2)}.$

To observe the equivalence between (1.1) and (1.4), as well as (1.2) and
(1.5), recall that the stereographic projection
$\phi:\mathbb{S}^{2}\to\mathbb{C}\cup\\{\infty\\}$ is given by

(1.8) $\phi(x)=\begin{cases}\dfrac{x_{1}+ix_{2}}{1-x_{3}},&\textup{if
}x\in\mathbb{S}^{2}\setminus{\rm{N}},\vskip 3.0pt plus 1.0pt minus 1.0pt\\\
\infty,&\textup{if
}x={\rm{N}},\end{cases}\qquad\phi^{-1}(z)=\Big{(}\frac{2\operatorname{Re}z}{1+|z|^{2}},\frac{2\operatorname{Im}z}{1+|z|^{2}},\frac{|z|^{2}-1}{1+|z|^{2}}\Big{)},$

where $x=(x_{1},x_{2},x_{3}).$ Using these, one can easily check that the
measures (1.1) and (1.2) are a pull-back of (1.4) and (1.5), whence the name
_spherical_ for the random matrix G was first coined [77, 64]. Furthermore,
the partition functions $\mathcal{Z}_{n}^{\mathbb{C}}$ and
$\mathcal{Z}_{n}^{\mathbb{H}}$ in (1.1) and (1.2) are also given by (1.7), up
to explicit constants. From this viewpoint, the potential (1.6) in the complex
plane can also be realised as point charge insertions at the origin and
infinity, cf. [93].

The potential $Q_{n}$ in (1.6) is indeed a prominent example of _weakly
confining potential_ , see e.g. [25] and references therein. Contrary to
Coulomb gas ensembles with a confining potential that makes the particles lie
in a compact set in the large-$n$ limit, the ensembles (1.4) and (1.5) tend to
be distributed in the whole complex plane. To be more precise, due to the
Laplacian growth property of the two-dimensional Coulomb gas ensemble, for
given fixed $\alpha$ and $c$, the ensembles (1.4) and (1.5) tend to occupy the
whole complex plane with the limiting density

(1.9) $\rho^{\rm sp}(z)=\frac{1}{(1+|z|^{2})^{2}}.$

As a consequence, using the inverse stereographic projection, we infer that as
$n\to\infty$, the ensembles (1.1) and (1.2) tend to be uniformly distributed
on the whole Riemann sphere $\mathbb{S}$, see Figure 1. We also mention that
the equilibrium measure problems associated with spherical Coulomb gases with
point charges have been recently studied in [83, 43].

(a) $\\{z_{j}\\}_{j=1}^{n}$, $z_{j}\in\mathbb{C}$

(b) $\\{x_{j}\\}_{j=1}^{n}$, $x_{j}\in\mathbb{S}$

Figure 1. The plot (A) displays eigenvalues of the spherical symplectic
ensembles, where $n=200$. The plot (B) shows its projection to the sphere.

###  1.2. Main results: large gap probabilities

In this work, we study the asymptotic behaviour of the large gap probabilities
as $n\to\infty$ of the ensembles (1.4) and (1.5):

(1.10)
$\displaystyle\begin{split}\mathcal{P}_{n}(R;\alpha,c)&:=\mathbb{P}\Big{[}\textup{there
is no eigenvalues in }\mathbb{D}(0,R)\Big{]}.\end{split}$

This problem was already considered a decade ago in the work [3], where the
leading term for the complex case with $\alpha=c=0$ was obtained (i.e. the
constant $A_{1}$ in Theorem 1.1).

More generally, obtaining large gap asymptotics is a classical and challenging
problem in random matrix theory with a long history, which has attracted
considerable attention over the years. The literature pertaining to this topic
will be reviewed in Subsection 1.3 below. Note that $\mathcal{P}_{n}$ can also
be written as

(1.11)
$\mathcal{P}_{n}(R;\alpha,c)=\mathbb{P}\Big{[}\min\\{|z_{1}|,\ldots,|z_{n}|\\}\geq
R\,\Big{]}.$

Hence $\mathcal{P}_{n}$ can be viewed as a heavy-tail distribution for the
distribution of the smallest moduli. Analogous extreme distributions have been
widely studied for various one-dimensional log-correlated point processes
(such as the Tracy-Widom distribution for the Airy point process); this will
be discussed in Subsection 1.3.4. In our case, the moduli
$\\{|z_{j}|\\}_{j=1}^{n}$ forms a permanantal point process (see e.g. [7]),
and the quantity $\mathcal{P}_{n}$ can also be interpreted as the large
deviation probability of this process. Furthermore, $\mathcal{P}_{n}$ is
intimately connected to an energy minimisation (electrostatics) problem, which
is a fundamental aspect of potential theory, see Subsection 1.3.2.
Additionally, it is equivalent to the free energy of a one-component plasma
confined by hard walls, see Subsection 1.3.3 for further details.

For the models (1.1) and (1.2) on the sphere, the probability (1.10) coincides
with the probability that there are no particles in a spherical cap (whose
center is the south pole) of area

(1.12) $4\pi\frac{R^{2}}{1+R^{2}},$

see Figure 3 for some illustrations. Note that instead of (1.10), one may also
consider

(1.13)
$\displaystyle\begin{split}\widetilde{\mathcal{P}}_{n}(R;\alpha,c)&:=\mathbb{P}\Big{[}\textup{there
is no eigenvalues in
}\mathbb{D}(0,R)^{c}\Big{]}=\mathbb{P}\Big{[}\max\\{|z_{1}|,\ldots,|z_{n}|\\}\leq
R\,\Big{]}.\end{split}$

Due to the sphere geometry, there is a duality relation between
$\mathcal{P}_{n}$ and $\widetilde{\mathcal{P}}_{n}$:

(1.14)
$\mathcal{P}_{n}(R;\alpha,c)=\widetilde{\mathcal{P}}_{n}(1/R;c,\alpha).$

Alternatively, (1.14) can also be directly obtained using Lemma 2.1 below.

In the sequel, we add the superscripts and write
$\mathcal{P}_{n}^{\mathbb{C}}$ and $\mathcal{P}_{n}^{\mathbb{H}}$ to
distinguish (1.4) and (1.5). To state our main results, we need some
elementary special functions. Recall that the complementary error function is
defined by [92, Chapter 7]

(1.15)
$\operatorname{erfc}(z):=\frac{2}{\sqrt{\pi}}\int_{z}^{\infty}e^{-t^{2}}\,dt$

and that the Barnes $G$-function is defined recursively by [92, Section 5.17]

(1.16) $G(z+1)=\Gamma(z)G(z),\qquad G(1)=1,$

where $\Gamma$ is the standard gamma function. We write

(1.17) $\displaystyle\mathcal{I}_{1}$
$\displaystyle=\sqrt{2}\int_{-\infty}^{0}\log\Big{(}\tfrac{1}{2}\operatorname{erfc}(t)\Big{)}\,dt+\sqrt{2}\int_{0}^{\infty}\bigg{[}\log\Big{(}\tfrac{1}{2}\operatorname{erfc}(t)\Big{)}+t^{2}+\log
t+\log(2\sqrt{\pi})\bigg{]}\,dt\approx-2.301,$ (1.18)
$\displaystyle\mathcal{I}_{2}$
$\displaystyle=\int_{-\infty}^{\infty}\bigg{[}\frac{(1+t^{2})e^{-t^{2}}}{\sqrt{\pi}\operatorname{erfc}(t)}-\Big{(}t^{3}+\frac{3}{2}t\Big{)}\mathbbm{1}_{(0,\infty)}(t)\bigg{]}\,dt\approx
0.928.$

We are now ready to state the asymptotics as $n\to\infty$ of the large gap
probabilities of the complex induced spherical ensemble (1.4).

######  Theorem 1.1 (Large gap probabilities of the complex ensemble).

Let $\alpha,c\in\mathbb{Z}_{\geq 0}$ and $R>0$ be fixed. Then as $n\to\infty$,
we have

(1.19)
$\displaystyle\log\mathcal{P}_{n}^{\mathbb{C}}(R;\alpha,c)=A_{1}n^{2}+A_{2}n\log
n+A_{3}n+A_{4}\sqrt{n}+A_{5}\log n+A_{6}+\mathcal{O}(n^{-\frac{1}{12}}),$

where

(1.20) $\displaystyle A_{1}$
$\displaystyle=\frac{1}{2}\Big{(}\frac{R^{2}}{1+R^{2}}-\log(1+R^{2})\Big{)},$
(1.21) $\displaystyle A_{2}$
$\displaystyle=-\frac{1}{2}\frac{R^{2}}{1+R^{2}},$ (1.22) $\displaystyle
A_{3}$
$\displaystyle=\frac{R^{2}}{1+R^{2}}\Big{(}1+\alpha+c-\frac{1}{2}\log(2\pi)-\log
R\Big{)}-\alpha\log(1+R^{2}),$ (1.23) $\displaystyle A_{4}$
$\displaystyle=\frac{R}{1+R^{2}}\,\mathcal{I}_{1},$ (1.24) $\displaystyle
A_{5}$
$\displaystyle=\frac{1+\alpha+c}{2(1+R^{2})}-\frac{1}{6}-\frac{\alpha+c^{2}}{2},$
(1.25)
$\displaystyle\begin{split}A_{6}&=\Big{(}\frac{1+\alpha+c}{1+R^{2}}-\frac{1}{3}-\alpha-c^{2}\Big{)}\log
R-\frac{\alpha^{2}-c^{2}}{2}\log(1+R^{2})\\\
&\quad+\Big{(}\frac{1}{2}-(1+\alpha+c)\frac{R^{2}}{1+R^{2}}\Big{)}\frac{\log(2\pi)}{2}+\frac{(\alpha+c)(1+\alpha+c)}{2}\frac{R^{2}}{1+R^{2}}\\\
&\quad-\frac{2}{3}\frac{1-R^{2}}{1+R^{2}}\,\mathcal{I}_{2}+\frac{1}{12}\Big{(}1+\frac{11}{1+R^{2}}\Big{)}-\zeta^{\prime}(-1)+\log
G(c+1).\end{split}$

Here, $\zeta$ is the Riemann zeta function, $G$ is the Barnes $G$-function,
and the constants $\mathcal{I}_{1}$ and $\mathcal{I}_{2}$ are given by (1.17)
and (1.18).

######  Remark 1.2 (Conjecture for the terms proportional to $n\log n$ and
$\sqrt{n}$).

In view of Theorem 1.1 and [34, Theorem 1.7], we formulate the following
conjecture for $A_{2}$ and $A_{4}$, valid for a general radially symmetric
potential $V$ such that $\\{z\in\mathbb{C}:|z|<R\\}$ is contained in the
droplet:

(1.26) $A_{2}=-\frac{1}{2}\int_{|z|<R}\,d\sigma_{V}(z)\qquad\mbox{and}\qquad
A_{4}=R\sqrt{\rho(R)}\,\mathcal{I}_{1},$

where $d\sigma_{V}=\Delta V\,dA=\rho\,dA$ is the equilibrium measure (1.39)
associated with the potential $V$, see also Subsection 1.3.2. This conjecture
is consistent with Theorem 1.1, and also with the result [34, Theorem 1.7] on
the Mittag-Leffler ensemble. In our case, $\rho$ is given by (1.9), whereas
for the Mittag-Leffler ensemble, $\rho$ is given by (1.36).

As previously mentioned, when $\alpha=c=0$, $A_{1}$ was already obtained in
[3, Proposition 3.1]. However, a more precise expansion has not been
discovered; indeed even the second term $A_{2}$ is new to our knowledge.

The symplectic counterpart of Theorem 1.1 is as follows.

######  Theorem 1.3 (Large gap probabilities of the symplectic ensemble).

Let $\alpha,c\in\mathbb{Z}_{\geq 0}$ and $R>0$ be fixed. Then as $n\to\infty$,
we have

(1.27)
$\displaystyle\log\mathcal{P}_{n}^{\mathbb{H}}(R;\alpha,c)=\widehat{A}_{1}n^{2}+\widehat{A}_{2}n\log
n+\widehat{A}_{3}n+\widehat{A}_{4}\sqrt{n}+\widehat{A}_{5}\log
n+\widehat{A}_{6}+\mathcal{O}(n^{-\frac{1}{12}}),$

where

(1.28) $\displaystyle\widehat{A}_{1}$
$\displaystyle=\frac{R^{2}}{1+R^{2}}-\log(1+R^{2}),$ (1.29)
$\displaystyle\widehat{A}_{2}$
$\displaystyle=-\frac{1}{2}\frac{R^{2}}{1+R^{2}},$ (1.30)
$\displaystyle\widehat{A}_{3}$
$\displaystyle=\frac{R^{2}}{1+R^{2}}\Big{(}2+2\alpha+2c-\frac{1}{2}\log(4\pi)-\log
R\Big{)}-\Big{(}2\alpha+\frac{1}{2}\Big{)}\log(1+R^{2}),$ (1.31)
$\displaystyle\widehat{A}_{4}$
$\displaystyle=\frac{1}{\sqrt{2}}\frac{R}{1+R^{2}}\,\mathcal{I}_{1},$ (1.32)
$\displaystyle\widehat{A}_{5}$
$\displaystyle=\frac{1+\alpha+c}{2(1+R^{2})}-\frac{5}{24}-\frac{\alpha+c}{2}-c^{2},$
(1.33)
$\displaystyle\begin{split}\widehat{A}_{6}&=\Big{(}\frac{1+\alpha+c}{1+R^{2}}-\frac{5}{12}-\alpha-c-2c^{2}\Big{)}\log
R-(\alpha-c)\Big{(}\frac{1}{2}+\alpha+c\Big{)}\log(1+R^{2})\\\
&\quad-\Big{(}c+(1+\alpha+c)\frac{R^{2}}{1+R^{2}}\Big{)}\frac{\log(4\pi)}{2}+\Big{(}c+\frac{1}{4}\Big{)}\log
2+\Big{(}\frac{1}{2}+\alpha+c\Big{)}(1+\alpha+c)\frac{R^{2}}{1+R^{2}}\\\
&\quad-\frac{1}{3}\frac{1-R^{2}}{1+R^{2}}\mathcal{I}_{2}+\frac{1}{24}\Big{(}1+\frac{11}{1+R^{2}}\Big{)}-2\zeta^{\prime}(-1)+\log\Big{(}G(c+1)G(c+\tfrac{3}{2})\Big{)}.\end{split}$

Here, the constants $\mathcal{I}_{1}$ and $\mathcal{I}_{2}$ are given by
(1.17) and (1.18).

The numerical verifications of Theorems 1.1 and 1.3 are provided in Figure 2.
Here, we present the plots for the case where $\alpha=c=0$, and similar
figures can be observed for other values of $\alpha$ and $c$.

In general, for Pfaffian point processes, the large gap problems have been
much less developed, even in dimension one, see Subsection 1.3.4. Indeed,
Theorem 1.3 provides one of the very few results on precise large gap
asymptotics for Pfaffian point processes.

(a) Complex

(b) Symplectic

Figure 2. The plot (A) shows $R\mapsto A_{6}$ (black full line) and its
comparison
$R\mapsto\log\mathcal{P}_{n}^{\mathbb{C}}(R;\alpha,c)-(A_{1}n^{2}+A_{2}n\log
n+A_{3}n+A_{4}\sqrt{n}+A_{5}\log n),$ where $\alpha=c=0$. Here, $n=10$ (red,
dotted line), $n=40$ (blue, dot-dashed line) and $n=160$ (purple, dashed
line), cf. Lemma 2.1. The plot (B) shows the same figure for the symplectic
case.

By comparing Theorems 1.1 and 1.3, one can observe

(1.34) $2\,A_{1}=\widehat{A}_{1},\qquad A_{2}=\widehat{A}_{2}.$

The agreement of the first two terms (up to a factor of 2 for the first term)
turns out to be a universal feature of free energy expansions of determinantal
and Pfaffian Coulomb gases. We refer the reader to [29], where such the same
phenomenon was established for partition functions with soft edges.

###  1.3. Related works and further results

In this subsection, we shall discuss the progress made regarding large gap
probabilities and related topics, providing further motivations for our study
and elucidating our contributions to this direction. In addition, we present
further consequences arising from our main results.

####  1.3.1. Large gap probabilities for the Ginibre ensemble

As previously mentioned, Ginibre matrices are non-Hermitian random matrices
with i.i.d. Gaussian entries, see [26, 27] for recent reviews. The joint
probability distributions of their eigenvalues are of the form (1.4) and
(1.5), where the external potential is instead given by $|z|^{2}$. Its two-
parameter extension, the Mittag-Leffler ensemble, has a potential of the form

(1.35) $Q^{\rm ML}(z)=|z|^{2b}-\frac{2c}{n}\log|z|,\qquad(b>0,\,c>-1).$

The associated limiting spectrum is a centred disc of radius
$b^{-\frac{1}{2b}}$ and the density is given by

(1.36) $\rho^{\rm ML}(z)=b^{2}|z|^{2b-2}.$

In particular, if $b=1$, this gives rise to the well-known circular law.

An early study on the expansion of large gap probabilities of the complex
Ginibre ensemble was due to Forrester [62], where he derived the first _four
terms_ in the expansion, i.e. $C_{j}$ ($j=1,2,3,4$) in (1.37) below. (See also
[70, 71] for the first two terms). Subsequently, using the theory of skew-
orthogonal polynomials, the symplectic counterpart was addressed in [8], where
the authors obtained the _first two terms_ , and partially conjectured the
third one. (See also [9] for the extension to the product of Ginibre
matrices.)

For a considerable period, obtaining a more precise expansion has been a
longstanding open problem. It is only now that the specialised methods of
asymptotic analysis required for this task have been mastered in the work [34]
of Charlier. Consequently, it has been established that the large gap
probabilities of the Mittag-Leffler ensemble take the form:

(1.37) $\exp\Big{(}C_{1}n^{2}+C_{2}n\log n+C_{3}n+C_{4}\sqrt{n}+C_{5}\log
n+C_{6}+\mathcal{O}(n^{-\frac{1}{12}})\Big{)}.$

We mention that even though the large gap probabilities were stated in [34]
only for the complex ensembles, the symplectic counterparts also follow by
combining the results in [34] with [27, Proposition 5.10].

The key ingredient for the analysis of Charlier is a uniform asymptotic
expansion of the incomplete gamma function due to Temme, see [98, Section
11.2.4] and [92, Section 8.12]. The basic concept of this method involves
dividing the required summation based on different asymptotic regimes of the
incomplete gamma function. Each division is meticulously chosen to minimise
error bounds during asymptotic analysis, ensuring the smallest possible
cumulative errors. Even though some of these ideas trace back to Forrester’s
work [62] more than 30 years ago, their implementation demands technical
mastery and a very systematic approach, particularly to achieve precision up
to the constants $C_{5}$ and $C_{6}$ [34]. This method was also recently used
in e.g. [33, 11] to obtain precise results on counting statistics.

Contrary to [34, 33, 11], the gap probability $\mathcal{P}_{n}$ is not
expressed in terms of the incomplete gamma function but in terms of the
_incomplete beta function_ (cf. Lemma 2.1). This difference with [34, 33, 11]
has far-reaching consequences in the proofs of Theorems 1.1 and 1.3. Indeed,
our analysis requires very detailed asymptotics of the incomplete beta
functions for various regimes of the parameters, and in particular we rely on
some results in [98, Section 11.3.3]. We illustrate our proofs’ strategy in
Figures 4 and 5 below. Interestingly, after a very detailed and intricate
analysis, the resulting error bounds in Theorems 1.1 and 1.3 are
$\mathcal{O}(n^{-\frac{1}{12}})$, i.e. of the same order as the bounds for the
Mittag-Leffler ensemble (1.37). As a side note, one of the additional
technical (or practical) difficulties in our work stems from a typo present in
Temme’s original book [98, Section 11.3.3]. This typo is also present in NIST
[92, Eq.(8.18.9)], and we are not aware of a literature where these typos have
been reported, see Lemma 2.2 for the corrected version.

####  1.3.2. Leading order asymptotics and balayage measure

The leading order asymptotic behaviour of the large gap probabilities can
alternatively be derived using a potential theoretic method [94] by
investigating the associated balayage measure. For this purpose, let us recall
that for a given probability measure $\mu$ on $\mathbb{C}$, the weighted
logarithmic energy $I_{V}[\mu]$ associated with the external potential $V$ is
given by

(1.38)
$I_{V}[\mu]:=\int_{\mathbb{C}^{2}}\log\frac{1}{|z-w|}\,d\mu(z)\,d\mu(w)+\int_{\mathbb{C}}V\,d\mu.$

Assuming that $V$ is lower semi-continuous and finite on some set of positive
capacity, $I_{V}[\mu]$ has a unique minimiser $\sigma_{V}$ with a support $S$
called the droplet. Furthermore, if $V$ is $C^{2}$-smooth in a neighbourhood
of $S$, then $\sigma_{V}$ is absolutely continuous with respect to $dA$, and
takes the form

(1.39) $d\sigma_{V}(z)=\Delta
V(z)\,\mathbbm{1}_{S_{V}}(z)\,dA(z),\qquad\Delta=\partial\bar{\partial}.$

This property is often called the Laplacian growth, especially in the context
of the Hele-Shaw flow. We now consider a gap $\Omega\subset\mathbb{C}$ and
redefine the potential by

(1.40) $V_{\Omega}(z)=\begin{cases}V(z)&\textup{if }z\not\in\Omega,\vskip
3.0pt plus 1.0pt minus 1.0pt\\\ +\infty&\textup{if }z\in\Omega.\end{cases}$

If $S\cap\Omega\neq\emptyset$, then the equilibrium measure associated with
$V_{\Omega}$ is no longer absolutely continuous with respect to $dA$, but
rather takes the form

(1.41) $d\sigma_{V_{\Omega}}(z)=\Delta
V(z)\,\mathbbm{1}_{S_{V}\setminus\Omega}(z)\,dA(z)+\nu(z)\,\mathbbm{1}_{\partial\Omega}(z)|dz|.$

As the term _balayage_ means _sweeping_ in French, from an electrostatic
perspective, the constraint that there are no particles in $\Omega$ pushes
them to the boundary of $\Omega$ with a non-trivial distribution $\nu(z)$, cf.
Figure 3. The balayage measure can be used to derive the leading order of the
large gap probabilities. In particular, the leading order of the (logarithm of
the) large gap probabilities is proportional to

(1.42) $I_{V}[\sigma_{V}]-I_{V_{\Omega}}[\sigma_{V_{\Omega}}],$

see e.g. [1, 44]. From a probabilistic point of view, the leading term (1.42)
is a large deviation rate function, see e.g. [95]. An advantage of this
approach is that it can be applied not only to the $\beta=2$ Coulomb gas
ensembles but also to general $\beta>0$. However, for a general shape of
domain, it is far from obvious to explicitly compute the balayage measure, and
some non-trivial examples have been recently obtained in [2, 35]. On one hand,
this approach usually has its limitations in deriving only the leading order
asymptotic expansion.

In our present case, one can consider the balayage measure associated with the
potential

(1.43) $Q^{R}_{n}(z):=\begin{cases}+\infty&\textup{if }|z|\leq R,\vskip 3.0pt
plus 1.0pt minus 1.0pt\\\ Q_{n}(z)&\textup{if }|z|>R,\end{cases}$

where $Q_{n}$ is given by (1.6). In this case, due to the rotational symmetry,
the balayage measure becomes uniform distribution on $|z|=R$, see Figure 3.
Then the leading order of our results also follows from computing the
associated energy, see [35, Eq.(2.9)].

(a) $\\{z_{j}\\}_{j=1}^{n}$, $z_{j}\in\mathbb{C}$

(b) $\\{x_{j}\\}_{j=1}^{n}$, $x_{j}\in\mathbb{S}$

Figure 3. The plot (A) displays eigenvalues of the spherical symplectic
ensembles conditioned to have a gap $\\{z\in\mathbb{C}:|z|<1\\}$, where
$n=200$. The plot (B) shows its projection to the sphere.

Let us also mention that beyond the two-dimensional Coulomb gases, the zeros
of Gaussian analytic functions have been studied in the literature, see e.g.
[68] and references therein. A characteristic feature when conditioning on a
hole event for this model is that it produces a macroscopic region outside the
hole region [69]. Such a forbidden region has been further investigated in
recent literature [90].

####  1.3.3. Partition functions for two-dimensional point processes

For a given external potential $V$, the partition functions of determinantal
and Pfaffian Coulomb gases are given by

(1.44) $\displaystyle Z_{n,V}^{\mathbb{C}}$
$\displaystyle:=\int_{\mathbb{C}^{n}}\prod_{j>k=1}^{n}|z_{j}-z_{k}|^{2}\prod_{j=1}^{n}e^{-n\,V(z_{j})}\,dA(z_{j}),$
(1.45) $\displaystyle Z_{n,V}^{\mathbb{H}}$
$\displaystyle:=\int_{\mathbb{C}^{n}}\prod_{j>k=1}^{n}|z_{j}-z_{k}|^{2}|z_{j}-\overline{z}_{k}|^{2}\prod_{j=1}^{n}|z_{j}-\overline{z}_{j}|^{2}e^{-2nV(z_{j})}\,dA(z_{j}),$

respectively. The asymptotic expansion of the partition function stands as one
of the cornerstones in Coulomb gas theory. For a regular potential $V$,
substantial progress has been made in this direction [17, 31, 107, 79].
Furthermore, for comprehensive literature and the latest advancements, we
refer the reader to [10, 29, 30, 95].

For the partition functions $Z_{n}^{\mathbb{C}}$ and $Z_{n}^{\mathbb{H}}$ of
induced spherical ensembles, their asymptotic expansions can be derived from
well-known formulas involving the Barnes $G$-function.

######  Proposition 1.4 (Free energy expansions of the induced spherical
ensembles).

As $n\to\infty$, the following asymptotic expansion holds.

  * •

(Complex ensemble) We have

(1.46) $\displaystyle\log
Z_{n}^{\mathbb{C}}=\mathcal{A}_{1}n^{2}+\mathcal{A}_{2}n\log
n+\mathcal{A}_{3}n+\mathcal{A}_{4}\sqrt{n}+\mathcal{A}_{5}\log
n+\mathcal{A}_{6}+\mathcal{O}(n^{-1}),$

where

$\displaystyle\mathcal{A}_{1}=-\frac{1}{2},\qquad\mathcal{A}_{2}=\frac{1}{2},\qquad\mathcal{A}_{3}=\frac{\log(2\pi)}{2}-1-\alpha-c,\qquad\mathcal{A}_{4}=0,\qquad\mathcal{A}_{5}=\frac{\alpha^{2}+c^{2}}{2}+\frac{1}{3},$
$\displaystyle\mathcal{A}_{6}=\frac{\log(2\pi)}{2}-\frac{1}{12}+2\zeta^{\prime}(-1)-\frac{1}{2}(\alpha+c)(\alpha+c+1-\log(2\pi))-\log\Big{(}G(\alpha+1)G(c+1)\Big{)}.$

  * •

(Symplectic ensemble) We have

(1.47) $\displaystyle\log
Z_{n}^{\mathbb{H}}=\widehat{\mathcal{A}}_{1}n^{2}+\widehat{\mathcal{A}}_{2}n\log
n+\widehat{\mathcal{A}}_{3}n+\widehat{\mathcal{A}}_{4}\sqrt{n}+\widehat{\mathcal{A}}_{5}\log
n+\widehat{\mathcal{A}}_{6}+\mathcal{O}(n^{-1}),$

where

$\displaystyle\widehat{\mathcal{A}}_{1}=-1,\qquad\widehat{\mathcal{A}}_{2}=\frac{1}{2},\qquad\widehat{\mathcal{A}}_{3}=\frac{\log(4\pi)}{2}-2-2\alpha-2c,\qquad\widehat{\mathcal{A}}_{4}=0,\qquad\widehat{\mathcal{A}}_{5}=\alpha^{2}+\frac{\alpha}{2}+c^{2}+\frac{c}{2}+\frac{5}{12},$
$\displaystyle\begin{split}\widehat{\mathcal{A}}_{6}&=\log(2\pi)-\frac{13}{24}+4\zeta^{\prime}(-1)-(\alpha+c)(\alpha+c+\tfrac{3}{2}-\log(2\pi))-\log\Big{(}G(\alpha+1)G(\alpha+\tfrac{3}{2})G(c+1)G(c+\tfrac{3}{2})\Big{)}.\end{split}$

For the complex case, we also refer to [61, 74]. See also [18] and references
therein for the asymptotic behaviours of the logarithmic energy of the
configurations.

Let us stress here that $\mathcal{A}_{4}=\widehat{\mathcal{A}}_{4}=0$. Such an
absence of the $\mathcal{O}(\sqrt{n})$-term is a general feature of the free
energy expansion for $\beta=2$ ensembles, see [29, 10] for the rotationally
symmetric case and [30] for a non-rotationally symmetric example. Furthermore,
note that for $\alpha=c=0$, we have

(1.48)
$\mathcal{A}_{5}\Big{|}_{\alpha=c=0}=\frac{1}{2}-\frac{\chi}{12},\qquad\widehat{\mathcal{A}}_{5}\Big{|}_{\alpha=c=0}=\frac{1}{2}-\frac{\chi}{24},$

where $\chi=2$ is the Euler index of the droplet, in our case, the Riemann
sphere. This form of the coefficient of the $\log n$ term was introduced in
the work of Jancovici et al. [72, 97] and is expected to hold for more general
droplets [29, 10, 30]. Note further that for $\alpha=c=0$, we have

(1.49)
$\mathcal{A}_{6}\Big{|}_{\alpha=c=0}=\frac{\log(2\pi)}{2}-\frac{1}{12}+\chi\,\zeta^{\prime}(-1),\qquad\widehat{\mathcal{A}}_{6}\Big{|}_{\alpha=c=0}=\frac{\log(2\pi)}{2}-\frac{13}{24}+\frac{\chi}{2}\Big{(}\frac{5\log
2}{12}+\zeta^{\prime}(-1)\Big{)}.$

Here, again the $\chi$-dependent terms are expected to be universal, see [29].
Other than these $\chi$-dependent terms, the other parts of the
$\mathcal{O}(1)$-term are related to the conformal geometric properties of the
equilibrium measure, see e.g. the introduction of [30]. For the ensembles on
the sphere with general radially symmetric potentials, the associated free
energy expansions will be addressed in a forthcoming work.

Contrary to the regular case, when $V$ contains certain singularities, there
has been less understanding of the asymptotic expansions. Among the various
types of singularities one may consider, the following types are particularly
noteworthy due to various motivations and have been extensively investigated
in the literature.

  * •

(Jump type singularity) This is the case where the weight function has a
discontinuity. This type of singularity naturally arises in the context of the
moment generating function of the disc counting function, see [33, 11, 12, 5,
38, 4] and references therein for recent progress. In particular, for the
spherical ensembles, this has been addressed in a recent work [89].

  * •

(Root type singularity) The pointwise root type singularity is equivalent to
the point charge insertions. This type of singularities also finds application
in the context of moments of characteristic polynomials [52, 102, 30]. We also
refer to [15, 78, 80, 81, 82, 20, 21, 16] for extensive studies on the
associated orthogonal polynomials. On the other hand, the circular (or non-
isolated in general) root type singularity naturally arises in the context of
the truncations of Haar unitary matrices [108, 13] or finite-rank perturbation
[67].

  * •

(Hard wall type singularity) This is the type we are exploring in our present
work, namely the potential takes the form (1.40). While one might consider it
as another jump type singularity, unlike the one seen in counting statistics,
in the context of gap probabilities, the potential reveals an extreme
discontinuity as it takes the value $+\infty$ in a certain region. As far as
we are aware, the only instance prior to our current work where the partition
function expansion, up to the constant term, was achieved for two-dimensional
point processes, is the work [34] of Charlier.

In the context we have discussed, our main results can then be stated as free
energy expansions of the partition functions (1.44) and (1.45), associated
with the potential (1.6), which exhibits both root and hard wall type
singularities.

######  Corollary 1.5 (Free energy expansions with singularities).

Let

(1.50)
$\mathsf{A}_{j}=A_{j}+\mathcal{A}_{j},\qquad\widehat{\mathsf{A}}_{j}=\widehat{A}_{j}+\widehat{\mathcal{A}}_{j},\qquad(j=1,2,\dots,6),$

where $A_{j}$ and $\widehat{A}_{j}$ are given in Theorems 1.1 and 1.3
respectively, and $\mathcal{A}_{j}$ and $\widehat{\mathcal{A}}_{j}$ are given
in Proposition 1.4. Let $V=Q^{R}$, where $Q^{R}$ is given by (1.43). Then as
$n\to\infty$, we have

(1.51) $\displaystyle\log Z_{n,V}^{\mathbb{C}}$
$\displaystyle=\mathsf{A}_{1}n^{2}+\mathsf{A}_{2}n\log
n+\mathsf{A}_{3}n+\mathsf{A}_{4}\sqrt{n}+\mathsf{A}_{5}\log
n+\mathsf{A}_{6}+\mathcal{O}(n^{-\frac{1}{12}}),$ (1.52) $\displaystyle\log
Z_{n,V}^{\mathbb{H}}$
$\displaystyle=\widehat{\mathsf{A}}_{1}n^{2}+\widehat{\mathsf{A}}_{2}n\log
n+\widehat{\mathsf{A}}_{3}n+\widehat{\mathsf{A}}_{4}\sqrt{n}+\widehat{\mathsf{A}}_{5}\log
n+\widehat{\mathsf{A}}_{6}+\mathcal{O}(n^{-\frac{1}{12}}).$

###### Proof.

Note that by the definition of $Q^{R}$ in (1.43), we have

(1.53)
$\mathcal{P}_{n}^{\mathbb{C}}(R;\alpha,c)=\frac{Z_{n,V}^{\mathbb{C}}}{Z_{n}^{\mathbb{C}}},\qquad\mathcal{P}_{n}^{\mathbb{H}}(R;\alpha,c)=\frac{Z_{n,V}^{\mathbb{H}}}{Z_{n}^{\mathbb{H}}}.$

Then the corollary immediately follows from Theorems 1.1, 1.3 and Proposition
1.4. ∎

Contrary to the regular case stated in Proposition 1.4 or in e.g. [29, 30,
10], in Corollary 1.5, a non-trivial coefficient of the
$\mathcal{O}(\sqrt{n})$-term does arise when considering the hard wall
constraints of the potential. This is due to the hard wall constraints of the
potential.

####  1.3.4. Large gap probabilities for one-dimensional point processes

Comparing with two-dimensional point processes, there have been far more
investigations into the gap probabilities for one-dimensional point processes.
For the reader’s convenience, let us list some of the literature.

Some early works where the leading order term of large gap probabilities for
the eigenvalues of classical Hermitian random matrices include [19, 46, 47,
73, 84, 85, 101, 66]. Some of these works also find interesting connections
with spin glass model as well as free fermionic systems. As in the two-
dimensional cases, the leading order can be obtained using potential-theoretic
computations, sometimes referred to as the Coulomb gas method.

For determinantal point processes, large gap problems can be interpreted as
asymptotic expansions for structured determinants. For one-dimensional point
processes, the Riemann-Hilbert approach has been particularly successful in
obtaining precise asymptotic behaviours. (On the other hand, in dimension two,
developing a Riemann-Hilbert approach to large gap problems breaking the
radial symmetry is still an outstanding open problem.) However, multiplicative
constants of large gap problems are in general very difficult to obtain and
are still open in some cases (see below).

  * •

(Hankel determinants) The form of the asymptotic expansion depends on the
external potential of the Hermitian random unitary matrices, particularly on
their edge behaviours of the associated equilibrium measure. For the
generalised Gaussian unitary ensembles, a general result can be found in [51,
36], and for the Laguerre and Jacobi types, these can be found in [37].
Furthermore, the large gap asymptotics of the Muttalib-Borodin model with
constant potential involve a generalisation of Hankel determinants and were
established in [32].

  * •

(Toeplitz determinants) For the circular unitary ensemble, the probability of
an arc being empty of eigenvalues, up to a multiplicative constant, was
obtained in [103]. This was later extended in [57, Proposition 1.1.3] to
situations where the gap region consists of multiple arcs of the same length
symmetrically located on the unit circle, and higher order correction terms
were obtained in [86].

  * •

(Fredholm determinants) In this case, the asymptotic expansions have been
investigated for different limiting point processes.

    * –

_(Sine point process)_ After the asymptotics of gap probabilities on a single
interval were conjectured in [53], they were proved simultaneously and
independently in [55, 75]. The case of several interval gaps was investigated
in [104], where later the results were made more explicit in [48]. In
particular, the oscillations were expressed in terms of Jacobi theta functions
in [48] for the first time. Nonetheless, the multiplicative constant was still
missing in [48]. For the two-interval case, this constant was recently
obtained in [58].

    * –

_(Airy point process)_ The conjecture on the large gap asymptotics made in
[99] was shown in [49, 14] up to and including the multiplicative constant.
For the case of two intervals, this was obtained in [22] without a
multiplicative constant, which was later obtained in [76]. The gap
probabilities in the bulk were also obtained in [23].

    * –

_(Bessel point process)_ The large gap asymptotics on a single interval were
obtained in [100] up to, but not including, the multiplicative constant. The
multiplicative constant conjectured in [100] was later established in [56, 50]
using different approaches. Recently, the case of several intervals has also
been addressed in [24] with the multiplicative constants left undetermined.

Beyond the standard limiting point processes, there have been further
developments on other types of point processes. These include the Wright’s
generalised Bessel and Meijer-$G$ point processes [41, 39, 40]; the Pearcey
point process [45]; the hard edge Pearcey point process [105]; the tacnode
point process [106]; and the Freud point process [42].

### Organisation of the paper

The rest of this paper is organised as follows. In Section 2, we provide an
outline of proofs together with necessary preliminaries. Section 3 is devoted
to the analysis of the complex case, where we prove Theorem 1.1. Section 4 is
for the symplectic counterpart, where we provide the proof of Theorem 1.3.
Additionally, this article contains an appendix where we compile some tail
asymptotic behaviours of the binomial distributions.

##  2\. Outline of proofs

In this section, we provide the preliminaries and outline of proofs of our
main results.

The key ingredients for the proofs are given as follows.

  * •

We first obtain the expression of gap probabilities in terms of products of
incomplete beta functions (Lemma 2.1). This follows from the
determinantal/Pfaffian structures of the underlying models and recent
development [6] on the planar skew-orthogonal polynomials.

  * •

We make use of the uniform asymptotic behaviours of the incomplete beta
function (Lemma 2.2). Although this formula is provided in Temme’s book [98,
Section 11.3.3.2], it turns out that there is a typo that can be corrected. We
note that the same typo appears in [92, Eq.(8.18.9)].

  * •

The remainder of the proof entails the asymptotic analysis of the finite-$n$
expression utilising the uniform asymptotic behaviours of the incomplete beta
function, see Figures 4 and 5. This constitutes the main aspect of our proofs,
and we apply the method developed by Charlier [34, 33] to our model.

$\log\mathcal{P}_{n}^{\mathbb{C}}=\sum_{k=0}^{n-1}(*)$$0$$k_{-}$$k_{+}$$k_{M}$$n-1$$k_{-}$$g_{-}$$g_{+}$$k_{+}$$S_{0}$
(Lem. 3.1)$S_{1}$ (Lem. 3.11)$S_{2}$ (Lem. 3.4)$S_{3}$ (Lem. 3.2)$S_{1}^{(1)}$
(Lem. 3.6)$S_{1}^{(2)}$ (Lem. 3.8)$S_{1}^{(3)}$ (Lem. 3.10) Figure 4.
Illustration of the divisions of the summations in (2.19) and (3.22), together
with lemmas on the asymptotic expansions in each sub-summation.
$\log\mathcal{P}_{n}^{\mathbb{H}}=\sum_{k=0}^{n-1}(*)$$0$$k_{-}$$k_{+}$$k_{M}$$n-1$$k_{-}$$g_{-}$$g_{+}$$k_{+}$$\widehat{S}_{0}$
(Lem. 4.1)$\widehat{S}_{1}$ (Lem. 4.10)$\widehat{S}_{2}$ (Lem.
4.4)$\widehat{S}_{3}$ (Lem. 4.2)$\widehat{S}_{1}^{(1)}$ (Lem.
4.6)$\widehat{S}_{1}^{(2)}$ (Lem. 4.7)$\widehat{S}_{1}^{(3)}$ (Lem. 4.9)
Figure 5. The analogue of Figure 4 for the symplectic counterpart, i.e.
illustration of the summations in (2.21) and (4.13).

Before introducing the strategy in more detail, let us first provide the proof
of Proposition 1.4.

###### Proof of Proposition 1.4.

By using the definition of the Barnes $G$-function, one can rewrite (1.7) as

(2.1) $\displaystyle Z_{n}^{\mathbb{C}}$
$\displaystyle=\frac{n!}{\Gamma(n+\alpha+c+1)^{n}}\frac{G(n+c+1)}{G(c+1)}\frac{G(n+\alpha+1)}{G(\alpha+1)},$
(2.2) $\displaystyle Z_{n}^{\mathbb{H}}$
$\displaystyle=n!\Big{(}\frac{2^{2n+2\alpha+2c+1}}{\pi\Gamma(2n+2\alpha+2c+2)}\Big{)}^{n}\frac{G(n+c+1)}{G(c+1)}\frac{G(n+c+3/2)}{G(c+3/2)}\frac{G(n+\alpha+1)}{G(\alpha+1)}\frac{G(n+\alpha+3/2)}{G(\alpha+3/2)}.$

Then the desired asymptotic expansion follows from the well-known asymptotic
behaviours of the gamma function [92, Eq.(5.11.1)]

(2.3) $\log N!=N\log N-N+\frac{1}{2}\log
N+\frac{1}{2}\log(2\pi)+\frac{1}{12N}+\mathcal{O}\Big{(}\frac{1}{N^{3}}\Big{)},\qquad(N\to\infty)$

and of the Barnes $G$-function [92, Eq.(5.17.5)]

(2.4) $\displaystyle\begin{split}\log G(z+1)=\frac{z^{2}\log
z}{2}-\frac{3}{4}z^{2}+\frac{\log(2\pi)z}{2}-\frac{\log
z}{12}+\zeta^{\prime}(-1)-\frac{1}{240z^{2}}+\mathcal{O}\Big{(}\frac{1}{z^{4}}\Big{)},\qquad(z\to\infty).\end{split}$

∎

Recall that the regularised incomplete beta function $I_{x}$ is given by

(2.5) $I_{x}(a,b):=\frac{B_{x}(a,b)}{B(a,b)},\qquad
B_{x}(a,b):=\int_{0}^{x}t^{a-1}(1-t)^{b-1}\,dt$

where

(2.6) $B(a,b)=B_{1}(a,b):=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$

is the beta function, see [92, Chapter 8]. Note that by definition, we have

(2.7) $I_{x}(a,b)=1-I_{1-x}(b,a).$

We also note that for integer values $n\geq m$, we have

(2.8) $I_{x}(m,n-m+1)=\sum_{j=m}^{n}\binom{n}{j}x^{j}(1-x)^{n-j}.$

This in turn implies that the regularised incomplete beta function is the
cumulative distribution function of the binomial distribution, i.e. for $X\sim
B(n,p)$,

(2.9) $\mathbb{P}(X\leq k)=I_{1-p}(n-k,k+1)=1-I_{p}(k+1,n-k).$

In the following lemma, we express the gap probabilities in terms of the
incomplete beta functions.

######  Lemma 2.1 (Evaluation of the gap probabilities at finite-$n$).

Let

(2.10) $x=\frac{1}{1+R^{2}}.$

Then we have

(2.11) $\displaystyle\mathcal{P}_{n}^{\mathbb{C}}(R;\alpha,c)$
$\displaystyle=\prod_{k=0}^{n-1}I_{x}(k+\alpha+1,n+c-k)=\prod_{k=0}^{n-1}I_{x}(n+\alpha-k,k+c+1),$
(2.12) $\displaystyle\mathcal{P}_{n}^{\mathbb{H}}(R;\alpha,c)$
$\displaystyle=\prod_{k=0}^{n-1}I_{x}(2k+2\alpha+2,2n+2c-2k)=\prod_{k=0}^{n-1}I_{x}(2n+2\alpha-2k,2k+2c+2).$

###### Proof.

Recall that $Q_{n}^{R}$ is given by (1.43). Let

(2.13) $\displaystyle
h_{k,m}:=\int_{\mathbb{C}}|z|^{2k}\,e^{-mQ_{n}(z)}\,dA(z),\qquad
h_{k,m}(R):=\int_{\mathbb{D}(0,R)^{c}}|z|^{2k}\,e^{-mQ_{n}(z)}\,dA(z).$

It follows from the general theory of determinantal and Pfaffian point
processes that the associated partition functions (1.44) and (1.45) can be
expressed in terms of (skew)-norms of the associated (skew)-orthogonal
polynomials. Combining this with (1.53), we have

(2.14)
$\mathcal{P}^{\mathbb{C}}_{n}(R)=\prod_{k=0}^{n-1}\frac{h_{k,n}(R)}{h_{k,n}},\qquad\mathcal{P}^{\mathbb{H}}_{n}(R)=\prod_{k=0}^{n-1}\frac{h_{2k+1,2n}(R)}{h_{2k+1,2n}},$

see e.g. [29, Eq.(1.37)]. In particular, we have used the construction of the
skew-orthogonal polynomials [6, Corollary 3.3]. On the other hand, by using
the Euler beta integrals, we have

$\displaystyle h_{k,m}(R)$
$\displaystyle=2\int_{R}^{\infty}\frac{r^{2k+1+2c\frac{m}{n}}}{(1+r^{2})^{(n+\alpha+c+1)\frac{m}{n}}}\,dr=\int_{R}^{\infty}\Big{(}\frac{r^{2}}{1+r^{2}}\Big{)}^{k+c\,\frac{m}{n}}\Big{(}\frac{1}{1+r^{2}}\Big{)}^{(n+\alpha+1)\,\frac{m}{n}-k-2}\frac{2r}{(1+r^{2})^{2}}\,dr$
$\displaystyle=\int_{0}^{x}t^{(n+\alpha+1)\,\frac{m}{n}-k-2}(1-t)^{k+c\,\frac{m}{n}}\,dt=B_{x}((n+\alpha+1)\,\tfrac{m}{n}-k-1,k+c\,\tfrac{m}{n}+1),$

which gives

$\displaystyle\frac{h_{k,n}(R)}{h_{k,n}}=I_{x}(n+\alpha-k,k+c+1),\qquad\frac{h_{2k+1,2n}(R)}{h_{2k+1,2n}}=I_{x}(2n+2\alpha-2k,2k+2c+2).$

This completes the proof. ∎

To analyse the expressions in Lemma 2.1, it is convenient to introduce the
notation

(2.15) $N=\begin{cases}n+\alpha+c&\textup{ for the complex case,}\vskip 3.0pt
plus 1.0pt minus 1.0pt\\\ n+\alpha+c+\frac{1}{2}&\textup{ for the symplectic
case.}\end{cases}$

We further define

(2.16) $k_{-}=\lfloor\delta N\rfloor-\alpha-1,\qquad
k_{+}=\lceil(1-\delta)N\rceil-\alpha-1,$

and

(2.17) $k_{M}=\begin{cases}N-\lceil M\rceil-\alpha-1&\textup{ for the complex
case,}\vskip 3.0pt plus 1.0pt minus 1.0pt\\\ N-\lceil
M\rceil-\alpha-\frac{3}{2}&\textup{ for the symplectic case.}\end{cases}$

Here, we set

(2.18) $M=N^{\frac{1}{3}}$

and choose $\delta\in(0,\min\\{x,1-x\\})$ small enough so that
$0<k_{-}<k_{+}<k_{M}<N$ holds for large $N$.

By using (2.16) and (2.17), we divide the sum in (2.11) by the following four
parts:

(2.19) $\log\mathcal{P}_{n}^{\mathbb{C}}(R;\alpha,c)=S_{0}+S_{1}+S_{2}+S_{3},$

where

(2.20) $\displaystyle S_{0}$ $\displaystyle=\sum_{k=0}^{k_{-}}\log
I_{x}(k+\alpha+1,n+c-k),\qquad$ $\displaystyle
S_{1}=\sum_{k=k_{-}+1}^{k_{+}-1}\log I_{x}(k+\alpha+1,n+c-k),$ $\displaystyle
S_{2}$ $\displaystyle=\sum_{k=k_{+}}^{k_{M}}\log
I_{x}(k+\alpha+1,n+c-k),\qquad$ $\displaystyle
S_{3}=\sum_{k=k_{M}+1}^{n-1}\log I_{x}(k+\alpha+1,n+c-k).$

Similarly, we divide the sum in (2.12) by

(2.21)
$\log\mathcal{P}_{n}^{\mathbb{H}}(R;\alpha,c)=\widehat{S}_{0}+\widehat{S}_{1}+\widehat{S}_{2}+\widehat{S}_{3},$

where

(2.22) $\displaystyle\widehat{S}_{0}$ $\displaystyle=\sum_{k=0}^{k_{-}}\log
I_{x}(2k+2\alpha+2,2n+2c-2k),\quad$
$\displaystyle\widehat{S}_{1}=\sum_{k=k_{-}+1}^{k_{+}-1}\log
I_{x}(2k+2\alpha+2,2n+2c-2k),$ $\displaystyle\widehat{S}_{2}$
$\displaystyle=\sum_{k=k_{+}}^{k_{M}}\log I_{x}(2k+2\alpha+2,2n+2c-2k),\quad$
$\displaystyle\widehat{S}_{3}=\sum_{k=k_{M}+1}^{n-1}\log
I_{x}(2k+2\alpha+2,2n+2c-2k).$

See Figures 4 and 5. As previously mentioned, the divisions in (2.20) and
(2.22) are chosen in a way to minimise cumulative errors. For the asymptotic
behaviours of $S_{0}$, $S_{2}$, and $S_{3}$, as well as $\widehat{S}_{0}$,
$\widehat{S}_{2}$, and $\widehat{S}_{3}$, we can apply various tail behaviours
of the binomial distribution (see Appendix A) together with a version of the
Euler-Maclaurin formula (Lemma 3.3). The most technical part involves $S_{1}$
and $\widehat{S}_{1}$, which, from the viewpoint of the asymptotic behaviour
of the incomplete beta function, are referred to as the _symmetric_ regime. In
this case, it is necessary to further divide the summations as in (3.22) and
(4.13). For this regime, the following uniform asymptotics of the incomplete
beta function play a key role.

######  Lemma 2.2 (Uniform asymptotics of the incomplete beta function).

Let $x_{0}=a/N$. Then as $N\to\infty$, we have

(2.23)
$I_{x}(a,N-a)=\frac{1}{2}\operatorname{erfc}\Big{(}-\eta\sqrt{\frac{N}{2}}\Big{)}+R_{a}(\eta),$

where

(2.24) $R_{a}(\eta)=e^{-\frac{1}{2}\eta^{2}N}\frac{1}{2\pi
i}\int_{-\infty}^{\infty}e^{-\frac{1}{2}Nu^{2}}g(u)\,du.$

Here, $\eta$ is given by

(2.25)
$-\frac{1}{2}\eta^{2}=x_{0}\log\Big{(}\frac{x}{x_{0}}\Big{)}+(1-x_{0})\log\Big{(}\frac{1-x}{1-x_{0}}\Big{)},\qquad\frac{\eta}{x-x_{0}}>0,$

and $g$ is given by

(2.26) $g(u)=\frac{dt}{du}\frac{1}{t-x}-\frac{1}{u+i\eta},$

where the variables $t$ and $u$ are correlated by the bijection

(2.27)
$\frac{1}{2}u^{2}=x_{0}\log\Big{(}\frac{t}{x_{0}}\Big{)}+(1-x_{0})\log\Big{(}\frac{1-t}{1-x_{0}}\Big{)},\qquad\frac{\operatorname{Im}t}{u}>0,\qquad
t\in\mathcal{L},$

on

(2.28)
$\mathcal{L}=\Big{\\{}t=\frac{\sin(\rho\sigma)}{\sin[(1+\rho)\sigma]}e^{i\sigma}:|\sigma|<\frac{\pi}{1+\rho}\Big{\\}}\mapsto
u\in\mathbb{R},\qquad\Big{(}\rho=\frac{x_{0}}{1-x_{0}}\Big{)}.$

Furthermore, we have

(2.29) $R_{a}(\eta)\sim\frac{1}{\sqrt{2\pi
N}}e^{-\frac{1}{2}\eta^{2}N}\sum_{k=0}^{\infty}\frac{(-1)^{k}c_{k}(\eta)}{N^{k}}$

uniformly for $x\in(0,1)$ and $x_{0}\in[\delta,1-\delta]$, where
$0<\delta<\frac{1}{2}$ is an arbitrarily fixed small constant. Here, the
coefficients $c_{n}(\eta)$ is given by

(2.30) $c_{n}(\eta)=-i(2n-1)!!g_{2n},\qquad
g_{n}=\frac{1}{n!}\frac{d^{n}}{du^{n}}g(u)\Big{|}_{u=0}.$

In particular, the first two coefficients $c_{0}$ and $c_{1}$ are given by

(2.31) $\displaystyle c_{0}(\eta)$
$\displaystyle=\frac{1}{\eta}-\frac{\sqrt{x_{0}(1-x_{0})}}{x-x_{0}},$ (2.32)
$\displaystyle c_{1}(\eta)$
$\displaystyle=-\frac{1}{\eta^{3}}+\frac{1-13x_{0}+13x_{0}^{2}}{12\sqrt{x_{0}(1-x_{0})}}\frac{1}{x-x_{0}}+\frac{(1-2x_{0})\sqrt{x_{0}(1-x_{0})}}{(x-x_{0})^{2}}+\frac{(x_{0}(1-x_{0}))^{3/2}}{(x-x_{0})^{3}}.$

All of the $c_{k}(\eta)$ are analytic at $\eta=0$.

######  Remark 2.3.

In [98, Section 11.3.3.2], the leading term of the right-hand side of (2.23)
is incorrectly written as
$\frac{1}{2}\operatorname{erfc}(-\eta\sqrt{\frac{N-a}{2}})$. The corrected
version follows from the integral

$e^{-\frac{1}{2}\eta^{2}N}\frac{1}{2\pi
i}\int_{-\infty}^{\infty}e^{-\frac{1}{2}Nu^{2}}\frac{1}{u+i\eta}\,du=\frac{1}{2}\operatorname{erfc}\Big{(}-\eta\sqrt{\frac{N}{2}}\Big{)}.$

This typo still appears in [92, Eq.(8.18.8)], where the first term of the
right-hand side there should be corrected as

(2.33)
$I_{x}(a,b)\sim\frac{1}{2}\operatorname{erfc}\Big{(}-\eta\sqrt{\frac{a+b}{2}}\Big{)}.$

The subleading terms in [98, 92] are correct. We note that the corrected
version (2.23) can also be easily verified numerically.

##  3\. Gap probabilities of the complex ensemble

In this section, we prove Theorem 1.1. As previously mentioned, we shall use
the splitting (2.19). In Subsection 3.1, we perform the analysis of $S_{0}$,
$S_{2}$ and $S_{3}$. Subsection 3.2 is devoted to the analysis of a more
complicated part $S_{1}$, where we further need the division (3.22).

###  3.1. Analysis of $S_{0}$, $S_{2}$ and $S_{3}$

We provide the asymptotic formulas of the $S_{j}\,(j=0,2,3)$ in increasing
order of difficulty: $S_{0}$ (Lemma 3.1), $S_{3}$ (Lemma 3.2), and then
$S_{2}$ (Lemma 3.4).

To simplify the notations, let us write

(3.1) $m_{k}:=k+\alpha+1$

so that the summand of (2.20) can be rewritten as

$I_{x}(k+\alpha+1,n+c-k)=I_{x}(m_{k},N+1-m_{k}).$

Recall that $S_{0}$ is given by (2.20).

######  Lemma 3.1 (Asymptotic behaviour of $S_{0}$).

There exists $C>0$ such that

(3.2) $S_{0}=\mathcal{O}(e^{-CN}),\qquad\text{as }N\to\infty.$

###### Proof.

Recall that $k_{-}$ is given by (2.16). Thus for $0\leq k\leq k_{-}$, we have
$\alpha+1\leq m_{k}\leq\lfloor\delta N\rfloor$. Let $X\sim B(N,x)$, where $B$
is the binomial distribution. Here $N$ and $x$ are given by (2.15) and (2.10).
Note that by (2.9), we have

(3.3) $\log I_{x}(m_{k},N+1-m_{k})=\log\Big{(}1-\mathbb{P}(X\leq
m_{k}-1)\Big{)}=\log\Big{(}1-\mathbb{P}(X<m_{k})\Big{)}.$

For every $\alpha+1\leq m\leq\lfloor\delta N\rfloor$, by using Lemma A.1, we
have

$\displaystyle\mathbb{P}(X<m)\leq\mathbb{P}(X<\delta
N)\leq\frac{1}{2}\operatorname{erfc}\Big{(}\frac{x-\delta}{\sqrt{2x}}\sqrt{N}\Big{)}\leq\frac{1}{2}e^{-\frac{(x-\delta)^{2}}{2x}N}=\mathcal{O}(e^{-CN}),$

where $C$ is independent of $m$. Hence, the summands of $S_{0}$ in (2.20) are
of order $\mathcal{O}(e^{-CN})$ uniformly, which completes the proof. ∎

Next, we analyse the sum $S_{3}$ in (2.20). For this purpose, let us write

(3.4) $\theta_{M}:=\lceil M\rceil-M,\qquad\theta_{\delta}:=\delta
N-\lfloor\delta N\rfloor,$

where $M$ is given by (2.18).

######  Lemma 3.2 (Asymptotic behaviour of $S_{3}$).

As $N\to\infty$, we have

(3.5) $S_{3}=D_{3}N+D_{5}\log
N+D_{6}-\frac{1}{6}\frac{M^{3}}{N}+\mathcal{O}\Big{(}\frac{1}{M}+\frac{M^{2}}{N}+\frac{M^{4}}{N^{2}}\Big{)},$

where

$\displaystyle D_{3}$ $\displaystyle=(M+\theta_{M}-c)\log x,\qquad
D_{5}=\frac{1}{2}(M^{2}+(2\theta_{M}-1)M+\theta_{M}^{2}-\theta_{M}-c^{2}+c),$
$\displaystyle D_{6}$
$\displaystyle=\frac{1}{2}\bigg{(}M^{2}+(2\theta_{M}-1)M+\theta_{M}^{2}-\theta_{M}-c^{2}+c\bigg{)}\log\Big{(}\frac{1-x}{x}\Big{)}-\Big{(}\frac{1}{2}M^{2}+M\theta_{M}+\frac{1}{2}\theta_{M}^{2}-\frac{1}{12}\Big{)}\log
M$ $\displaystyle\quad+\frac{3}{4}M^{2}+\Big{(}\theta_{M}-\frac{1}{2}\log
2\pi\Big{)}M-\Big{(}\frac{1}{2}\log
2\pi\Big{)}\theta_{M}-\zeta^{\prime}(-1)+\log G(c+1).$

Here, $\theta_{M}$ is given by (3.4), $\zeta$ is the Riemann zeta function and
$G$ is the Barnes $G$-function.

###### Proof.

By the definitions (3.1), (2.17), and (2.15), as well as (2.9), one can write

(3.6) $S_{3}=\sum_{\ell=c}^{\lceil M\rceil-1}\log\mathbb{P}(Y\leq\ell),\qquad
Y\sim B(N,1-x).$

Observe that

(3.7) $\displaystyle\begin{split}&\quad\log\mathbb{P}(Y\leq\ell)=(N-\ell)\log
x+\log\bigg{[}\sum_{j=0}^{\ell}\binom{N}{j}(1-x)^{j}x^{\ell-j}\bigg{]}\\\
&=(N-\ell)\log
x+\log\bigg{[}\binom{N}{\ell}(1-x)^{\ell}\bigg{]}+\log\bigg{(}1+\sum_{j=0}^{\ell-1}\Big{(}\frac{x}{1-x}\Big{)}^{\ell-j}\frac{(j+1)\cdots\ell}{(N-\ell+1)\cdots(N-j)}\bigg{)}.\end{split}$

Notice here that the last term in the second line is of order
$\mathcal{O}(\frac{\ell}{N})$. Note also that

(3.8) $\log\binom{N}{\ell}=\ell\log
N+\sum_{j=0}^{\ell-1}\log\Big{(}1-\frac{j}{N}\Big{)}-\log\ell!=\ell\log
N-\frac{1}{N}\frac{\ell(\ell-1)}{2}-\log\ell!+\mathcal{O}\Big{(}\frac{\ell^{3}}{N^{2}}\Big{)}.$

Then, by combining (3.7) and (3.8), we obtain

(3.9) $\log\mathbb{P}(Y\leq\ell)=N\log x+\ell\log
N+\ell\log\Big{(}\frac{1-x}{x}\Big{)}-\log\ell!-\frac{1}{N}\frac{\ell^{2}}{2}+\mathcal{O}\Big{(}\frac{\ell}{N}+\frac{\ell^{3}}{N^{2}}\Big{)}.$

By taking the sum of (3.9) over $c\leq\ell\leq\lceil M\rceil-1$, it follows
that

(3.10) $\displaystyle\begin{split}S_{3}&=N(\lceil M\rceil-c)\log
x+\Big{(}\frac{\lceil M\rceil^{2}-\lceil
M\rceil}{2}-\frac{c^{2}-c}{2}\Big{)}\log\Big{(}\frac{1-x}{x}N\Big{)}\\\
&\quad-\log G(\lceil M\rceil+1)+\log G(c+1)-\frac{1}{N}\frac{\lceil
M\rceil^{3}}{6}+\mathcal{O}\Big{(}\frac{M^{2}}{N}+\frac{M^{4}}{N^{2}}\Big{)}.\end{split}$

Then the desired asymptotic formula (3.5) follows from (2.4) and $\lceil
M\rceil=M+\theta_{M}$. ∎

Next, we analyse $S_{2}$. For this, we need the following version of the
Euler-Maclaurin formula taken from [34, Lemma 3.4].

######  Lemma 3.3 (cf. Lemma 3.4 in [34]).

Let $A$, $a_{0}$, $B$ and $b_{0}$ be bounded functions of $n\in\mathbb{Z}^{+}$
such that

$a_{n}:=An+a_{0},\qquad b_{n}:=Bn+b_{0}$

are integer-valued. Furthermore, assume that $B-A$ is positive and bounded
away from 0. Let $f$ be a function independent of $n$, and which is
$C^{3}([\min\\{\frac{a_{n}}{n},A\\},\max\\{\frac{b_{n}}{n},B\\}])$ for all
$n\in\mathbb{Z}^{+}$. Then as $n\to\infty$, we have

(3.11)
$\displaystyle\begin{split}\sum_{j=a_{n}}^{b_{n}}f\Big{(}\frac{j}{n}\Big{)}&=n\int_{A}^{B}f(x)\,dx+\frac{(1-2a_{0})f(A)+(1+2b_{0})f(B)}{2}\\\
&\quad+\frac{(-1+6a_{0}-6a_{0}^{2})f^{\prime}(A)+(1+6b_{0}+6b_{0}^{2})f^{\prime}(B)}{12n}\\\
&\quad+\mathcal{O}\bigg{(}\frac{\mathfrak{m}_{A,n}(f^{\prime\prime})+\mathfrak{m}_{B,n}(f^{\prime\prime})}{n^{2}}+\sum_{j=a_{n}}^{b_{n}-1}\frac{\mathfrak{m}_{j,n}(f^{\prime\prime\prime})}{n^{3}}\bigg{)}.\end{split}$

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

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

We now derive the asymptotic behaviour of $S_{2}$. Compared to $S_{0}$ and
$S_{3}$, the asymptotics of $S_{2}$ contribute to the leading order of the
total summation (2.19).

######  Lemma 3.4 (Asymptotic behaviour of $S_{2}$).

As $N\to\infty$, we have

(3.12) $S_{2}=C_{1}N^{2}+C_{2}N\log N+C_{3}N+C_{5}\log
N+C_{6}+\mathcal{O}\Big{(}\frac{1}{M}+\frac{M}{N}\Big{)},$

where

$\displaystyle C_{1}$ $\displaystyle=-\int_{M/N}^{\delta}f_{1}(t)\,dt,\qquad
C_{2}=-\frac{\delta}{2},\qquad
C_{3}=\Big{(}\theta_{M}-\frac{1}{2}\Big{)}f_{1}\Big{(}\frac{M}{N}\Big{)}+\Big{(}\theta_{\delta}-\frac{1}{2}\Big{)}f_{1}(\delta)-\int_{M/N}^{\delta}f_{2}(t)\,dt,$
$\displaystyle C_{5}$
$\displaystyle=\frac{1}{2}(M+\theta_{\delta}+\theta_{M}-1)-\frac{1}{12},$
$\displaystyle C_{6}$
$\displaystyle=\frac{1-6\theta_{M}+6\theta_{M}^{2}}{12}f_{1}^{\prime}\Big{(}\frac{M}{N}\Big{)}-\frac{1-6\theta_{\delta}+6\theta_{\delta}^{2}}{12}f_{1}^{\prime}(\delta)+\Big{(}\theta_{M}-\frac{1}{2}\Big{)}f_{2}\Big{(}\frac{M}{N}\Big{)}+\Big{(}\theta_{\delta}-\frac{1}{2}\Big{)}f_{2}(\delta)$
$\displaystyle\quad+\frac{1}{12}\log
M+\Big{[}\frac{1}{12}\Big{(}\delta-\log\Big{(}\frac{\delta}{1-\delta}\Big{)}\Big{)}-\frac{\delta
x}{1-x-\delta}-x\log\Big{(}\frac{1-x-\delta}{1-x}\Big{)}\Big{]},$

where $\theta_{M}$ and $\theta_{\delta}$ are given by (3.4), and the functions
$f_{1}$, $f_{2}$ are defined as

(3.13)
$f_{1}(t):=t\log\Big{(}\frac{t}{1-x}\Big{)}+(1-t)\log\Big{(}\frac{1-t}{x}\Big{)},\qquad
f_{2}(t):=\log\Big{(}\frac{1-x-t}{(1-x)(1-t)}\Big{)}+\frac{1}{2}\log\Big{(}2\pi
t(1-t)\Big{)},$

for $0<t<1-x$.

###### Proof.

Let $X\sim B(N,x)$. Then by (2.16) and (2.17), we have

(3.14) $S_{2}=\sum_{k=k_{+}}^{k_{M}}\log\mathbb{P}(X\geq
k+\alpha+1)=\sum_{j=\lceil M\rceil}^{\lfloor\delta
N\rfloor}\log\mathbb{P}(X\geq N-j).$

Let $t=j/N$. By applying Lemma A.2, we have

$\log\mathbb{P}(X\geq tN)=-Nf_{1}(t)-\frac{1}{2}\log
N-f_{2}(t)+\frac{1}{N}f_{3}(t)+\mathcal{O}\Big{(}\frac{1}{t^{3}N^{3}}+\frac{1}{N^{2}}\Big{)},$

where

(3.15) $f_{3}(t)=\frac{1}{12}-\frac{tx}{(1-x-t)^{2}}-\frac{1}{12t(1-t)}.$

Note that

$\sum_{j=\lceil M\rceil}^{\lfloor\delta
N\rfloor}\mathcal{O}\Big{(}\frac{1}{j^{3}}+\frac{1}{N^{2}}\Big{)}=\mathcal{O}\Big{(}\frac{1}{M^{2}}+\frac{1}{N}\Big{)}.$

By using Lemma 3.3 with $A(N)=M/N$, $a_{0}(N)=\theta_{M}$, $B(N)=\delta$ and
$b_{0}(N)=-\theta_{\delta}$, we obtain

$\displaystyle-N\sum_{j=\lceil M\rceil}^{\lfloor\delta
N\rfloor}f_{1}\Big{(}\frac{j}{N}\Big{)}$
$\displaystyle=-N^{2}\int_{M/N}^{\delta}f_{1}(t)\,dt+\Big{[}\Big{(}\theta_{M}-\frac{1}{2}\Big{)}f_{1}\Big{(}\frac{M}{N}\Big{)}+\Big{(}\theta_{\delta}-\frac{1}{2}\Big{)}f_{1}(\delta)\Big{]}N$
$\displaystyle\quad+\frac{1-6\theta_{M}+6\theta_{M}^{2}}{12}f_{1}^{\prime}\Big{(}\frac{M}{N}\Big{)}-\frac{1-6\theta_{\delta}+6\theta_{\delta}^{2}}{12}f_{1}^{\prime}(\delta)+\mathcal{O}\Big{(}\frac{1}{M}\Big{)},$
$\displaystyle-\sum_{j=\lceil M\rceil}^{\lfloor\delta
N\rfloor}f_{2}\Big{(}\frac{j}{N}\Big{)}$
$\displaystyle=-N\int_{M/N}^{\delta}f_{2}(t)\,dt+\Big{(}\theta_{M}-\frac{1}{2}\Big{)}f_{2}\Big{(}\frac{M}{N}\Big{)}+\Big{(}\theta_{\delta}-\frac{1}{2}\Big{)}f_{2}(\delta)+\mathcal{O}\Big{(}\frac{1}{M}+\frac{1}{N}\Big{)},$
$\displaystyle\frac{1}{N}\sum_{j=\lceil M\rceil}^{\lfloor\delta
N\rfloor}f_{3}\Big{(}\frac{j}{N}\Big{)}$
$\displaystyle=\int_{M/N}^{\delta}f_{3}(t)\,dt+\mathcal{O}\Big{(}\frac{1}{M}\Big{)}$
$\displaystyle=\frac{1}{12}\Big{(}\delta-\log\Big{(}\frac{\delta}{1-\delta}\Big{)}\Big{)}-\frac{\delta
x}{1-x-\delta}-x\log\Big{(}\frac{1-x-\delta}{1-x}\Big{)}+\frac{1}{12}\log\Big{(}\frac{M}{N}\Big{)}+\mathcal{O}\Big{(}\frac{1}{M}+\frac{M}{N}\Big{)}.$

Combining all of the above, the proof is complete. ∎

We conclude this subsection by presenting the asymptotics of
$S_{0}+S_{2}+S_{3}$, derived from the previous lemmas.

######  Lemma 3.5 (Asymptotic behaviour of $S_{0}+S_{2}+S_{3}$).

As $N\to\infty$, we have

(3.16) $S_{0}+S_{2}+S_{3}=E_{1}N^{2}+E_{2}N\log N+E_{3}N+E_{5}\log
N+E_{6}+\mathcal{O}\Big{(}\frac{1}{M}+\frac{M^{2}}{N}+\frac{M^{4}}{N^{2}}\Big{)},$

where

$\displaystyle E_{1}$ $\displaystyle=\frac{1}{2}\log
x-\frac{1}{2}\delta^{2}\log\Big{(}\frac{\delta}{1-x}\Big{)}+\frac{1}{2}(1-\delta)^{2}\log\Big{(}\frac{1-\delta}{x}\Big{)}+\frac{\delta}{2},\qquad
E_{2}=-\frac{\delta}{2},$ $\displaystyle E_{3}$
$\displaystyle=\Big{(}\frac{1}{2}-c\Big{)}\log
x+(1-x-\delta)\log\Big{(}\frac{1-x-\delta}{1-x}\Big{)}+\delta-\frac{1}{2}\delta\log(2\pi\delta)-\frac{1}{2}(1-\delta)\log(1-\delta)-\Big{(}\frac{1}{2}-\theta_{\delta}\Big{)}f_{1}(\delta),$
$\displaystyle E_{5}$
$\displaystyle=\frac{1}{2}\Big{(}\theta_{\delta}+c-c^{2}\Big{)}-\frac{5}{12},$
$\displaystyle E_{6}$
$\displaystyle=\Big{(}\frac{1}{12}-\frac{c-c^{2}}{2}\Big{)}\log\Big{(}\frac{x}{1-x}\Big{)}-\frac{1}{4}\log(2\pi)-\zeta^{\prime}(-1)+\log
G(c+1)$
$\displaystyle\quad-\frac{1-6\theta_{\delta}+6\theta_{\delta}^{2}}{12}f_{1}^{\prime}(\delta)-\Big{(}\frac{1}{2}-\theta_{\delta}\Big{)}f_{2}(\delta)+\frac{1}{12}\delta-\frac{\delta
x}{1-x-\delta}-x\log\Big{(}\frac{1-x-\delta}{1-x}\Big{)}-\frac{1}{12}\log\Big{(}\frac{\delta}{1-\delta}\Big{)}.$

Here, $f_{1}$ and $f_{2}$ are defined as (3.13).

###### Proof.

First, note that since $S_{0}$ is exponentially small due to Lemma 3.1, it
suffices to compute $S_{2}+S_{3}.$ Note that by definition (3.13), we have

(3.17) $\int
f_{1}(t)\,dt=\frac{1}{2}t^{2}\log\Big{(}\frac{t}{1-x}\Big{)}-\frac{1}{2}(1-t)^{2}\log\Big{(}\frac{1-t}{x}\Big{)}-\frac{t}{2},\qquad
f_{1}^{\prime}(t)=\log\Big{(}\frac{tx}{(1-t)(1-x)}\Big{)},$

and

(3.18) $\displaystyle\begin{split}\int
f_{2}(t)\,dt=-(1-x-t)\log(1-x-t)+t\Big{(}\log\Big{(}\frac{\sqrt{2\pi}}{1-x}\Big{)}-1\Big{)}+\frac{1}{2}\Big{(}t\log
t+(1-t)\log(1-t)\Big{)},\end{split}$

up to integration constants. Combining these formulas with Lemma 3.4, we have

$\displaystyle C_{1}N^{2}$
$\displaystyle=\bigg{[}-\frac{1}{2}\delta^{2}\log\Big{(}\frac{\delta}{1-x}\Big{)}+\frac{1}{2}(1-\delta)^{2}\log\Big{(}\frac{1-\delta}{x}\Big{)}+\frac{\delta}{2}+\frac{1}{2}\log
x\bigg{]}N^{2}-(M\log x)N-\frac{1}{2}M^{2}\log N$
$\displaystyle\quad+\bigg{[}\frac{1}{2}M^{2}\log
M+\frac{1}{2}M^{2}\log\Big{(}\frac{x}{1-x}\Big{)}-\frac{3}{4}M^{2}\bigg{]}+\frac{M^{3}}{6N}+\mathcal{O}\Big{(}\frac{M^{4}}{N^{2}}\Big{)},$
$\displaystyle C_{3}N$
$\displaystyle=\bigg{[}(1-x-\delta)\log(1-x-\delta)-\delta\log\Big{(}\frac{\sqrt{2\pi}}{1-x}\Big{)}+\delta-\frac{1}{2}\delta\log\delta-\frac{1}{2}(1-\delta)\log(1-\delta)-\Big{(}\frac{1}{2}-\theta_{\delta}\Big{)}f_{1}(\delta)\bigg{]}N$
$\displaystyle\quad+\bigg{[}\Big{(}\frac{1}{2}-\theta_{M}\Big{)}\log
x-(1-x)\log(1-x)\bigg{]}N-M\theta_{M}\log N+M\theta_{M}\log M$
$\displaystyle\quad+\bigg{[}\frac{1}{2}\log(2\pi)-\theta_{M}+\Big{(}\frac{1}{2}-\theta_{M}\Big{)}\log\Big{(}\frac{1-x}{x}\Big{)}\bigg{]}M+\mathcal{O}\Big{(}\frac{M^{2}}{N}\Big{)},$
$\displaystyle C_{6}$
$\displaystyle=\Big{(}\frac{1}{6}-\frac{1}{2}\theta_{M}^{2}\Big{)}\log
N+\Big{(}\frac{1}{2}\theta_{M}^{2}-\frac{1}{12}\Big{)}\log
M+\Big{(}\frac{1}{2}\theta_{M}-\frac{1}{4}\Big{)}\log(2\pi)+\frac{1-6\theta_{M}+6\theta_{M}^{2}}{12}\log\Big{(}\frac{x}{1-x}\Big{)}+\frac{1}{12}\delta$
$\displaystyle\quad-\frac{1}{12}\log\Big{(}\frac{\delta}{1-\delta}\Big{)}-\frac{1-6\theta_{\delta}+6\theta_{\delta}^{2}}{12}f_{1}^{\prime}(\delta)-\Big{(}\frac{1}{2}-\theta_{\delta}\Big{)}f_{2}(\delta)-\frac{\delta
x}{1-x-\delta}-x\log\Big{(}\frac{1-x-\delta}{1-x}\Big{)}+\mathcal{O}\Big{(}\frac{M}{N}\Big{)}.$

Then after lengthy but straightforward computations, one can observe that all
terms depending on $M$ in $D_{j}$ and $C_{j}$ from Lemmas 3.2 and 3.4 cancel
each other out, up to an error term
$\mathcal{O}(M^{-1}+M^{2}N^{-1}+M^{4}N^{-2})$. The only terms remaining are
those stated in the desired asymptotic formula (3.16). ∎

###  3.2. Analysis of $S_{1}$

By Lemma 3.5 in the previous subsection, it remains to derive the asymptotic
behaviour of $S_{1}$. The aim of this subsection is to perform this remaining
task and demonstrate Lemma 3.11 below. This requires to further divide the
summation $S_{1}$. First, recall that

(3.19) $S_{1}=\sum_{k=k_{-}+1}^{k_{+}-1}\log
I_{x}(k+\alpha+1,n+c-k)=\sum_{m=\lfloor\delta
N\rfloor+1}^{\lceil(1-\delta)N\rceil-1}\log I_{x}(m,N+1-m).$

We denote

(3.20) $N_{1}=N+1,\qquad M_{1}=N_{1}^{\frac{1}{12}},\qquad
x_{k}=\frac{m_{k}}{N_{1}},$

where $N$ and $m_{k}$ are given by (2.15) and (3.1), respectively. Define

(3.21) $g_{-}=\lceil N_{1}x-M_{1}\sqrt{N_{1}}-\alpha-1\rceil,\qquad
g_{+}=\lfloor N_{1}x+M_{1}\sqrt{N_{1}}-\alpha-1\rfloor.$

We split $S_{1}$ into the following three parts:

(3.22) $S_{1}=S_{1}^{(1)}+S_{1}^{(2)}+S_{1}^{(3)},$

where

(3.23) $\displaystyle\begin{split}S_{1}^{(1)}&=\sum_{k=k_{-}+1}^{g_{-}-1}\log
I_{x}(m_{k},N_{1}-m_{k}),\\\ S_{1}^{(2)}&=\sum_{k=g_{-}}^{g_{+}}\log
I_{x}(m_{k},N_{1}-m_{k}),\\\ S_{1}^{(3)}&=\sum_{k=g_{+}+1}^{k_{+}-1}\log
I_{x}(m_{k},N_{1}-m_{k}).\end{split}$

Note that the sizes of $x_{k}$ in $S_{1}^{(1)}$, $S_{1}^{(2)}$ and
$S_{1}^{(3)}$ are roughly $[\delta,x-\frac{M_{1}}{\sqrt{N_{1}}})$,
$[x-\frac{M_{1}}{\sqrt{N_{1}}},x+\frac{M_{1}}{\sqrt{N_{1}}}]$ and
$(x+\frac{M_{1}}{\sqrt{N_{1}}},1-\delta]$, respectively. On the other hand,
since $k_{\pm}$ is defined using $\lfloor\delta N\rfloor$, not $\lfloor\delta
N_{1}\rfloor$, one cannot guarantee $\delta\leq x_{k}\leq 1-\delta$ on the
boundary of $S_{1}$. Nevertheless, this distinction is not important, as the
approximate size of $x_{k}$ determines the asymptotic behaviour.

We denote the $\eta$ in (2.25) corresponding to $x_{0}=x_{k}$ by $\eta_{k}$,
i.e.

(3.24)
$-\frac{1}{2}\eta_{k}^{2}=x_{k}\log\Big{(}\frac{x}{x_{k}}\Big{)}+(1-x_{k})\log\Big{(}\frac{1-x}{1-x_{k}}\Big{)},\qquad\frac{\eta_{k}}{x-x_{k}}>0.$

Note that $\eta_{k}>0$ for $x_{k}<x$, whereas $\eta_{k}<0$ for $x_{k}>x$. Then
by Lemma 2.2, we have

(3.25)
$I_{x}(m_{k},N_{1}-m_{k})=\frac{1}{2}\operatorname{erfc}\Big{(}-\eta_{k}\sqrt{\frac{N_{1}}{2}}\Big{)}+\frac{1}{\sqrt{2\pi
N_{1}}}e^{-\frac{1}{2}\eta_{k}^{2}N_{1}}\Big{(}c_{0}(\eta_{k})+\frac{1}{N}c_{1}(\eta_{k})+\mathcal{O}\Big{(}\frac{1}{N_{1}^{2}}\Big{)}\Big{)}.$

We first compute the asymptotic behaviour of $S_{1}^{(1)}$.

######  Lemma 3.6.

There exists a constant $C>0$ such that

(3.26) $S_{1}^{(1)}=\mathcal{O}(e^{-CM_{1}^{2}}),\qquad\text{ as }N\to\infty.$

###### Proof.

By the Taylor expansion of $\eta_{k}^{2}$ at $x_{k}=x$, we have

$-\frac{1}{2}\eta_{k}^{2}=-\frac{(x_{k}-x)^{2}}{2x(1-x)}+\mathcal{O}((x_{k}-x)^{3}),\qquad\text{
as }x_{k}\to x.$

This in turn implies that

$-\frac{1}{2}\eta_{k}^{2}N_{1}=-\frac{1}{2x(1-x)}M_{1}^{2}+o(1),\qquad\text{
as }x_{k}\to x-\frac{M_{1}}{\sqrt{N_{1}}}.$

Therefore, it follows that $-\frac{1}{2}\eta_{k}^{2}N_{1}\to-\infty$ for all
$k\in[k_{-}+1,g_{-}-1]$.

Recall that by [92, Eq.(7.12.1)], as $x\to\infty$,

(3.27)
$\displaystyle\operatorname{erfc}x\sim\frac{e^{-x^{2}}}{\sqrt{\pi}x}\sum_{m=0}^{\infty}(-1)^{m}\frac{(2m-1)!!}{(2x^{2})^{m}},\qquad\operatorname{erfc}(-x)\sim
2-\frac{e^{-x^{2}}}{\sqrt{\pi}x}\sum_{m=0}^{\infty}(-1)^{m}\frac{(2m-1)!!}{(2x^{2})^{m}}.$

Using this, (3.25) can be rewritten as

$I_{x}(m_{k},N_{1}-m_{k})=1-\frac{1}{\sqrt{2\pi
N_{1}}\eta_{k}}e^{-\frac{1}{2}\eta_{k}^{2}N_{1}}\mathcal{O}(1)+\frac{1}{\sqrt{2\pi
N_{1}}}e^{-\frac{1}{2}\eta_{k}^{2}N_{1}}\mathcal{O}(1)=1+\mathcal{O}(e^{-cM^{2}}),$

which completes the proof. ∎

We denote

(3.28) $M_{k}:=\sqrt{N_{1}}(x_{k}-x),$

where $x_{k}$ is given by (3.20). We acknowledge that this is a slight abuse
of notation, as we already use $M_{1}$ in (3.20). However, since $M_{k}$ with
$k=1$ will not be used further in the sequel, we allow this abuse of notation.

To analyse $S_{1}^{(2)}$, let us introduce the following lemma.

######  Lemma 3.7.

Let $h\in C^{2}(\mathbb{R})$. Then as $N\to\infty$,

(3.29)
$\sum_{k=g_{-}}^{g_{+}}h(M_{k})=\sqrt{N_{1}}\int_{-M_{1}}^{M_{1}}h(t)\,dt+\Big{(}\frac{1}{2}-\theta_{-}\Big{)}h(-M_{1})+\Big{(}\frac{1}{2}-\theta_{+}\Big{)}h(M_{1})+\mathcal{O}\Big{(}\frac{\mathfrak{m}(h^{\prime})}{\sqrt{N_{1}}}+\frac{M_{1}\mathfrak{m}(h^{\prime\prime})}{\sqrt{N_{1}}}\Big{)},$

where

(3.30)
$\theta_{-}=g_{-}-(N_{1}x-M_{1}\sqrt{N_{1}}-\alpha-1),\qquad\theta_{+}=(N_{1}x+M_{1}\sqrt{N_{1}}-\alpha-1)-g_{+},$

and

(3.31) $\mathfrak{m}(f):=\max\\{f(t):M_{g_{-}}\leq t\leq M_{g_{+}}\\}.$

###### Proof.

Note that by (3.28), we have $M_{k+1}-M_{k}=\frac{1}{\sqrt{N_{1}}}$.
Therefore, it follows that

$\displaystyle\int_{M_{g_{-}}}^{M_{g_{+}}}h(t)\,dt$
$\displaystyle=\sum_{k=g_{-}}^{g_{+}-1}\int_{M_{k}}^{M_{k+1}}h(t)\,dt$
$\displaystyle=\sum_{k=g_{-}}^{g_{+}-1}\Big{[}h(M_{k})(M_{k+1}-M_{k})+h^{\prime}(M_{k})\frac{(M_{k+1}-M_{k})^{2}}{2}\Big{]}+\mathcal{O}\Big{(}\sum_{k=M_{g_{-}}}^{M_{g_{+}}}(M_{k+1}-M_{k})^{3}\mathfrak{m}(h^{\prime\prime})\Big{)}$
$\displaystyle=\frac{1}{\sqrt{N_{1}}}\sum_{k=g_{-}}^{g_{+}-1}h(M_{k})+\frac{1}{2N_{1}}\sum_{k=g_{-}}^{g_{+}-1}h^{\prime}(M_{k})+\mathcal{O}\Big{(}\frac{M_{1}}{N_{1}}\mathfrak{m}(h^{\prime\prime})\Big{)},$

which leads to

$\sum_{k=g_{-}}^{g_{+}-1}h(M_{k})=\sqrt{N_{1}}\int_{M_{g_{-}}}^{M_{g_{+}}}h(t)\,dt-\frac{1}{2\sqrt{N_{1}}}\sum_{k=g_{-}}^{g_{+}-1}h^{\prime}(M_{k})+\mathcal{O}\Big{(}\frac{M_{1}}{\sqrt{N_{1}}}\mathfrak{m}(h^{\prime\prime})\Big{)}.$

By replacing $h(t)$ above by $h^{\prime}(t)$, we obtain

$\sum_{k=g_{-}}^{g_{+}-1}h^{\prime}(M_{k})=\sqrt{N_{1}}\Big{(}h(M_{g_{+}})-h(M_{g_{-}})\Big{)}+\mathcal{O}\Big{(}M_{1}\mathfrak{m}(h^{\prime\prime})\Big{)}.$

Combining these two equations, it follows that

$\sum_{k=g_{-}}^{g_{+}}h(M_{k})=\sqrt{N_{1}}\int_{M_{g_{-}}}^{M_{g_{+}}}h(t)\,dt+\frac{1}{2}\Big{(}h(M_{g_{+}})+h(M_{g_{-}})\Big{)}+\mathcal{O}\Big{(}\frac{M_{1}}{\sqrt{N_{1}}}\mathfrak{m}(h^{\prime\prime})\Big{)}.$

Note here that by (3.21) and (3.28), we have

$M_{g_{+}}=M_{1}-\frac{\theta_{+}}{\sqrt{N_{1}}},\qquad
M_{g_{-}}=-M_{1}+\frac{\theta_{-}}{\sqrt{N_{1}}}.$

By using this, we obtain

$\int_{M_{g_{-}}}^{M_{g_{+}}}h(t)\,dt=\int_{-M_{1}}^{M_{1}}h(t)\,dt-h(-{M_{1}})\frac{\theta_{-}}{\sqrt{N_{1}}}-h(M_{1})\frac{\theta_{+}}{\sqrt{N}_{1}}+\mathcal{O}\Big{(}\frac{\mathfrak{m}(h^{\prime})}{N_{1}}\Big{)}$

and

$\frac{1}{2}(h(M_{g_{+}})+h(M_{g_{-}}))=\frac{1}{2}(h(M_{1})+h(-M_{1}))+\mathcal{O}\Big{(}\frac{\mathfrak{m}(h^{\prime})}{\sqrt{N_{1}}}\Big{)}.$

Combining all of the above, the lemma follows. ∎

We now compute the asymptotic behaviour of $S_{1}^{(2)}$. For this, it is
convenient to introduce the notation

(3.32) $\sigma:=\sqrt{x(1-x)}.$

Let us also define

(3.33)
$g(t)=\frac{1-2x}{3\sigma^{3}\sqrt{2\pi}\operatorname{erfc}(\frac{t}{\sqrt{2}\sigma})}(t^{2}+2\sigma^{2})e^{-\frac{t^{2}}{2\sigma^{2}}}.$

######  Lemma 3.8.

As $N\to\infty$, we have

(3.34)
$S_{1}^{(2)}=B_{4}^{(2)}\sqrt{N_{1}}+B_{6}^{(2)}+\mathcal{O}\Big{(}\frac{M_{1}^{5}}{\sqrt{N_{1}}}+\frac{M_{1}^{10}}{N_{1}}\Big{)},$

where

$B_{4}^{(2)}=\int_{-M_{1}}^{M_{1}}\log\Big{[}\frac{1}{2}\operatorname{erfc}\Big{(}\frac{t}{\sqrt{2}\sigma}\Big{)}\Big{]}\,dt,\qquad
B_{6}^{(2)}=\int_{-M_{1}}^{M_{1}}g(t)\,dt+\Big{(}\frac{1}{2}-\theta_{+}\Big{)}\log\Big{[}\frac{1}{2}\operatorname{erfc}\Big{(}\frac{M_{1}}{\sqrt{2}\sigma}\Big{)}\Big{]}.$

Here $g$ is given by (3.33).

###### Proof.

Note that $|x_{k}-x|<{M_{1}}/\sqrt{N_{1}}\to 0$ as $N_{1}\to\infty$. By
(3.24), as $N_{1}\to\infty$, we have

(3.35)
$\displaystyle\begin{split}-\frac{1}{2}\eta_{k}^{2}&=-\frac{1}{2\sigma^{2}}\frac{M_{k}^{2}}{N_{1}}\Big{(}1-\frac{1-2x}{3\sigma^{2}}\frac{M_{k}}{\sqrt{N_{1}}}+\frac{1-3\sigma^{2}}{6\sigma^{4}}\frac{M_{k}^{2}}{N_{1}}+\mathcal{O}\Big{(}\frac{M_{k}^{3}}{N_{1}\sqrt{N_{1}}}\Big{)}\Big{)},\\\
\eta_{k}&=-\frac{M_{k}}{\sigma\sqrt{N_{1}}}\Big{(}1-\frac{1-2x}{6\sigma^{2}}\frac{M_{k}}{\sqrt{N_{1}}}+\frac{5-14\sigma^{2}}{72\sigma^{4}}\frac{M_{k}^{2}}{N_{1}}+\mathcal{O}\Big{(}\frac{M_{k}^{3}}{N_{1}\sqrt{N_{1}}}\Big{)}\Big{)},\\\
-\eta_{k}\sqrt{\frac{N_{1}}{2}}&=\frac{M_{k}}{\sqrt{2}\sigma}\Big{(}1-\frac{1-2x}{6\sigma^{2}}\frac{M_{k}}{\sqrt{N_{1}}}+\frac{5-14\sigma^{2}}{72\sigma^{4}}\frac{M_{k}^{2}}{N_{1}}+\mathcal{O}\Big{(}\frac{M_{k}^{3}}{N_{1}\sqrt{N_{1}}}\Big{)}\Big{)}.\end{split}$

Define

$\Delta_{k}:=-\eta_{k}\sqrt{\frac{N_{1}}{2}}-\frac{M_{k}}{\sqrt{2}\sigma}=-\frac{1-2x}{6\sqrt{2}\sigma^{3}}\frac{M_{k}^{2}}{\sqrt{N_{1}}}+\frac{5-14\sigma^{2}}{72\sqrt{2}\sigma^{5}}\frac{M_{k}^{3}}{N_{1}}+\mathcal{O}\Big{(}\frac{M_{k}^{4}}{N_{1}\sqrt{N_{1}}}\Big{)}.$

Note that by using the Taylor expansion

$\operatorname{erfc}(a+x)=\operatorname{erfc}a-\frac{2}{\sqrt{\pi}}e^{-a^{2}}\Big{(}x-ax^{2}+\frac{2a^{2}-1}{3}x^{3}+\mathcal{O}(a^{3}x^{4})\Big{)},$

as $x\to 0$, we obtain

$\displaystyle\quad\operatorname{erfc}\Big{(}-\eta_{k}\sqrt{\frac{N}{2}}\Big{)}=\operatorname{erfc}\Big{(}\frac{M_{k}}{\sqrt{2}\sigma}\Big{)}-\frac{2}{\sqrt{\pi}}e^{-\frac{M_{k}^{2}}{2\sigma^{2}}}\Big{[}\Delta_{k}-\frac{M_{k}}{\sqrt{2}\sigma}\Delta_{k}^{2}+\mathcal{O}((1+M_{k}^{2})\Delta_{k}^{3})\Big{]}$
$\displaystyle=\operatorname{erfc}\Big{(}\frac{M_{k}}{\sqrt{2}\sigma}\Big{)}-\sqrt{\frac{2}{\pi}}e^{-\frac{M_{k}^{2}}{2\sigma^{2}}}\bigg{[}-\frac{1-2x}{6\sigma^{3}}\frac{M_{k}^{2}}{\sqrt{N_{1}}}+\frac{5-14\sigma^{2}}{72\sigma^{5}}\frac{M_{k}^{3}}{N_{1}}-\frac{1-4\sigma^{2}}{72\sigma^{7}}\frac{M_{k}^{5}}{N_{1}}+\mathcal{O}\Big{(}\frac{M_{k}^{4}+M_{k}^{8}}{N_{1}\sqrt{N_{1}}}\Big{)}\bigg{]}.$

On the other hand, by (2.31), we have

$c_{0}(\eta_{k})=\frac{1}{\eta_{k}}-\frac{\sqrt{x_{k}(1-x_{k})}}{x-x_{k}}=\frac{1-2x}{3\sigma}-\frac{1+\sigma^{2}}{12\sigma^{3}}\frac{M_{k}}{\sqrt{N_{1}}}+\mathcal{O}\Big{(}\frac{M_{k}^{2}}{N_{1}}\Big{)}.$

Note also that

$e^{-\frac{1}{2}\eta_{k}^{2}N_{1}}=e^{-\frac{M_{k}^{2}}{2\sigma^{2}}}\Big{[}1+\frac{1-2x}{6\sigma^{4}}\frac{M_{k}^{3}}{\sqrt{N_{1}}}+\mathcal{O}\Big{(}\frac{M_{k}^{4}+M_{k}^{6}}{N_{1}}\Big{)}\Big{]}.$

Then by Lemma 2.2, it follows that

$\displaystyle I_{x}(m_{k},N_{1}-m_{k})$
$\displaystyle=\frac{1}{2}\operatorname{erfc}\Big{(}-\eta_{k}\sqrt{\frac{N_{1}}{2}}\Big{)}+\frac{1}{\sqrt{2\pi{N_{1}}}}e^{-\frac{1}{2}\eta_{k}^{2}{N_{1}}}\Big{(}c_{0}(\eta_{k})+\mathcal{O}\Big{(}\frac{1}{N_{1}}\Big{)}\Big{)}$
$\displaystyle=u_{1}(M_{k})+\frac{1}{\sqrt{N_{1}}}u_{2}(M_{k})+\frac{1}{N_{1}}u_{3}(M_{k})+\mathcal{O}\Big{(}\frac{1+M_{k}^{8}}{N_{1}\sqrt{N_{1}}}e^{-\frac{M_{k}^{2}}{2\sigma^{2}}}\Big{)},$

where

$\displaystyle u_{1}(t)$
$\displaystyle=\frac{1}{2}\operatorname{erfc}\Big{(}\frac{t}{\sqrt{2}\sigma}\Big{)},\qquad
u_{2}(t)=\frac{1-2x}{\sqrt{2\pi}}e^{-\frac{t^{2}}{2\sigma^{2}}}\Big{[}\frac{1}{6\sigma^{3}}t^{2}+\frac{1}{3\sigma}\Big{]},$
$\displaystyle u_{3}(t)$
$\displaystyle=\frac{1}{\sqrt{2\pi}}e^{-\frac{t^{2}}{2\sigma^{2}}}\Big{[}\frac{1-4\sigma^{2}}{72\sigma^{7}}t^{5}-\frac{1+2\sigma^{2}}{72\sigma^{5}}t^{3}-\frac{1+\sigma^{2}}{12\sigma^{3}}t\Big{]}.$

Therefore we obtain

$\displaystyle\log I_{x}(m_{k},N_{1}-m_{k})$ $\displaystyle=\log
u_{1}(M_{k})+\frac{1}{\sqrt{N_{1}}}g(M_{k})+\frac{1}{N_{1}}g_{1}(M_{k})$
$\displaystyle\quad+\mathcal{O}\bigg{(}\frac{1}{N_{1}\sqrt{N}_{1}}\Big{(}\frac{u_{2}(M_{k})^{3}}{u_{1}(M_{k})^{3}}+\frac{u_{2}(M_{k})u_{3}(M_{k})}{u_{1}(M_{k})^{2}}+\frac{1+M_{k}^{8}}{u_{1}(M_{k})}e^{-\frac{M_{k}^{2}}{2\sigma^{2}}}\Big{)}\bigg{)},$

where

$g(t)=\frac{u_{2}(t)}{u_{1}(t)},\qquad
g_{1}(t)=-\frac{u_{2}(t)^{2}}{2u_{1}(t)^{2}}+\frac{u_{3}(t)}{u_{1}(t)}.$

Recall that $S_{1}^{(2)}$ is given by (3.23). Notice here that as
$t\to\infty,$ we have

$u_{1}(t)=\mathcal{O}(t^{-1}e^{-\frac{1}{2}xt^{2}}),\qquad
u_{2}(t)=\mathcal{O}(t^{2}e^{-\frac{1}{2}xt^{2}}),\qquad
u_{3}(t)=\mathcal{O}(t^{5}e^{-\frac{1}{2}xt^{2}}).$

Then by applying Lemma 3.7 with $h(t)=t^{9}$, we have

$\sum_{k=g_{-}}^{g_{+}}\mathcal{O}\bigg{(}\frac{1}{N\sqrt{N}}\Big{(}\frac{u_{2}(M_{k})^{3}}{u_{1}(M_{k})^{3}}+\frac{u_{2}(M_{k})u_{3}(M_{k})}{u_{1}(M_{k})^{2}}+\frac{1+M_{k}^{8}}{u_{1}(M_{k})}e^{-\frac{M_{k}^{2}}{2\sigma^{2}}}\Big{)}\bigg{)}=\mathcal{O}\Big{(}\frac{1+M_{1}^{10}}{N_{1}}\Big{)}.$

Furthermore, by using Lemma 3.7 again with $h(t)=\log u_{1}(t)$, $h(t)=g(t)$
and $h(t)=g_{1}(t)$, the desired lemma follows. Here, we have used
$g_{1}(t)=\mathcal{O}(t^{4})$ as $t\to\infty$. ∎

Finally, we compute $S_{1}^{(3)}$ in (3.23), where $g_{+}$ and $k_{+}$ are
given as (3.21) and (2.16). For the analysis of $S_{1}^{(3)}$, we shall apply
the following lemma.

######  Lemma 3.9.

Let $h\in C^{3}((x,1-\delta))$. Then as $N\to\infty$,

(3.36)
$\displaystyle\begin{split}\sum_{k=g_{+}+1}^{k_{+}-1}h(x_{k})&=N_{1}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}h(t)\,dt+\Big{(}\theta_{+}-\frac{1}{2}\Big{)}h\Big{(}x+\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}+\Big{(}\theta_{\delta}^{\prime}-\frac{1}{2}\Big{)}h(1-\delta)\\\
&\quad-\frac{1-6\theta_{+}+6\theta_{+}^{2}}{12N_{1}}h^{\prime}\Big{(}x+\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}+\frac{1-6\theta_{\delta}^{\prime}+6\theta_{\delta}^{\prime
2}}{12N_{1}}h^{\prime}(1-\delta)+\mathcal{O}\Big{(}\frac{\mathfrak{m}(h^{\prime\prime\prime})+\mathfrak{m}(h^{\prime\prime})}{N_{1}^{2}}\Big{)},\end{split}$

where $\theta_{\delta}^{\prime}:=\theta_{\delta}-(1-\delta)$ and

(3.37) $\mathfrak{m}(f):=\max\Big{\\{}f(t):x+\frac{M_{1}}{\sqrt{N_{1}}}\leq
t\leq 1-\delta\Big{\\}}.$

Here, $\theta_{+}$ and $\theta_{\delta}$ are given by (3.30) and (3.4).

###### Proof.

Let $A=x+\frac{M_{1}}{\sqrt{N_{1}}}$, $a_{0}=1-\theta_{+}$, $B=1-\delta$ and
$b_{0}=\theta_{\delta}^{\prime}-1$. Then we have

$\displaystyle a_{n}$
$\displaystyle=AN_{1}+a_{0}=\Big{(}x+\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}N_{1}+1-\theta_{+}=g_{+}+\alpha+2,$
$\displaystyle b_{n}$
$\displaystyle=BN_{1}+b_{0}=(1-\delta)N_{1}+\theta_{\delta}^{\prime}-1=(1-\delta)N+\theta_{\delta}-1=k_{+}+\alpha.$

Then the lemma follows from Lemma 3.3. ∎

Define

(3.38)
$T(t)=2\bigg{(}t\log\Big{(}\frac{t}{x}\Big{)}+(1-t)\log\Big{(}\frac{1-t}{1-x}\Big{)}\bigg{)}.$

Note that $T(x_{k})=\eta_{k}^{2}$, where $\eta_{k}$ is given by (3.24).

######  Lemma 3.10.

As $N\to\infty$, we have

(3.39)
$\displaystyle\begin{split}S_{1}^{(3)}&=B_{1}^{(3)}N_{1}^{2}+B_{2}^{(3)}N_{1}\log
N_{1}+B_{3}^{(3)}N_{1}+B_{4}^{(3)}\sqrt{N_{1}}+B_{5}^{(3)}\log
N_{1}+B_{6}^{(3)}\\\
&\quad+B_{7}^{(3)}\frac{1}{N_{1}}+B_{8}^{(3)}\frac{1}{N_{1}^{2}}+\mathcal{O}\Big{(}\frac{M_{1}}{\sqrt{N_{1}}}+\frac{1}{M_{1}^{2}}+\frac{\sqrt{N_{1}}}{M_{1}^{7}}\Big{)},\end{split}$

where

$\displaystyle B_{1}^{(3)}$
$\displaystyle=-\frac{1}{2}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}T(t)\,dt,\qquad
B_{2}^{(3)}=-\frac{1}{2}(1-\delta-x),$ $\displaystyle B_{3}^{(3)}$
$\displaystyle=\frac{1}{2}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}h_{1}(t)\,dt-\frac{1}{2}(1-\delta-x)\log(2\pi)+\frac{1}{2}\Big{[}\Big{(}\frac{1}{2}-\theta_{+}\Big{)}T_{M_{1}}+\Big{(}\frac{1}{2}-\theta_{\delta}^{\prime}\Big{)}T_{1-\delta}\Big{]},$
$\displaystyle B_{4}^{(3)}$ $\displaystyle=\frac{1}{2}M_{1}\log(2\pi
N_{1}),\qquad B_{5}^{(3)}=\frac{1}{2}(1-\theta_{+}-\theta_{\delta}^{\prime}),$
$\displaystyle B_{6}^{(3)}$
$\displaystyle=\frac{1}{2}(1-\theta_{+}-\theta_{\delta}^{\prime})\log(2\pi)+\frac{1-6\theta_{+}+6\theta_{+}^{2}}{24}T_{M_{1}}^{\prime}-\frac{1-6\theta_{\delta}^{\prime}+6\theta_{\delta}^{\prime
2}}{24}T_{1-\delta}^{\prime}$
$\displaystyle\quad+\frac{1}{2}\Big{(}\theta_{+}-\frac{1}{2}\Big{)}h_{1}\Big{(}x+\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}+\frac{1}{2}\Big{(}\theta_{\delta}^{\prime}-\frac{1}{2}\Big{)}h_{1}(1-\delta)-\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}h_{2}(t)\,dt,$
$\displaystyle B_{7}^{(3)}$
$\displaystyle=\frac{5}{2}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}\frac{1}{T(t)^{2}}\,dt,\qquad
B_{8}^{(3)}=-\frac{37}{3}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}\frac{1}{T(t)^{3}}\,dt.$

Here $T$ is given by (3.38) and

(3.40) $T_{M_{1}}=T(x+\tfrac{M_{1}}{\sqrt{N_{1}}}),\qquad
T_{1-\delta}=T(1-\delta),\qquad
T_{M_{1}}^{\prime}=T^{\prime}(x+\tfrac{M_{1}}{\sqrt{N_{1}}}),\qquad
T_{1-\delta}^{\prime}=T^{\prime}(1-\delta)$

and

(3.41) $h_{1}(t)=\log\Big{(}\frac{t(1-t)}{(t-x)^{2}}\Big{)},\qquad
h_{2}(t)=\frac{1-13t+13t^{2}}{12t(1-t)}-\frac{1-2t}{t-x}+\frac{t(1-t)}{(t-x)^{2}}.$

###### Proof.

By Lemma 2.2, we have

(3.42)
$I_{x}(m_{k},N_{1}-m_{k})=\frac{1}{2}\operatorname{erfc}\Big{(}-\eta_{k}\sqrt{\frac{N_{1}}{2}}\Big{)}+\frac{1}{\sqrt{2\pi
N_{1}}}e^{-\frac{1}{2}\eta_{k}^{2}N_{1}}\Big{(}c_{0}(\eta_{k})+\mathcal{O}\Big{(}\frac{1}{N_{1}}\Big{)}\Big{)}.$

Let us write $x_{k}=x+d$ with $d\ll 1$. Note that

$-\eta_{k}\sqrt{\frac{N_{1}}{2}}=\sqrt{\frac{N_{1}}{2}}\frac{d}{\sigma}\Big{(}1-\frac{1-2x}{6\sigma^{2}}d+\mathcal{O}(d^{2})\Big{)}\to\infty.$

Then by (3.27), we obtain

(3.43) $I_{x}(m_{k},N_{1}-m_{k})=-\frac{1}{\eta_{k}}\,\frac{1}{\sqrt{2\pi
N_{1}}}e^{-\frac{1}{2}\eta_{k}^{2}N_{1}}\Big{[}\alpha_{k}-\beta_{k}\frac{1}{N_{1}}+\frac{3}{\eta_{k}^{4}N_{1}^{2}}-\frac{15}{\eta_{k}^{6}N_{1}^{3}}+\mathcal{O}\Big{(}\frac{1}{\eta_{k}^{8}N_{1}^{4}}+\frac{\eta_{k}}{N_{1}^{2}}\Big{)}\Big{]},$

where

(3.44)
$\alpha_{k}:=1-\eta_{k}c_{0}(\eta_{k})=-\eta_{k}\frac{\sqrt{x_{k}(1-x_{k})}}{x_{k}-x},\qquad\beta_{k}:=\frac{1}{\eta_{k}^{2}}+\eta_{k}c_{1}(\eta_{k}).$

Here $c_{0}$ and $c_{1}$ are given by (2.31) and (2.32).

Note that as $x_{k}\to x$,

$\alpha_{k}=1+\mathcal{O}(x_{k}-x),\qquad\beta_{k}=\frac{\sigma^{2}}{(x_{k}-x)^{2}}+\frac{1-2x}{3}\frac{1}{x_{k}-x}-\frac{1-\sigma^{2}}{18\sigma^{2}}+\mathcal{O}(x_{k}-x).$

Notice also that $\alpha_{k}$ is bounded away from 0 in
$x_{k}\in[x,1-\delta]$. Therefore, we have

(3.45) $\displaystyle\begin{split}\log
I_{x}(m_{k},N_{1}-m_{k})&=-\frac{1}{2}\eta_{k}^{2}N_{1}-\frac{1}{2}\log
N_{1}-\frac{1}{2}\log\eta_{k}^{2}-\frac{1}{2}\log(2\pi)+\log\alpha_{k}-\frac{\beta_{k}}{\alpha_{k}N_{1}}\\\
&\quad+\frac{5}{2\eta_{k}^{4}N_{1}^{2}}-\frac{37}{3\eta_{k}^{6}N_{1}^{3}}+\mathcal{O}\Big{(}\frac{1}{\eta_{k}^{8}N_{1}^{4}}+\frac{1}{\eta_{k}^{2}N_{1}^{2}(x_{k}-x)}+\frac{x_{k}-x}{\eta_{k}^{4}N_{1}^{2}}+\frac{x_{k}-x}{\eta_{k}^{6}N_{1}^{3}}\Big{)}.\end{split}$

We compute the summation of each term of (3.45) using Lemma 3.9. For this
purpose, let us write

$\displaystyle\mathcal{S}_{0}$
$\displaystyle=-\frac{1}{2}\sum_{k=g_{+}+1}^{k_{+}-1}\Big{(}\log
N_{1}+\log(2\pi)\Big{)},\qquad\mathcal{S}_{1}=-\frac{N_{1}}{2}\sum_{k=g_{+}+1}^{k_{+}-1}\eta_{k}^{2},\qquad\mathcal{S}_{2}=\sum_{k=g_{+}+1}^{k_{+}-1}\Big{(}-\frac{1}{2}\log\eta_{k}^{2}+\log\alpha_{k}\Big{)},$
$\displaystyle\mathcal{S}_{3}$
$\displaystyle=-\frac{1}{N_{1}}\sum_{k=g_{+}+1}^{k_{+}-1}\frac{\beta_{k}}{\alpha_{k}},\qquad\mathcal{S}_{4}=\frac{5}{2N_{1}^{2}}\sum_{k=g_{+}+1}^{k_{+}-1}\frac{1}{\eta_{k}^{4}},\qquad\mathcal{S}_{5}=-\frac{37}{3N_{1}^{3}}\sum_{k=g_{+}+1}^{k_{+}-1}\frac{1}{\eta_{k}^{6}}.$

First, observe that since

$\sum_{k=g_{+}+1}^{k_{+}-1}1=k_{+}-g_{+}-1=(1-\delta-x)N_{1}-M_{1}\sqrt{N_{1}}+\theta_{+}+\theta_{\delta}^{\prime}-1,$

we have

(3.46)
$\displaystyle\begin{split}\mathcal{S}_{0}&=-\frac{1}{2}(1-\delta-x)N_{1}\log
N_{1}-\frac{1}{2}(1-\delta-x)\log(2\pi)N_{1}+\frac{1}{2}M_{1}\sqrt{N_{1}}\log
N_{1}\\\
&\quad+\frac{1}{2}\log(2\pi)M_{1}\sqrt{N_{1}}+\frac{1}{2}(1-\theta_{+}-\theta_{\delta}^{\prime})\log
N_{1}+\frac{1}{2}(1-\theta_{+}-\theta_{\delta}^{\prime})\log(2\pi).\end{split}$

We take account of the following variation of Lemma 3.9 which can be obtained
by putting $h\circ T$ instead of $h$:

$\displaystyle\sum_{k=g_{+}+1}^{k_{+}-1}h(\eta_{k}^{2})$
$\displaystyle=N_{1}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}h(T(t))\,dt+\Big{(}\theta_{+}-\frac{1}{2}\Big{)}h(T_{M_{1}})+\Big{(}\theta_{\delta}^{\prime}-\frac{1}{2}\Big{)}h(T_{1-\delta})$
$\displaystyle\quad-\frac{1-6\theta_{+}+6\theta_{+}^{2}}{12N_{1}}h^{\prime}(T_{M_{1}})T_{M_{1}}^{\prime}+\frac{1-6\theta_{\delta}^{\prime}+6\theta_{\delta}^{\prime
2}}{12N_{1}}h^{\prime}(T_{1-\delta})T_{1-\delta}^{\prime}+\mathcal{O}\Big{(}\frac{\mathfrak{m}((h\circ
T)^{\prime\prime})+\mathfrak{m}((h\circ
T)^{\prime\prime\prime})}{N_{1}^{2}}\Big{)}.$

By using this formula, we obtain

(3.47)
$\displaystyle\begin{split}\mathcal{S}_{1}&=-\frac{N_{1}^{2}}{2}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}T(t)\,dt+\frac{N_{1}}{2}\Big{[}\Big{(}\frac{1}{2}-\theta_{+}\Big{)}T_{M_{1}}+\Big{(}\frac{1}{2}-\theta_{\delta}^{\prime}\Big{)}T_{1-\delta}\Big{]}\\\
&\quad+\frac{1-6\theta_{+}+6\theta_{+}^{2}}{24}T_{M_{1}}^{\prime}-\frac{1-6\theta_{\delta}^{\prime}+6\theta_{\delta}^{\prime
2}}{24}T_{1-\delta}^{\prime}+\mathcal{O}\Big{(}\frac{1}{N_{1}}\Big{)},\\\
\mathcal{S}_{4}&=\frac{5}{2N_{1}}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}\frac{1}{T(t)^{2}}\,dt+\mathcal{O}\Big{(}\frac{1}{M_{1}^{4}}\Big{)},\qquad\mathcal{S}_{5}=-\frac{37}{3N_{1}^{2}}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}\frac{1}{T(t)^{3}}\,dt+\mathcal{O}\Big{(}\frac{1}{M_{1}^{6}}\Big{)}.\end{split}$

In order to analyse $\mathcal{S}_{2}$, note that

$\frac{\alpha_{k}}{\eta_{k}}=\frac{1-\eta_{k}c_{0}(\eta_{k})}{\eta_{k}}=-\frac{\sqrt{x_{k}(1-x_{k})}}{x_{k}-x}.$

Then by using Lemma 3.9 with $h_{1}$ in (3.41), we have

(3.48)
$\mathcal{S}_{2}=\frac{N_{1}}{2}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}h_{1}(t)\,dt+\frac{1}{2}\Big{(}\theta_{+}-\frac{1}{2}\Big{)}h_{1}\Big{(}x+\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}+\frac{1}{2}\Big{(}\theta_{\delta}^{\prime}-\frac{1}{2}\Big{)}h_{1}(1-\delta)+\mathcal{O}\Big{(}\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}.$

Finally, note that by (2.32), we have

$\frac{\beta_{k}}{\alpha_{k}}=\frac{x_{k}(1-x_{k})}{(x_{k}-x)^{2}}-\frac{1-2x_{k}}{x_{k}-x}+\frac{1-13x_{k}+13x_{k}^{2}}{12x_{k}(1-x_{k})}=h_{2}(x_{k}).$

Then it follows from Lemma 3.9 that

(3.49)
$\mathcal{S}_{3}=-\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}h_{2}(t)\,dt+\mathcal{O}\Big{(}\frac{1}{M_{1}^{2}}\Big{)}.$

Note also that Lemma 3.9 shows

$\sum_{k=g_{+}+1}^{k_{+}-1}\mathcal{O}\Big{(}\frac{1}{\eta_{k}^{8}N_{1}^{4}}+\frac{1}{\eta_{k}^{2}N_{1}^{2}(x_{k}-x)}+\frac{x_{k}-x}{\eta_{k}^{4}N_{1}^{2}}+\frac{x_{k}-x}{\eta_{k}^{6}N_{1}^{3}}\Big{)}=\mathcal{O}\Big{(}\frac{\sqrt{N_{1}}}{M_{1}^{7}}+\frac{1}{M_{1}^{2}}\Big{)}.$

Combining (3.46), (3.47), (3.48) and (3.49), the lemma follows. ∎

To state the asymptotic behaviour of $S_{1}$, we define

(3.50)
$\mathcal{J}_{1}^{(\delta)}=\int_{x}^{1-\delta}\Big{[}h_{1}(t)-2\log\Big{(}\frac{\sigma}{t-x}\Big{)}\Big{]}\,dt,\qquad\mathcal{J}_{2}^{(\delta)}=\int_{x}^{1-\delta}\Big{[}h_{2}(t)-\frac{\sigma^{2}}{(t-x)^{2}}\Big{]}\,dt,$

where $h_{1}$ and $h_{2}$ are defined by (3.41).

######  Lemma 3.11 (Asymptotic behaviour of $S_{1}$).

As $N\to\infty$, we have

(3.51) $S_{1}=B_{1}N_{1}^{2}+B_{2}N_{1}\log
N_{1}+B_{3}N_{1}+B_{4}\sqrt{N_{1}}+B_{5}\log
N_{1}+B_{6}+\mathcal{O}\bigg{(}\frac{M_{1}^{5}}{\sqrt{N_{1}}}+\frac{1}{M_{1}^{2}}+\frac{\sqrt{N_{1}}}{M_{1}^{7}}\bigg{)},$

where

$\displaystyle B_{1}$
$\displaystyle=-\frac{1}{2}\int_{x}^{1-\delta}T(t)\,dt,\qquad
B_{2}=-\frac{1}{2}(1-x-\delta),$ $\displaystyle B_{3}$
$\displaystyle=\frac{1}{2}\mathcal{J}_{1}^{(\delta)}-\frac{1}{2}\Big{(}\theta_{\delta}^{\prime}-\frac{1}{2}\Big{)}T_{1-\delta}+(1-x-\delta)\Big{[}1+\log\Big{(}\frac{\sigma}{\sqrt{2\pi}(1-x-\delta)}\Big{)}\Big{]},$
$\displaystyle B_{4}$ $\displaystyle=\sigma\,\mathcal{I}_{1},\qquad
B_{5}=\frac{1}{4}-\frac{1}{2}\theta_{\delta}^{\prime},$ $\displaystyle B_{6}$
$\displaystyle=\frac{2(1-2x)}{3}\mathcal{I}_{2}-\mathcal{J}_{2}^{(\delta)}+\Big{(}\frac{1}{2}-\theta_{\delta}^{\prime}\Big{)}\log\sqrt{2\pi}-\frac{1-6\theta_{\delta}^{\prime}+6\theta_{\delta}^{\prime
2}}{24}T_{1-\delta}^{\prime}+\frac{1}{2}\Big{(}\theta_{\delta}^{\prime}-\frac{1}{2}\Big{)}h_{1}(1-\delta)+\frac{\sigma^{2}}{1-x-\delta}.$

Here, $T$ and $h_{1}$ are given by (3.38) and (3.41); $T_{1-\delta}$,
$T_{1-\delta}^{\prime}$ are given by (3.40); and
$\mathcal{I}_{1},\mathcal{I}_{2},\mathcal{J}_{1}^{(\delta)}$ and
$\mathcal{J}_{2}^{(\delta)}$ are given by (1.17), (1.18) and (3.50).

###### Proof.

By combining Lemmas 3.6, 3.8 and 3.10, we obtain

(3.52)
$\displaystyle\begin{split}S_{1}&=B_{1}^{(3)}N_{1}^{2}+B_{2}^{(3)}N_{1}\log
N_{1}+B_{3}^{(3)}N_{1}+B^{\prime}_{4}\sqrt{N_{1}}+B_{5}^{(3)}\log
N_{1}+B^{\prime}_{6}\\\
&\quad+B_{7}^{(3)}\frac{1}{N_{1}}+B_{8}^{(3)}\frac{1}{N_{1}^{2}}+\mathcal{O}\bigg{(}\frac{\sqrt{N_{1}}}{M_{1}^{7}}+\frac{1}{M_{1}^{2}}+\frac{M_{1}^{5}}{\sqrt{N_{1}}}+\frac{M_{1}^{10}}{N_{1}}\bigg{)},\end{split}$

where

(3.53) $B^{\prime}_{4}=B_{4}^{(2)}+B_{4}^{(3)},\qquad
B^{\prime}_{6}=B_{6}^{(2)}+B_{6}^{(3)}.$

We claim that every $M_{1}$-dependent terms cancel out.

Let us begin with the leading order term. Note that as $t\to x$, we have

$T(t)=\frac{1}{\sigma^{2}}(t-x)^{2}-\frac{1-2x}{3\sigma^{4}}(t-x)^{3}+\frac{1-3\sigma^{2}}{6\sigma^{6}}(t-x)^{4}+\mathcal{O}((t-x)^{5}).$

Then we obtain

(3.54)
$\displaystyle\begin{split}B_{1}^{(3)}N_{1}^{2}&=-\frac{1}{2}N_{1}^{2}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}T(t)\,dt=-\frac{1}{2}N_{1}^{2}\int_{x}^{1-\delta}T(t)\,dt+\frac{1}{2}N_{1}^{2}\int_{x}^{x+\frac{M_{1}}{\sqrt{N_{1}}}}T(t)\,dt\\\
&=-\frac{1}{2}N_{1}^{2}\int_{x}^{1-\delta}T(t)\,dt+\frac{1}{6\sigma^{2}}M_{1}^{3}\sqrt{N_{1}}-\frac{1-2x}{24\sigma^{4}}M_{1}^{4}+\mathcal{O}\bigg{(}\frac{M_{1}^{5}}{\sqrt{N_{1}}}\bigg{)}.\end{split}$

Note that by Lemma 3.10,

(3.55) $\displaystyle B_{2}^{(3)}N_{1}\log N_{1}$
$\displaystyle=-\frac{1}{2}(1-\delta-x)N_{1}\log N_{1},$ (3.56) $\displaystyle
B_{5}^{(3)}\log N_{1}$
$\displaystyle=\frac{1}{2}(1-\theta_{+}-\theta_{\delta}^{\prime})\log N_{1}.$

For the term $B_{3}^{(3)}N_{1}$, note that

$h_{1}(t)=2\log\Big{(}\frac{\sigma}{t-x}\Big{)}+\frac{1-2x}{\sigma^{2}}(t-x)+\mathcal{O}((t-x)^{2}),\qquad\text{as
}t\to x.$

This gives rise to

$\displaystyle\quad\frac{1}{2}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}h_{1}(t)\,dt=\frac{1}{2}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}\Big{(}h_{1}(t)-2\log\Big{(}\frac{\sigma}{t-x}\Big{)}\Big{)}\,dt+\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}\log\Big{(}\frac{\sigma}{t-x}\Big{)}\,dt$
$\displaystyle=\frac{\mathcal{J}_{1}^{(\delta)}}{2}-\frac{1}{2}\int_{x}^{x+\frac{M_{1}}{\sqrt{N_{1}}}}\Big{[}\frac{1-2x}{\sigma^{2}}(t-x)+\mathcal{O}((t-x)^{2})\Big{]}\,dt+\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}\log\Big{(}\frac{\sigma}{t-x}\Big{)}\,dt$
$\displaystyle=\frac{\mathcal{J}_{1}^{(\delta)}}{2}+(1-\delta-x)\Big{[}1+\log\Big{(}\frac{\sigma}{1-\delta-x}\Big{)}\Big{]}+\frac{M_{1}}{\sqrt{N_{1}}}\log\Big{(}\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}-\frac{M_{1}(1+\log\sigma)}{\sqrt{N_{1}}}-\frac{1-2x}{4\sigma^{2}}\frac{M_{1}^{2}}{N_{1}}+\mathcal{O}\Big{(}\frac{M_{1}^{3}}{N_{1}\sqrt{N_{1}}}\Big{)}.$

Using this and
$T_{M_{1}}=\frac{M_{1}^{2}}{\sigma^{2}N_{1}}+\mathcal{O}(M_{1}^{3}N_{1}^{-3/2})$,
we obtain

(3.57)
$\displaystyle\begin{split}B_{3}^{(3)}N_{1}&=N_{1}\bigg{[}\frac{1}{2}\mathcal{J}_{1}^{(\delta)}-\frac{1}{2}\Big{(}\theta_{\delta}^{\prime}-\frac{1}{2}\Big{)}T_{1-\delta}+(1-\delta-x)\Big{[}1+\log\Big{(}\frac{\sigma}{\sqrt{2\pi}(1-\delta-x)}\Big{)}\Big{]}\bigg{]}\\\
&\quad+\sqrt{N_{1}}\bigg{[}M_{1}\log M_{1}-\frac{1}{2}M_{1}\log
N_{1}-M_{1}(1+\log\sigma)\bigg{]}-\frac{\theta_{+}-x}{2\sigma^{2}}M_{1}^{2}+\mathcal{O}\Big{(}\frac{M_{1}^{3}}{\sqrt{N_{1}}}\Big{)}.\end{split}$

Recall that by Lemmas 3.8 and 3.10,

$B^{\prime}_{4}=-\int_{-M_{1}}^{M_{1}}\log\Big{[}\frac{1}{2}\operatorname{erfc}\Big{(}\frac{t}{\sqrt{2}\sigma}\Big{)}\Big{]}\,dt+\frac{1}{2}M_{1}\log(2\pi
N_{1}).$

Here, we have

$\log\Big{[}\frac{1}{2}\operatorname{erfc}\Big{(}\frac{t}{\sqrt{2}\sigma}\Big{)}\Big{]}=-\frac{t^{2}}{2\sigma^{2}}-\frac{1}{2}\log\Big{(}1+\frac{t^{2}}{2\sigma^{2}}\Big{)}-\frac{1}{2}\log(4\pi)+\frac{3\sigma^{4}}{2(1+t^{4})}-\frac{11\sigma^{6}}{1+t^{6}}+\mathcal{O}\Big{(}\frac{1}{t^{8}}\Big{)}\qquad\text{
as }t\to+\infty,$

and

$\log\Big{[}\frac{1}{2}\operatorname{erfc}\Big{(}\frac{t}{\sqrt{2}\sigma}\Big{)}\Big{]}=\mathcal{O}\Big{(}\frac{1}{|t|}e^{-\frac{t^{2}}{2\sigma^{2}}}\Big{)},\qquad\text{
as }t\to-\infty.$

Define

$w(t)=\log\Big{[}\frac{1}{2}\operatorname{erfc}\Big{(}\frac{t}{\sqrt{2}\sigma}\Big{)}\Big{]}+\Big{[}\frac{t^{2}}{2\sigma^{2}}+\frac{1}{2}\log\Big{(}1+\frac{t^{2}}{2\sigma^{2}}\Big{)}+\frac{1}{2}\log(4\pi)\Big{]}\mathbbm{1}_{(0,\infty)}(t).$

We also write

(3.58)
$\widetilde{\mathcal{I}}_{1}=\int_{-\infty}^{\infty}\bigg{[}\log\Big{(}\tfrac{1}{2}\operatorname{erfc}(t)\Big{)}+\Big{(}t^{2}+\frac{1}{2}\log(1+t^{2})+\frac{1}{2}\log(4\pi)\Big{)}\mathbbm{1}_{(0,\infty)}(t)\bigg{]}\,dt.$

Note that $\widetilde{\mathcal{I}}_{1}$ is related to (1.17) as

(3.59)
$\mathcal{I}_{1}=\sqrt{2}\Big{(}\widetilde{\mathcal{I}}_{1}-\frac{\pi}{2}\Big{)}.$

Then we have

$\displaystyle\quad\int_{-M_{1}}^{M_{1}}\log\Big{[}\frac{1}{2}\operatorname{erfc}\Big{(}\frac{t}{\sqrt{2}\sigma}\Big{)}\Big{]}\,dt=\int_{-M_{1}}^{M_{1}}w(t)\,dt-\int_{0}^{M_{1}}\Big{(}\frac{t^{2}}{2\sigma^{2}}+\frac{1}{2}\log\Big{(}1+\frac{t^{2}}{2\sigma^{2}}\Big{)}+\frac{1}{2}\log(4\pi)\Big{)}\,dt$
$\displaystyle=\sqrt{2}\sigma\widetilde{\mathcal{I}}_{1}-\int_{M_{1}}^{\infty}\Big{(}\frac{3\sigma^{4}}{2(1+t^{4})}-\frac{11\sigma^{6}}{1+t^{6}}\Big{)}\,dt-\int_{0}^{M_{1}}\Big{(}\frac{t^{2}}{2\sigma^{2}}+\frac{1}{2}\log\Big{(}1+\frac{t^{2}}{2\sigma^{2}}\Big{)}+\log\sqrt{4\pi}\Big{)}\,dt+\mathcal{O}\Big{(}\frac{1}{M_{1}^{7}}+\frac{e^{-M_{1}^{2}}}{M_{1}}\Big{)}.$

Then by direct computations, it follows that

(3.60)
$\displaystyle\begin{split}B^{\prime}_{4}\sqrt{N_{1}}&=-\frac{M_{1}^{3}\sqrt{N_{1}}}{6\sigma^{2}}+\frac{1}{2}M_{1}\sqrt{N_{1}}\log
N_{1}-M_{1}\log M_{1}\sqrt{N_{1}}+(1+\log\sigma)M_{1}\sqrt{N_{1}}\\\
&\quad+\sigma\,\mathcal{I}_{1}\sqrt{N_{1}}+\sigma^{2}\frac{\sqrt{N_{1}}}{M_{1}}-\frac{5\sigma^{4}}{6}\frac{\sqrt{N_{1}}}{M_{1}^{3}}+\frac{37\sigma^{6}}{15}\frac{\sqrt{N_{1}}}{M_{1}^{5}}+\mathcal{O}\Big{(}\frac{\sqrt{N_{1}}}{M_{1}^{7}}\Big{)}.\end{split}$

By employing similar computations as above, one can obtain

(3.61) $\displaystyle B_{6}^{(2)}$
$\displaystyle=\frac{1-2x}{24\sigma^{4}}M_{1}^{4}+\frac{\theta_{+}-x}{2\sigma^{2}}M_{1}^{2}+\Big{(}\theta_{+}-\frac{1}{2}\Big{)}\log\Big{(}\frac{\sqrt{2\pi}M_{1}}{\sigma}\Big{)}+\frac{2(1-2x)}{3}\mathcal{I}_{2}+\mathcal{O}\Big{(}\frac{1}{M_{1}^{2}}\Big{)},$
(3.62)
$\displaystyle\begin{split}B_{6}^{(3)}&=-\frac{\sigma^{2}}{M_{1}}\sqrt{N_{1}}+\frac{1}{2}\Big{(}\theta_{+}-\frac{1}{2}\Big{)}\log
N_{1}-\Big{(}\theta_{+}-\frac{1}{2}\Big{)}\log\Big{(}\frac{\sqrt{2\pi}M_{1}}{\sigma}\Big{)}+\Big{(}\frac{1}{2}-\theta_{\delta}^{\prime}\Big{)}\log\sqrt{2\pi}\\\
&\quad-\frac{1-6\theta_{\delta}^{\prime}+6\theta_{\delta}^{\prime
2}}{24}T_{1-\delta}^{\prime}+\frac{1}{2}\Big{(}\theta_{\delta}^{\prime}-\frac{1}{2}\Big{)}h_{1}(1-\delta)+\frac{\sigma^{2}}{1-\delta-x}-\mathcal{J}_{2}^{(\delta)}+\mathcal{O}\Big{(}\frac{M_{1}}{\sqrt{N_{1}}}\Big{)},\end{split}$
(3.63) $\displaystyle\frac{1}{N_{1}}B_{7}^{(3)}$
$\displaystyle=\frac{5}{6}\sigma^{4}\frac{\sqrt{N_{1}}}{M_{1}^{3}}+\mathcal{O}\Big{(}\frac{1}{M_{1}^{2}}+\frac{1}{N_{1}}+\frac{M_{1}}{N_{1}\sqrt{N_{1}}}\Big{)},$
(3.64) $\displaystyle\frac{1}{N_{1}^{2}}B_{8}^{(3)}$
$\displaystyle=-\frac{37}{15}\sigma^{6}\frac{\sqrt{N_{1}}}{M_{1}^{5}}+\mathcal{O}\Big{(}\frac{1}{M_{1}^{4}}+\frac{1}{N_{1}^{2}}+\frac{M_{1}}{N_{1}^{2}\sqrt{N_{1}}}\Big{)}.$

Here, we have used the asymptotic behaviour

$g(t)=\frac{1-2x}{6\sigma^{4}}\Big{[}t^{3}+3\sigma^{2}t+\mathcal{O}\Big{(}\frac{1}{t^{3}}\Big{)}\Big{]}$

as $t\to+\infty$, as well as the behaviours

$h_{2}(t)=\frac{\sigma^{2}}{(t-x)^{2}}+\mathcal{O}(1),\qquad\frac{1}{T(t)^{2}}=\frac{\sigma^{4}}{(t-x)^{4}}+\mathcal{O}((t-x)^{-3}),\qquad\frac{1}{T(t)^{3}}=\frac{\sigma^{6}}{(t-x)^{6}}+\mathcal{O}((t-x)^{-5})$

as $t\to x$. Combining all of the above, after straightforward
simplifications, one can observe that all the terms depending on $M_{1}$
cancel out, leading to the desired asymptotic behaviour. ∎

###  3.3. Proof of Theorem 1.1

This subsection culminates the proof of Theorem 1.1. The remaining task is to
combine Lemmas 3.5 and 3.11, and then collect all the coefficients.

###### Proof of Theorem 1.1.

Recall that $N_{1}=N+1$. Thus we have

$\displaystyle S_{1}$ $\displaystyle=B_{1}N^{2}+B_{2}N\log
N+(2B_{1}+B_{3})N+B_{4}\sqrt{N}$ $\displaystyle\quad+(B_{2}+B_{5})\log
N+B_{1}+B_{2}+B_{3}+B_{6}+\mathcal{O}\Big{(}\frac{1}{\sqrt{N}}\Big{)}.$

By combining (2.19) with Lemmas 3.5 and 3.11, we have

$\displaystyle\log\mathcal{P}_{n}^{\mathbb{C}}(R;\alpha,c)$
$\displaystyle=A_{1}^{\prime}N^{2}+A_{2}^{\prime}N\log
N+A_{3}^{\prime}N+A_{4}^{\prime}\sqrt{N}+A_{5}^{\prime}\log N+A_{6}^{\prime}$
$\displaystyle\quad+\mathcal{O}\Big{(}\frac{1}{M}+\frac{M^{2}}{N}+\frac{M^{4}}{N^{2}}+\frac{\sqrt{N_{1}}}{M_{1}^{7}}+\frac{1}{M_{1}^{2}}+\frac{M_{1}^{5}}{\sqrt{N_{1}}}+\frac{M_{1}^{10}}{N_{1}}\Big{)}$

as $N\to\infty$, where

$\displaystyle A_{1}^{\prime}$ $\displaystyle=B_{1}+E_{1},\qquad
A_{2}^{\prime}=B_{2}+E_{2},\qquad A_{3}^{\prime}=2B_{1}+B_{3}+E_{3},$
$\displaystyle A_{4}^{\prime}$ $\displaystyle=B_{4},\qquad
A_{5}^{\prime}=B_{2}+B_{5}+E_{5},\qquad
A_{6}^{\prime}=B_{1}+B_{2}+B_{3}+B_{6}+E_{6}.$

Note that since

$B_{1}=-\frac{1}{2}\int_{x}^{1-\delta}T(t)\,dt=\frac{1}{2}\delta^{2}\log\Big{(}\frac{\delta}{1-x}\Big{)}-\frac{1}{2}(1-\delta)^{2}\log\Big{(}\frac{1-\delta}{x}\Big{)}+\frac{1}{2}(1-x-\delta),$

we have

(3.65) $A_{1}^{\prime}=\frac{1}{2}(1-x+\log x),\qquad
A_{2}^{\prime}=-\frac{1}{2}(1-x).$

For $A_{3}^{\prime}$, recall the definitions of $f_{1}$, $T_{1-\delta}$ from
(3.13), Lemma 3.10 and $\theta_{\delta}^{\prime}:=\theta_{\delta}-(1-\delta)$.
Note also that $\mathcal{J}_{1}^{(\delta)}$ in (3.50) is computed as

(3.66)
$\displaystyle\begin{split}\mathcal{J}_{1}^{(\delta)}&:=\int_{x}^{1-\delta}\Big{(}h_{1}(t)-2\log\Big{(}\frac{\sigma}{t-x}\Big{)}\Big{)}\,dt=\int_{x}^{1-\delta}\log\Big{(}\frac{t(1-t)}{\sigma^{2}}\Big{)}\,dt\\\
&\,=-\delta\log\delta+(1-\delta)\log(1-\delta)-2(1-x-\delta)+\log(1-x)-2(1-\delta)\log\sigma.\end{split}$

Then after straightforward computations, we obtain

(3.67)
$A_{3}^{\prime}=(1-x)\Big{[}1-\frac{1}{2}\log(2\pi)+\frac{1}{2}\log\Big{(}\frac{x}{1-x}\Big{)}\Big{]}-c\log
x.$

The $\mathcal{O}(\sqrt{N})$ term immediately follows from $B_{4}$, and
$A_{5}^{\prime}$ is readily obtained from $B_{2}$, $B_{5}$, $E_{5}$ with the
fact that $\theta_{\delta}^{\prime}=\theta_{\delta}-(1-\delta)$. Consequently,
we have

(3.68) $A_{4}^{\prime}=B_{4}=\sqrt{x(1-x)}\,\mathcal{I}_{1},\qquad
A_{5}^{\prime}=\frac{1}{2}x-\frac{1-3c+3c^{2}}{6}.$

For the $\mathcal{O}(1)$ term, notice that

$\displaystyle\mathcal{J}_{2}^{(\delta)}$
$\displaystyle:=\int_{x}^{1-\delta}\Big{(}h_{2}(t)-\frac{\sigma^{2}}{(t-x)^{2}}\Big{)}\,dt=\frac{1}{12}\int_{x}^{1-\delta}\Big{(}\frac{1}{t(1-t)}-1\Big{)}\,dt$
$\displaystyle=\frac{1}{12}\log\Big{(}\frac{(1-\delta)(1-x)}{\delta
x}\Big{)}-\frac{1}{12}(1-x-\delta).$

Then by using

$f_{1}(t)=\frac{1}{2}T(1-t),\qquad
T^{\prime}(t)=2\log\Big{(}\frac{(1-x)t}{x(1-t)}\Big{)},$

after some computations, one can show that

(3.69)
$A_{6}^{\prime}=\Big{(}\frac{1-3c+3c^{2}}{6}-\frac{x}{2}\Big{)}\log\Big{(}\frac{x}{1-x}\Big{)}+\Big{(}\frac{x}{2}-\frac{1}{4}\Big{)}\log(2\pi)+\frac{2(1-2x)}{3}\mathcal{I}_{2}+\frac{1+11x}{12}-\zeta^{\prime}(-1)+\log
G(c+1).$

Finally, the facts that $N=n+\alpha+c$, $M=N^{\frac{1}{3}}$ and
$M_{1}=N_{1}^{\frac{1}{12}}$ infer that

(3.70)
$\displaystyle\begin{split}\log\mathcal{P}_{n}^{\mathbb{C}}(R;\alpha,c)&=A_{1}^{\prime}n^{2}+A_{2}^{\prime}n\log
n+(2(\alpha+c)A_{1}^{\prime}+A_{3}^{\prime})n+A_{4}^{\prime}\sqrt{n}\\\
&\quad+((\alpha+c)A_{2}^{\prime}+A_{5}^{\prime})\log
n+(\alpha+c)^{2}A_{1}^{\prime}+(\alpha+c)(A_{2}^{\prime}+A_{3}^{\prime})+A_{6}^{\prime}+\mathcal{O}(n^{-\frac{1}{12}}).\end{split}$

Using (3.65), (3.67), (3.68), (3.69), together with $x=1/(1+R^{2})$, one can
observe that the coefficients in (3.70) are expressed as in Theorem 1.1. This
completes the proof. ∎

##  4\. Gap probabilities of the symplectic ensemble

This section is organised in parallel with the previous section, and we
provide the proof of Theorem 1.3. As before, we shall use the division (2.21)
of $\log\mathcal{P}_{n}^{\mathbb{H}}$, as shown in Figure 5. Subsection 4.1 is
devoted to the analysis of $\widehat{S}_{0}$, $\widehat{S}_{2}$, and
$\widehat{S}_{3}$, while Subsection 4.2 focuses on $\widehat{S}_{1}$. During
the proof, it is convenient to redefine some of the notations in Section 3,
such as $N$, $M$, $N_{1}$, $M_{1}$, $\theta_{M}$, $\theta_{\delta}$, and
$\theta_{\delta}^{\prime}$, which allows us to reuse certain computations.

###  4.1. Analysis of $\widehat{S}_{0}$, $\widehat{S}_{2}$ and
$\widehat{S}_{3}$

In this subsection, we provide the asymptotic behaviours of $\widehat{S}_{0}$,
$\widehat{S}_{2}$ and $\widehat{S}_{3}$.

######  Lemma 4.1 (Asymptotic behaviour of $\widehat{S}_{0}$).

There exists $C>0$ such that

(4.1) $\widehat{S}_{0}=\mathcal{O}(e^{-CN}),\qquad\text{as }N\to\infty.$

###### Proof.

Recall that $m_{k}$ is given by (3.1). Note that the sum $\widehat{S}_{0}$ in
(2.21) corresponds to $\alpha+1\leq m_{k}\leq\lfloor\delta N\rfloor$. Let
$X\sim B(2N,x)$, where $N$ is given by (2.15) and $x=1/(1+R^{2})$ as in Lemma
2.1. Then since

$\displaystyle\log I_{x}(2k+2\alpha+2,2n+2c-2k)=\log
I_{x}(2m_{k},2N+1-2m_{k})=\log(1-\mathbb{P}(X<2m_{k})),$

it follows from Lemma A.1 that

$\mathbb{P}(X<2m_{k})\leq\mathbb{P}(X<2\delta
N)=\frac{1}{2}\operatorname{erfc}\Big{(}\frac{x-\delta}{\sqrt{2x}}\sqrt{2N}\Big{)}\leq\frac{1}{2}e^{-\frac{(x-\delta)^{2}}{x}N}\leq
e^{-CN}.$

This completes the proof. ∎

We write

(4.2) $\theta_{M}:=\lceil M\rceil-M,\qquad\theta_{\delta}:=\delta
N-\lfloor\delta N-\tfrac{1}{2}\rfloor.$

Then we have the following.

######  Lemma 4.2 (Asymptotic behaviour of $\widehat{S}_{3}$).

As $N\to\infty$, we have

(4.3) $\widehat{S}_{3}=\widehat{D}_{3}N+\widehat{D}_{5}\log
N+\widehat{D}_{6}-\frac{1}{3}\frac{M^{3}}{N}+\mathcal{O}\Big{(}\frac{1}{M}+\frac{M^{2}}{N}+\frac{M^{4}}{N^{2}}\Big{)},$

where

$\displaystyle\widehat{D}_{3}$ $\displaystyle=2(M+\theta_{M}-c)\log
x,\quad\widehat{D}_{5}=M^{2}+2\theta_{M}M+\theta_{M}^{2}-c^{2},$
$\displaystyle\widehat{D}_{6}$ $\displaystyle=-M^{2}\log
M+\Big{(}\frac{3}{2}+\log\Big{(}\frac{1-x}{x}\Big{)}\Big{)}M^{2}-\Big{(}2\theta_{M}+\frac{1}{2}\Big{)}M\log
M$
$\displaystyle\quad+\Big{[}\frac{1}{2}-\frac{1}{2}\log(4\pi)+\Big{(}2+2\log\Big{(}\frac{1-x}{x}\Big{)}\Big{)}\theta_{M}\Big{]}M-\Big{(}\theta_{M}^{2}+\frac{1}{2}\theta_{M}-\frac{1}{24}\Big{)}\log
M-\Big{(}\frac{1}{4}+\theta_{M}\Big{)}\log 2$
$\displaystyle\quad+(\theta_{M}^{2}-c^{2})\log\Big{(}\frac{1-x}{x}\Big{)}-\frac{1+2c+2\theta_{M}}{4}\log\pi-2\zeta^{\prime}(-1)+\log\Big{(}G(c+1)G(c+\tfrac{3}{2})\Big{)}.$

###### Proof.

By letting $X\sim B(2N,x)$ and $Y\sim B(2N,1-x)$, the summand of $S_{3}$ is
expressed as

$\log I_{x}(2m_{k},2N+1-2m_{k})=\mathbb{P}(X\geq 2m_{k})=\mathbb{P}(Y\leq
2N-2m_{k}).$

Note that by (3.1), the indices of the summation of $S_{3}$ correspond to

$2c+1\leq 2N-2m_{k}\leq 2\lceil M\rceil-1.$

By the argument used to derive (3.7), (3.8) and (3.9), we obtain

(4.4) $\log\mathbb{P}(Y\leq 2\ell+1)=2N\log
x+(2\ell+1)\log(2N)+(2\ell+1)\log\Big{(}\frac{1-x}{x}\Big{)}-\log[(2\ell+1)!]-\frac{\ell^{2}}{N}+\mathcal{O}\Big{(}\frac{\ell}{N}+\frac{\ell^{3}}{N^{2}}\Big{)},$

which leads to

(4.5) $\displaystyle\begin{split}&\quad\widehat{S}_{3}=\sum_{\ell=c}^{\lceil
M\rceil-1}\log\mathbb{P}(Y\leq 2\ell+1)\\\ &=[2(\lceil M\rceil-c)\log
x]N+(\lceil
M\rceil^{2}-c^{2})\log\Big{(}2N\frac{1-x}{x}\Big{)}-\sum_{\ell=c}^{\lceil
M\rceil-1}\log[(2\ell+1)!]-\frac{M^{3}}{3N}+\mathcal{O}\Big{(}\frac{M^{2}}{N}+\frac{M^{4}}{N^{2}}\Big{)}.\end{split}$

Here, by using the well-known duplication formula for the gamma function and
the definition of the Barnes $G$-function (1.16),

$\prod_{\ell=0}^{n-1}(2\ell+1)!=\prod_{\ell=1}^{n}\Gamma(2\ell)=\prod_{\ell=1}^{n}\frac{2^{2\ell-1}}{\sqrt{\pi}}\Gamma(\ell)\Gamma(\ell+\tfrac{1}{2})=\frac{2^{n^{2}}}{\pi^{n/2}}\frac{G(n+1)G(n+\frac{3}{2})}{G(\frac{3}{2})},$

which implies

$\displaystyle\sum_{\ell=c}^{\lceil M\rceil-1}\log[(2\ell+1)!]$
$\displaystyle=(\lceil M\rceil^{2}-c^{2})\log 2-\frac{\lceil
M\rceil-c}{2}\log\pi+\log\bigg{(}\frac{G(\lceil M\rceil+1)G(\lceil
M\rceil+\frac{3}{2})}{G(c+1)G(c+\frac{3}{2})}\bigg{)}.$

Then by using the asymptotic formula (2.4), we obtain

(4.6) $\displaystyle\begin{split}\sum_{\ell=c}^{\lceil
M\rceil-1}&\log[(2\ell+1)!]=\lceil M\rceil^{2}\log\lceil M\rceil+\Big{(}\log
2-\frac{3}{2}\Big{)}\lceil M\rceil^{2}+\frac{1}{2}\lceil M\rceil\log\lceil
M\rceil+\frac{\log(4\pi)-1}{2}\lceil M\rceil\\\ &\quad-\frac{1}{24}\log\lceil
M\rceil+\Big{(}\frac{1}{4}-c^{2}\Big{)}\log
2+\frac{1+2c}{4}\log\pi+2\zeta^{\prime}(-1)-\log\Big{(}G(c+1)G(c+\tfrac{3}{2})\Big{)}+\mathcal{O}\Big{(}\frac{1}{M}\Big{)}.\end{split}$

Now putting (4.6) and $\lceil M\rceil=M+\theta_{M}$ to (4.5) yields the lemma.
∎

Next, we analyse the summation $\widehat{S}_{2}$. For this, we shall use a
slightly modified version of Lemma 3.3.

######  Lemma 4.3.

Let $A,B,a_{0},b_{0},a_{n},b_{n},\mathfrak{m}_{A,n},\mathfrak{m}_{B,n}$ and
$\mathfrak{m}_{j,n}$ be as in Lemma 3.3. Then as $n\to\infty$, we have

(4.7)
$\displaystyle\begin{split}\sum_{j=a_{n}}^{b_{n}}f\Big{(}\frac{2j+1}{2n}\Big{)}&=n\int_{A}^{B}f(x)\,dx-
a_{0}f(A)+(1+b_{0})f(B)+\frac{(1-12a_{0}^{2})f^{\prime}(A)-(1-12(1+b_{0})^{2})f^{\prime}(B)}{24n}\\\
&\quad+\mathcal{O}\bigg{(}\frac{\mathfrak{m}_{A,n}(f^{\prime\prime})+\mathfrak{m}_{B,n}(f^{\prime\prime})}{n^{2}}+\sum_{j=a_{n}}^{b_{n}}\frac{\mathfrak{m}_{j,n}(f^{\prime\prime\prime})}{n^{3}}\bigg{)}.\end{split}$

###### Proof.

This immediately follows from Lemma 3.3 with
$\widetilde{f}(t):=f(t+\frac{1}{2n})$ and simplifications. ∎

######  Lemma 4.4 (Asymptotic behaviour of $\widehat{S}_{2}$).

As $N\to\infty$, we have

(4.8) $\widehat{S}_{2}=\widehat{C}_{1}N^{2}+\widehat{C}_{2}N\log
N+\widehat{C}_{3}N+\widehat{C}_{5}\log
N+\widehat{C}_{6}+\mathcal{O}\Big{(}\frac{1}{M}+\frac{M}{N}\Big{)},$

where

$\displaystyle\widehat{C}_{1}$
$\displaystyle=-2\int_{M/N}^{\delta}f_{1}(t)\,dt,\quad\widehat{C}_{2}=-\frac{\delta}{2},\quad\widehat{C}_{3}=2\theta_{M}f_{1}\Big{(}\frac{M}{N}\Big{)}-2(1-\theta_{\delta})f_{1}(\delta)-\frac{\delta}{2}\log
2-\int_{M/N}^{\delta}f_{2}(t)\,dt$ $\displaystyle\widehat{C}_{5}$
$\displaystyle=\frac{1}{2}(M+\theta_{\delta}+\theta_{M}-1)-\frac{1}{24},$
$\displaystyle\widehat{C}_{6}$
$\displaystyle=\Big{(}\theta_{M}^{2}-\frac{1}{12}\Big{)}f_{1}^{\prime}\Big{(}\frac{M}{N}\Big{)}+\Big{(}\frac{1}{12}-(1-\theta_{\delta})^{2}\Big{)}f_{1}^{\prime}(\delta)+\theta_{M}f_{2}\Big{(}\frac{M}{N}\Big{)}-(1-\theta_{\delta})f_{2}(\delta)$
$\displaystyle\quad+\frac{1}{24}\log
M+\frac{1}{24}\Big{(}\delta-\log\Big{(}\frac{\delta}{1-\delta}\Big{)}\Big{)}-\frac{\delta
x}{2(1-x-\delta)}-\frac{x}{2}\log\Big{(}\frac{1-x-\delta}{1-x}\Big{)}+\frac{\log
2}{2}(M+\theta_{\delta}+\theta_{M}-1).$

Here, $\theta_{M}$ and $\theta_{\delta}$ are given by (4.2) and the functions
$f_{1}$ and $f_{2}$ are given by (3.13).

###### Proof.

Let $X\sim B(2N,x)$. Then as in (3.14), we have

(4.9) $\widehat{S}_{2}=\sum_{k=k_{+}}^{k_{M}}\log
I_{x}(2m_{k},2N+1-2m_{k})=\sum_{k=k_{+}}^{k_{M}}\log\mathbb{P}(X\geq
2m_{k})=\sum_{j=\lceil M\rceil}^{\lfloor\delta
N-\frac{1}{2}\rfloor}\log\mathbb{P}(X\geq 2N-2j-1).$

Then by using Lemma A.2, we have

$\log\mathbb{P}(X\geq
2N(1-t))=-2Nf_{1}(t)-\frac{1}{2}\log(2N)-f_{2}(t)+\frac{1}{2N}f_{3}(t)+\mathcal{O}\Big{(}\frac{1}{t^{3}N^{3}}+\frac{1}{N^{2}}\Big{)},$

where $f_{3}$ is given by (3.15). By employing Lemma 4.3 with $A(N)=M/N$,
$a_{0}(N)=\theta_{M}$, $B(N)=\delta$ and $b_{0}(N)=-\theta_{\delta}$, we
obtain

$\displaystyle-2N\sum_{j=\lceil M\rceil}^{\lfloor\delta
N-\frac{1}{2}\rfloor}f_{1}\Big{(}\frac{2j+1}{2N}\Big{)}$
$\displaystyle=-2N^{2}\int_{M/N}^{\delta}f_{1}(t)\,dt+\Big{[}2\theta_{M}f_{1}\Big{(}\frac{M}{N}\Big{)}-2(1-\theta_{\delta})f_{1}(\delta)\Big{]}N$
$\displaystyle\quad+\Big{(}\theta_{M}^{2}-\frac{1}{12}\Big{)}f_{1}^{\prime}\Big{(}\frac{M}{N}\Big{)}+\Big{(}\frac{1}{12}-(1-\theta_{\delta})^{2}\Big{)}f_{1}^{\prime}(\delta)+\mathcal{O}\Big{(}\frac{1}{M}\Big{)},$
$\displaystyle-\sum_{j=\lceil M\rceil}^{\lfloor\delta
N-\frac{1}{2}\rfloor}f_{2}\Big{(}\frac{2j+1}{2N}\Big{)}$
$\displaystyle=-N\int_{M/N}^{\delta}f_{2}(t)\,dt+\theta_{M}f_{2}\Big{(}\frac{M}{N}\Big{)}-(1-\theta_{\delta})f_{2}(\delta)+\mathcal{O}\Big{(}\frac{1}{M}+\frac{1}{N}\Big{)},$
$\displaystyle\frac{1}{2N}\sum_{j=\lceil M\rceil}^{\lfloor\delta
N-\frac{1}{2}\rfloor}f_{3}\Big{(}\frac{2j+1}{2N}\Big{)}$
$\displaystyle=\frac{1}{24}\bigg{(}\delta-\log\Big{(}\frac{\delta}{1-\delta}\Big{)}\bigg{)}-\frac{\delta
x}{2(1-x-\delta)}-\frac{x}{2}\log\Big{(}\frac{1-x-\delta}{1-x}\Big{)}$
$\displaystyle\quad+\frac{1}{24}\log\Big{(}\frac{M}{N}\Big{)}+\mathcal{O}\Big{(}\frac{1}{M}+\frac{M}{N}\Big{)}.$

Combining all of the above, we obtain the desired asympototic formula. ∎

As a counterpart of Lemma 3.5, we have the following.

######  Lemma 4.5 (Asymptotic behaviour of
$\widehat{S}_{0}+\widehat{S}_{2}+\widehat{S}_{3}$).

As $N\to\infty$, we have

(4.10)
$\widehat{S}_{0}+\widehat{S}_{2}+\widehat{S}_{3}=\widehat{E}_{1}N^{2}+\widehat{E}_{2}N\log
N+\widehat{E}_{3}N+\widehat{E}_{5}\log
N+\widehat{E}_{6}+\mathcal{O}\Big{(}\frac{1}{M}+\frac{M^{2}}{N}+\frac{M^{4}}{N^{2}}\Big{)},$

where

$\displaystyle\widehat{E}_{1}$ $\displaystyle=\log
x-\delta^{2}\log\Big{(}\frac{\delta}{1-x}\Big{)}+(1-\delta)^{2}\log\Big{(}\frac{1-\delta}{x}\Big{)}+\delta,\qquad\widehat{E}_{2}=-\frac{\delta}{2},$
$\displaystyle\widehat{E}_{3}$ $\displaystyle=-2c\log
x+(1-x-\delta)\log\Big{(}\frac{1-x-\delta}{1-x}\Big{)}+\delta-\frac{1}{2}\delta\log(4\pi\delta)-\frac{1}{2}(1-\delta)\log(1-\delta)-2(1-\theta_{\delta})f_{1}(\delta),$
$\displaystyle\widehat{E}_{5}$
$\displaystyle=\frac{1}{2}\theta_{\delta}-c^{2}-\frac{11}{24},$
$\displaystyle\widehat{E}_{6}$
$\displaystyle=\Big{(}c^{2}-\frac{1}{12}\Big{)}\log\Big{(}\frac{x}{1-x}\Big{)}-\frac{1}{4}\log(2\pi)-\frac{c}{2}\log\pi-2\zeta^{\prime}(-1)+\log\Big{(}G(c+1)G(c+\tfrac{3}{2})\Big{)}$
$\displaystyle\quad+\Big{(}\frac{1}{12}-(1-\theta_{\delta})^{2}\Big{)}f_{1}^{\prime}(\delta)-(1-\theta_{\delta})f_{2}(\delta)+\frac{1}{24}\Big{(}\delta-\log\Big{(}\frac{\delta}{1-\delta}\Big{)}\Big{)}$
$\displaystyle\quad-\frac{\delta
x}{2(1-x-\delta)}-\frac{x}{2}\log\Big{(}\frac{1-x-\delta}{1-x}\Big{)}-\frac{1-\theta_{\delta}}{2}\log
2.$

Here, $f_{1}$ and $f_{2}$ are given by (3.13).

###### Proof.

Note that by Lemma 4.1, one can ignore the exponentially small
$\widehat{S}_{0}$. Using (3.17) and (3.18) again, the terms in Lemma 4.4 can
be rewritten as

$\displaystyle\widehat{C}_{1}N^{2}$
$\displaystyle=\Big{[}-\delta^{2}\log\Big{(}\frac{\delta}{1-x}\Big{)}+(1-\delta)^{2}\log\Big{(}\frac{1-\delta}{x}\Big{)}+\delta+\log
x\Big{]}N^{2}-(2M\log x)N-M^{2}\log N$ $\displaystyle\quad+\Big{[}M^{2}\log
M+M^{2}\log\Big{(}\frac{x}{1-x}\Big{)}-\frac{3}{2}M^{2}\Big{]}+\frac{M^{3}}{3N}+\mathcal{O}\Big{(}\frac{M^{4}}{N^{2}}\Big{)},$
$\displaystyle\widehat{C}_{3}N$
$\displaystyle=\Big{[}(1-x-\delta)\log(1-x-\delta)-\delta\log\Big{(}\frac{\sqrt{4\pi}}{1-x}\Big{)}+\delta-\frac{1}{2}\delta\log\delta-\frac{1}{2}(1-\delta)\log(1-\delta)-2(1-\theta_{\delta})f_{1}(\delta)\Big{]}N$
$\displaystyle\quad-(2\theta_{M}\log
x+(1-x)\log(1-x))N-\Big{(}2\theta_{M}+\frac{1}{2}\Big{)}M\log
N+\Big{(}2\theta_{M}+\frac{1}{2}\Big{)}M\log M$
$\displaystyle\quad+\Big{[}\frac{1}{2}\log(2\pi)-2\theta_{M}+2\theta_{M}\log\Big{(}\frac{x}{1-x}\Big{)}-\frac{1}{2}\Big{]}M+\mathcal{O}\Big{(}\frac{M^{2}}{N}\Big{)},$
$\displaystyle\widehat{C}_{6}$
$\displaystyle=\Big{(}\frac{1}{12}-\frac{1}{2}\theta_{M}-\theta_{M}^{2}\Big{)}\log
N+\frac{\log
2}{2}M+\Big{(}-\frac{1}{24}+\frac{1}{2}\theta_{M}+\theta_{M}^{2}\Big{)}\log
M+\frac{1}{2}\theta_{M}\log(4\pi)$
$\displaystyle\quad+\Big{(}\theta_{M}^{2}-\frac{1}{12}\Big{)}\log\Big{(}\frac{x}{1-x}\Big{)}-\Big{(}\frac{1}{12}-(1-\theta_{\delta})^{2}\Big{)}f_{1}^{\prime}(\delta)-(1-\theta_{\delta})f_{2}(\delta)$
$\displaystyle\quad+\frac{1}{24}\Big{(}\delta-\log\Big{(}\frac{\delta}{1-\delta}\Big{)}\Big{)}-\frac{\delta
x}{2(1-x-\delta)}-\frac{x}{2}\log\Big{(}\frac{1-x-\delta}{1-x}\Big{)}-\frac{1-\theta_{\delta}}{2}\log
2+\mathcal{O}\Big{(}\frac{M}{N}\Big{)}.$

Then these terms cancel out with the terms of Lemma 4.2 up to the error
$\mathcal{O}(M^{-1}+M^{2}N^{-1}+M^{4}N^{-2})$. After simplifications, only the
coefficients depending on $\delta$ remain, and the lemma follows. ∎

###  4.2. Analysis of $\widehat{S}_{1}$

In this subsection, we derive the asymptotic behaviour of $\widehat{S}_{1}$,
which corresponds to the symmetric regime of the symplectic ensemble. We shall
follow the strategy used in Subsection 3.2. Recall that $N$ is given by
(2.15). We denote

(4.11) $N_{1}=N+\frac{1}{2},\qquad M_{1}=N_{1}^{\frac{1}{12}},\qquad
x_{k}=\frac{m_{k}}{N_{1}}.$

Notice here that $N_{1}$ is an integer. It is convenient to redefine

(4.12)
$g_{-}=\Big{\lceil}N_{1}\Big{(}x+\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}-\alpha-1\Big{\rceil},\qquad
g_{+}=\Big{\lfloor}N_{1}\Big{(}x+\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}-\alpha-1\Big{\rfloor}.$

As an analogue of (3.22), we divide $\widehat{S}_{1}$ by

(4.13)
$\widehat{S}_{1}=\widehat{S}_{1}^{(1)}+\widehat{S}_{1}^{(2)}+\widehat{S}_{1}^{(3)},$

where

(4.14)
$\displaystyle\begin{split}\widehat{S}_{1}^{(1)}&=\sum_{k=k_{-}+1}^{g_{-}-1}\log
I_{x}(2m_{k},2N_{1}-2m_{k}),\\\
\widehat{S}_{1}^{(2)}&=\sum_{k=g_{-}}^{g_{+}}\log
I_{x}(2m_{k},2N_{1}-2m_{k}),\\\
\widehat{S}_{1}^{(3)}&=\sum_{k=g_{+}+1}^{k_{+}-1}\log
I_{x}(2m_{k},2N_{1}-2m_{k}).\end{split}$

With our new notations, we can efficiently reuse most of the calculations from
the complex case. In particular, by setting $\eta_{k}$ as defined in (3.24),
Lemma 2.2 yields

(4.15)
$I_{x}(2m_{k},2N_{1}-2m_{k})=\frac{1}{2}\operatorname{erfc}(-\eta_{k}\sqrt{N_{1}})+\frac{1}{\sqrt{4\pi
N_{1}}}e^{-\eta_{k}^{2}N_{1}}\Big{(}c_{0}(\eta_{k})+\frac{1}{2N_{1}}c_{1}(\eta_{k})+\mathcal{O}\Big{(}\frac{1}{N_{1}^{2}}\Big{)}\Big{)}.$

This also follows by putting $2N_{1}$ to $N_{1}$ in (3.25).

######  Lemma 4.6.

There exists a constant $C>0$ such that

(4.16) $\widehat{S}_{1}^{(1)}=\mathcal{O}(e^{-CM_{1}^{2}}),\qquad\text{ as
}N\to\infty.$

###### Proof.

An argument similar to Lemma 3.6 shows

$I_{x}(2m_{k},2N_{1}-2m_{k})=1-\frac{1}{\sqrt{4\pi
N_{1}}\eta_{k}}e^{-\eta_{k}^{2}N_{1}}\mathcal{O}(1)+\frac{1}{\sqrt{4\pi
N_{1}}}e^{-\eta_{k}^{2}N_{1}}\mathcal{O}(1)=1+\mathcal{O}(e^{-cM_{1}^{2}}),$

which verifies the lemma. ∎

We write

(4.17)
$\theta_{-}=g_{-}-\Big{(}N_{1}\Big{(}x-\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}-\alpha-1\Big{)},\qquad\theta_{+}=\Big{(}N_{1}\Big{(}x+\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}-\alpha-1\Big{)}-g_{+}.$

Define $M_{k}=\sqrt{N_{1}}(x_{k}-x)$. Note that Lemma 3.7 is still valid for
the new $N_{1}$ and $M_{1}$ since $M_{k},\theta_{-}$ and $\theta_{+}$ are
defined in the same way as in the complex case, and its proof did not use any
relations between $N_{1},M_{1}$, and $n$.

######  Lemma 4.7.

As $N\to\infty$,

(4.18)
$\widehat{S}_{1}^{(2)}=\widehat{B}_{4}^{(2)}\sqrt{N_{1}}+\widehat{B}_{6}^{(2)}+\mathcal{O}\Big{(}\frac{M_{1}^{5}}{\sqrt{N_{1}}}+\frac{M_{1}^{10}}{N_{1}}\Big{)},$

where

$\displaystyle\widehat{B}_{4}^{(2)}=\int_{-M_{1}}^{M_{1}}\log\Big{[}\frac{1}{2}\operatorname{erfc}\Big{(}\frac{t}{\sigma}\Big{)}\Big{]}\,dt,\qquad\widehat{B}_{6}^{(2)}=\int_{-M_{1}}^{M_{1}}\widehat{g}(t)\,dt+\Big{(}\frac{1}{2}-\theta_{+}\Big{)}\log\Big{[}\frac{1}{2}\operatorname{erfc}\Big{(}\frac{M_{1}}{\sigma}\Big{)}\Big{]}.$

Here, $\widehat{g}$ is a rescaled version of $g$ defined as (3.33):

$\widehat{g}(t)=\frac{1}{\sqrt{2}}g(\sqrt{2}t)=\frac{1-2x}{3\sigma^{3}\sqrt{\pi}\operatorname{erfc}(t/\sigma)}(t^{2}+\sigma^{2})e^{-\frac{t^{2}}{\sigma^{2}}}.$

###### Proof.

By using (3.35), we have

$\Delta_{k}:=-\eta_{k}\sqrt{N_{1}}-\frac{M_{k}}{\sigma}=-\frac{1-2x}{6\sigma^{3}}\frac{M_{k}^{2}}{\sqrt{N_{1}}}+\frac{5-14\sigma^{2}}{72\sigma^{5}}\frac{M_{k}^{3}}{N_{1}}+\mathcal{O}\Big{(}\frac{M_{k}^{4}}{N_{1}\sqrt{N_{1}}}\Big{)}.$

This in turn implies that

$\displaystyle\quad\operatorname{erfc}(-\eta_{k}\sqrt{N_{1}})=\operatorname{erfc}\Big{(}\frac{M_{k}}{\sigma}\Big{)}-\frac{2}{\sqrt{\pi}}e^{-\frac{M_{k}^{2}}{\sigma^{2}}}\Big{[}\Delta_{k}-\frac{M_{k}}{\sigma}\Delta_{k}^{2}+\mathcal{O}((1+M_{k}^{2})\Delta_{k}^{3})\Big{]}$
$\displaystyle=\operatorname{erfc}\Big{(}\frac{M_{k}}{\sigma}\Big{)}-\frac{2}{\sqrt{\pi}}e^{-\frac{M_{k}^{2}}{\sigma^{2}}}\Big{[}-\frac{1-2x}{6\sigma^{3}}\frac{M_{k}^{2}}{\sqrt{N_{1}}}+\frac{5-14\sigma^{2}}{72\sigma^{5}}\frac{M_{k}^{3}}{N_{1}}-\frac{1-4\sigma^{2}}{36\sigma^{7}}\frac{M_{k}^{5}}{N_{1}}+\mathcal{O}\Big{(}\frac{M_{k}^{4}+M_{k}^{8}}{N_{1}\sqrt{N_{1}}}\Big{)}\Big{]}.$

Following the argument in Lemma 3.8, we can then obtain

$\displaystyle I_{x}(2m_{k},2N-2m_{k})$
$\displaystyle=\frac{1}{2}\operatorname{erfc}(-\eta_{k}\sqrt{N_{1}})+\frac{1}{2\sqrt{\pi{N_{1}}}}e^{-\eta_{k}^{2}{N_{1}}}\Big{(}c_{0}(\eta_{k})+\mathcal{O}\Big{(}\frac{1}{N_{1}}\Big{)}\Big{)}$
$\displaystyle=\widehat{u}_{1}(M_{k})+\frac{1}{\sqrt{N_{1}}}\widehat{u}_{2}(M_{k})+\frac{1}{N_{1}}\widehat{u}_{3}(M_{k})+\mathcal{O}\Big{(}\frac{1+M_{k}^{8}}{N_{1}\sqrt{N_{1}}}e^{-\frac{M_{k}^{2}}{\sigma^{2}}}\Big{)},$

where

$\displaystyle\widehat{u}_{1}(t)$
$\displaystyle=\frac{1}{2}\operatorname{erfc}\Big{(}\frac{t}{\sigma}\Big{)},\qquad\widehat{u}_{2}(t)=\frac{1-2x}{6\sigma\sqrt{\pi}}e^{-\frac{t^{2}}{\sigma^{2}}}\bigg{(}\frac{t^{2}}{\sigma^{2}}+1\bigg{)},$
$\displaystyle\widehat{u}_{3}(t)$
$\displaystyle=\frac{1}{2\sqrt{\pi}}e^{-\frac{t^{2}}{\sigma^{2}}}\bigg{(}\frac{1-4\sigma^{2}}{18\sigma^{7}}t^{5}-\frac{1-2\sigma^{2}}{36\sigma^{5}}t^{3}+\frac{1+\sigma^{2}}{12\sigma^{3}}t\bigg{)}.$

Therefore, it follows that

$\displaystyle\log I_{x}(2m_{k},2N_{1}-2m_{k})$
$\displaystyle=\log\widehat{u}_{1}(M_{k})+\frac{1}{\sqrt{N_{1}}}\widehat{g}(M_{k})+\frac{1}{N_{1}}\widehat{g}_{1}(M_{k})$
$\displaystyle\quad+\mathcal{O}\bigg{(}\frac{1}{N_{1}\sqrt{N}_{1}}\bigg{(}\frac{\widehat{u}_{2}(M_{k})^{3}}{\widehat{u}_{1}(M_{k})^{3}}+\frac{\widehat{u}_{2}(M_{k})\widehat{u}_{3}(M_{k})}{\widehat{u}_{1}(M_{k})^{2}}+\frac{1+M_{k}^{8}}{\widehat{u}_{1}(M_{k})}e^{-\frac{M_{k}^{2}}{\sigma^{2}}}\bigg{)}\bigg{)},$

where

$\widehat{g}(t)=\frac{\widehat{u}_{2}(t)}{\widehat{u}_{1}(t)},\qquad\widehat{g}_{1}(t)=-\frac{\widehat{u}_{2}(t)^{2}}{2\widehat{u}_{1}(t)^{2}}+\frac{\widehat{u}_{3}(t)}{\widehat{u}_{1}(t)}.$

Note that $\widehat{g}_{1}(t)=\mathcal{O}(t^{4})$ as $t\to+\infty$. Then by
applying Lemma 3.7 to $\log\widehat{u}_{1}(t)$, $\widehat{g}(t)$,
$\widehat{g}_{1}(t)$ and the error term, the lemma follows. ∎

It now suffices to analyse $\widehat{S}_{1}^{(3)}$. We need a modified version
of Lemma 3.9 since the definition of $\theta_{\delta}$ is different from the
complex case.

######  Lemma 4.8.

Let $h\in C^{3}((x,1-\delta))$. Then as $N\to\infty$,

$\displaystyle\sum_{k=g_{+}+1}^{k_{+}-1}h(x_{k})$
$\displaystyle=N_{1}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}h(t)\,dt+\Big{(}\theta_{+}-\frac{1}{2}\Big{)}h\Big{(}x+\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}+\Big{(}\theta_{\delta}^{\prime}-\frac{1}{2}\Big{)}h(1-\delta)$
$\displaystyle\quad-\frac{1-6\theta_{+}+6\theta_{+}^{2}}{12N_{1}}h^{\prime}\Big{(}x+\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}+\frac{1-6\theta_{\delta}^{\prime}+6\theta_{\delta}^{\prime
2}}{12N_{1}}h^{\prime}(1-\delta)+\mathcal{O}\Big{(}\frac{\mathfrak{m}(h^{\prime\prime\prime})+\mathfrak{m}(h^{\prime\prime})}{N_{1}^{2}}\Big{)},$

where $\theta_{\delta}^{\prime}:=\theta_{\delta}-(1-\frac{\delta}{2})$ and
$\mathfrak{m}$ is given by (3.37). Here, $\theta_{+}$ and $\theta_{\delta}$
are given by (4.17) and (4.2).

###### Proof.

Let $A=x+\frac{M_{1}}{\sqrt{N_{1}}}$, $a_{0}=1-\theta_{+}$, $B=1-\delta$ and
$b_{0}=\theta_{\delta}^{\prime}-1$. Then we have

$\displaystyle a_{n}$
$\displaystyle=AN_{1}+a_{0}=\Big{(}x+\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}N_{1}+1-\theta_{+}=g_{+}+\alpha+2,$
$\displaystyle b_{n}$
$\displaystyle=BN_{1}+b_{0}=(1-\delta)N_{1}+\theta_{\delta}^{\prime}-1=(1-\delta)N+\theta_{\delta}-\frac{3}{2}=k_{+}+\alpha.$

Now Lemma 3.3 completes the proof. ∎

######  Lemma 4.9.

As $N\to\infty$, we have

(4.19)
$\displaystyle\begin{split}\widehat{S}_{1}^{(3)}&=\widehat{B}_{1}^{(3)}N_{1}^{2}+\widehat{B}_{2}^{(3)}N_{1}\log
N_{1}+\widehat{B}_{3}^{(3)}N_{1}+\widehat{B}_{4}^{(3)}\sqrt{N_{1}}+\widehat{B}_{5}^{(3)}\log
N_{1}+\widehat{B}_{6}^{(3)}\\\
&\quad+\widehat{B}_{7}^{(3)}\frac{1}{N_{1}}+\widehat{B}_{8}^{(3)}\frac{1}{N_{1}^{2}}+\mathcal{O}\Big{(}\frac{M_{1}}{\sqrt{N_{1}}}+\frac{1}{M_{1}^{2}}+\frac{\sqrt{N_{1}}}{M_{1}^{7}}\Big{)},\end{split}$

where

$\displaystyle\widehat{B}_{1}^{(3)}$
$\displaystyle=-\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}T(t)\,dt,\qquad\widehat{B}_{2}^{(3)}=-\frac{1}{2}(1-\delta-x),$
$\displaystyle\widehat{B}_{3}^{(3)}$
$\displaystyle=\frac{1}{2}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}h_{1}(t)\,dt-\frac{1}{2}(1-\delta-x)\log(4\pi)+\Big{(}\frac{1}{2}-\theta_{+}\Big{)}T_{M_{1}}+\Big{(}\frac{1}{2}-\theta_{\delta}^{\prime}\Big{)}T_{1-\delta},$
$\displaystyle\widehat{B}_{4}^{(3)}$ $\displaystyle=\frac{1}{2}M_{1}\log(4\pi
N_{1}),\qquad\widehat{B}_{5}^{(3)}=\frac{1}{2}(1-\theta_{+}-\theta_{\delta}^{\prime}),$
$\displaystyle\widehat{B}_{6}^{(3)}$
$\displaystyle=\frac{1}{2}(1-\theta_{+}-\theta_{\delta}^{\prime})\log(4\pi)+\frac{1-6\theta_{+}+6\theta_{+}^{2}}{12}T_{M_{1}}^{\prime}-\frac{1-6\theta_{\delta}^{\prime}+6\theta_{\delta}^{\prime
2}}{12}T_{1-\delta}^{\prime}$
$\displaystyle\quad+\frac{1}{2}\Big{(}\theta_{+}-\frac{1}{2}\Big{)}h_{1}\Big{(}x+\frac{M_{1}}{\sqrt{N_{1}}}\Big{)}+\frac{1}{2}\Big{(}\theta_{\delta}^{\prime}-\frac{1}{2}\Big{)}h_{1}(1-\delta)-\frac{1}{2}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}h_{2}(t)\,dt,$
$\displaystyle\widehat{B}_{7}^{(3)}$
$\displaystyle=\frac{5}{8}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}\frac{1}{T(t)^{2}}\,dt,\qquad\widehat{B}_{8}^{(3)}=-\frac{37}{24}\int_{x+\frac{M_{1}}{\sqrt{N_{1}}}}^{1-\delta}\frac{1}{T(t)^{3}}\,dt.$

Here $T$ is given by (3.38),

(4.20) $T_{M_{1}}=T(x+\tfrac{M_{1}}{\sqrt{N_{1}}}),\qquad
T_{1-\delta}=T(1-\delta),\qquad
T_{M_{1}}^{\prime}=T^{\prime}(x+\tfrac{M_{1}}{\sqrt{N_{1}}}),\qquad
T_{1-\delta}^{\prime}=T^{\prime}(1-\delta).$

Also, $h_{1}$ and $h_{2}$ are given by (3.41), $\theta_{+}$ is given by
(4.17), and $\theta_{\delta}^{\prime}$ is given by Lemma 4.8.

###### Proof.

By putting $2N_{1}$ to $N_{1}$ in (3.45), we have

(4.21)
$\displaystyle\begin{split}I_{x}(2m_{k},2N_{1}-2m_{k})&=-\eta_{k}^{2}N_{1}-\frac{1}{2}\log
N_{1}-\frac{1}{2}\log\eta_{k}^{2}-\frac{1}{2}\log(4\pi)+\log\alpha_{k}-\frac{\beta_{k}}{2\alpha_{k}N_{1}}\\\
&\quad+\frac{5}{8\eta_{k}^{4}N_{1}^{2}}-\frac{37}{24\eta_{k}^{6}N_{1}^{3}}+\mathcal{O}\Big{(}\frac{1}{\eta_{k}^{8}N_{1}^{4}}+\frac{1}{\eta_{k}^{2}N_{1}^{2}(x_{k}-x)}+\frac{x_{k}-x}{\eta_{k}^{4}N_{1}^{2}}+\frac{x_{k}-x}{\eta_{k}^{6}N_{1}^{3}}\Big{)},\end{split}$

where $\alpha_{k}$ and $\beta_{k}$ are defined by (3.44). Let us write

$\displaystyle\widehat{\mathcal{S}}_{0}$
$\displaystyle=-\frac{1}{2}\sum_{k=g_{+}+1}^{k_{+}-1}\Big{(}\log
N_{1}+\log(4\pi)\Big{)},\qquad\widehat{\mathcal{S}}_{1}=-N_{1}\sum_{k=g_{+}+1}^{k_{+}-1}\eta_{k}^{2},\qquad\widehat{\mathcal{S}}_{2}=\sum_{k=g_{+}+1}^{k_{+}-1}\Big{(}-\frac{1}{2}\log\eta_{k}^{2}+\log\alpha_{k}\Big{)},$
$\displaystyle\widehat{\mathcal{S}}_{3}$
$\displaystyle=-\frac{1}{2N_{1}}\sum_{k=g_{+}+1}^{k_{+}-1}\frac{\beta_{k}}{\alpha_{k}},\qquad\widehat{\mathcal{S}}_{4}=\frac{5}{8N_{1}^{2}}\sum_{k=g_{+}+1}^{k_{+}-1}\frac{1}{\eta_{k}^{4}},\qquad\widehat{\mathcal{S}}_{5}=-\frac{37}{24N_{1}^{3}}\sum_{k=g_{+}+1}^{k_{+}-1}\frac{1}{\eta_{k}^{6}}.$

Here, each sum can then be computed using Lemma 4.8 and expressed with
formulas similar to those in Lemma 3.10. We leave the details to the
interested reader. Then the desired asymptotic formula follows. ∎

We are now ready to state the asymptotic behaviour of $\widehat{S}_{1}.$

######  Lemma 4.10 (Asymptotic behaviour of $\widehat{S}_{1}$).

As $N\to\infty$, we have

(4.22) $\widehat{S}_{1}=\widehat{B}_{1}N_{1}^{2}+\widehat{B}_{2}N_{1}\log
N_{1}+\widehat{B}_{3}N_{1}+\widehat{B}_{4}\sqrt{N_{1}}+\widehat{B}_{5}\log
N_{1}+\widehat{B}_{6}+\mathcal{O}\Big{(}\frac{M_{1}^{5}}{\sqrt{N_{1}}}+\frac{1}{M_{1}^{2}}+\frac{\sqrt{N_{1}}}{M_{1}^{7}}\Big{)},$

where

$\displaystyle\widehat{B}_{1}$
$\displaystyle=-\int_{x}^{1-\delta}T(t)\,dt,\qquad\widehat{B}_{2}=-\frac{1}{2}(1-x-\delta),$
$\displaystyle\widehat{B}_{3}$
$\displaystyle=\frac{1}{2}\mathcal{J}_{1}^{(\delta)}-\Big{(}\theta_{\delta}^{\prime}-\frac{1}{2}\Big{)}T_{1-\delta}+(1-\delta-x)\Big{(}1+\log\Big{(}\frac{\sigma}{\sqrt{4\pi}(1-\delta-x)}\Big{)}\Big{)},$
$\displaystyle\widehat{B}_{4}$
$\displaystyle=\frac{\sigma}{\sqrt{2}}\,\mathcal{I}_{1},\qquad\widehat{B}_{5}=\frac{1}{4}-\frac{1}{2}\theta_{\delta}^{\prime},$
$\displaystyle\widehat{B}_{6}$
$\displaystyle=\frac{1-2x}{3}\mathcal{I}_{2}-\frac{1}{2}\mathcal{J}_{2}^{(\delta)}+\Big{(}\frac{1}{4}-\frac{1}{2}\theta_{\delta}^{\prime}\Big{)}\log(4\pi)-\frac{1-6\theta_{\delta}^{\prime}+6\theta_{\delta}^{\prime
2}}{12}T_{1-\delta}^{\prime}+\Big{(}\frac{1}{2}\theta_{\delta}^{\prime}-\frac{1}{4}\Big{)}h_{1}(1-\delta)+\frac{\sigma^{2}}{2(1-\delta-x)}.$

Here $T$, $h_{1}$, $T_{1-\delta}$ and $T_{1-\delta}^{\prime}$ are given by
(3.38), (3.41) and (4.20), whereas
$\mathcal{I}_{1},\mathcal{I}_{2},\mathcal{J}_{1}^{(\delta)}$ and
$\mathcal{J}_{2}^{(\delta)}$ are given by (1.17), (1.18) and (3.50).

###### Proof.

Combining Lemmas 4.6, 4.7 and 4.9, we have

(4.23)
$\displaystyle\begin{split}\widehat{S}_{1}&=\widehat{B}_{1}^{(3)}N_{1}^{2}+\widehat{B}_{2}^{(3)}N_{1}\log
N_{1}+\widehat{B}_{3}^{(3)}N_{1}+\widehat{B}_{4}^{\prime}\sqrt{N_{1}}+\widehat{B}_{5}^{(3)}\log
N_{1}+\widehat{B}_{6}^{\prime}\\\
&\quad+\widehat{B}_{7}^{(3)}\frac{1}{N_{1}}+\widehat{B}_{8}^{(3)}\frac{1}{N_{1}^{2}}+\mathcal{O}\Big{(}\frac{\sqrt{N_{1}}}{M_{1}^{7}}+\frac{1}{M_{1}^{2}}+\frac{M_{1}^{5}}{\sqrt{N_{1}}}+\frac{M_{1}^{10}}{N_{1}}\Big{)},\end{split}$

where

$\widehat{B}_{4}^{\prime}=\widehat{B}_{4}^{(2)}+\widehat{B}_{4}^{(3)},\qquad\widehat{B}_{6}^{\prime}=\widehat{B}_{6}^{(2)}+\widehat{B}_{6}^{(3)}.$

Through long but straightforward computations similar to those in Lemma 3.11,
we obtain

$\displaystyle\widehat{B}_{1}^{(3)}N_{1}^{2}$
$\displaystyle=-N_{1}^{2}\int_{x}^{1-\delta}T(t)\,dt+\frac{1}{3\sigma^{2}}M_{1}^{3}\sqrt{N_{1}}-\frac{1-2x}{12\sigma^{4}}M_{1}^{4}+\mathcal{O}\Big{(}\frac{M_{1}^{5}}{\sqrt{N_{1}}}\Big{)},$
$\displaystyle\widehat{B}_{2}^{(3)}N_{1}\log N_{1}$
$\displaystyle=-\frac{1}{2}(1-\delta-x)N_{1}\log N_{1},$
$\displaystyle\widehat{B}_{3}^{(3)}N_{1}$
$\displaystyle=\Big{[}\frac{1}{2}\mathcal{J}_{1}^{(\delta)}-\Big{(}\theta_{\delta}^{\prime}-\frac{1}{2}\Big{)}T_{1-\delta}+(1-x-\delta)\Big{(}1+\log\Big{(}\frac{\sigma}{\sqrt{4\pi}(1-x-\delta)}\Big{)}\Big{)}\Big{]}N_{1}$
$\displaystyle\quad+\Big{(}M_{1}\log M_{1}-\frac{1}{2}M_{1}\log
N_{1}-M_{1}(1+\log\sigma)\Big{)}\sqrt{N_{1}}+\frac{1+2x-4\theta_{+}}{4\sigma^{2}}M_{1}^{2}+\mathcal{O}\Big{(}\frac{M_{1}^{3}}{\sqrt{N_{1}}}\Big{)},$
$\displaystyle\widehat{B}_{4}^{\prime}\sqrt{N_{1}}$
$\displaystyle=-\frac{M_{1}^{3}\sqrt{N_{1}}}{3\sigma^{2}}+\frac{1}{2}M_{1}\sqrt{N_{1}}\log
N_{1}-M_{1}\log M_{1}\sqrt{N_{1}}+(1+\log\sigma)M_{1}\sqrt{N_{1}}$
$\displaystyle\quad+\frac{1}{\sqrt{2}}\sigma\,\mathcal{I}_{1}\sqrt{N_{1}}+\frac{\sigma^{2}}{2}\frac{\sqrt{N_{1}}}{M_{1}}-\frac{5\sigma^{4}}{24}\frac{\sqrt{N_{1}}}{M_{1}^{3}}+\frac{37\sigma^{6}}{120}\frac{\sqrt{N_{1}}}{M_{1}^{5}}+\mathcal{O}\Big{(}\frac{\sqrt{N_{1}}}{M_{1}^{7}}\Big{)},$
$\displaystyle\widehat{B}_{5}^{(3)}\log N_{1}$
$\displaystyle=\frac{1}{2}(1-\theta_{+}-\theta_{\delta}^{\prime})\log N_{1},$
$\displaystyle\widehat{B}_{6}^{(2)}$
$\displaystyle=\frac{1-2x}{12\sigma^{4}}M_{1}^{4}-\frac{1+2x-4\theta_{+}}{4\sigma^{2}}M_{1}^{2}+\Big{(}\theta_{+}-\frac{1}{2}\Big{)}\log\Big{(}\frac{\sqrt{4\pi}M_{1}}{\sigma}\Big{)}+\frac{1-2x}{3}\mathcal{I}_{2}+\mathcal{O}\Big{(}\frac{1}{M_{1}^{2}}\Big{)},$
$\displaystyle\widehat{B}_{6}^{(3)}$
$\displaystyle=-\frac{\sigma^{2}}{2}\frac{\sqrt{N_{1}}}{M_{1}}+\Big{(}\frac{\theta_{+}}{2}-\frac{1}{4}\Big{)}\log
N_{1}-\Big{(}\theta_{+}-\frac{1}{2}\Big{)}\log\Big{(}\frac{\sqrt{4\pi}M_{1}}{\sigma}\Big{)}+\Big{(}\frac{1}{4}-\frac{1}{2}\theta_{\delta}^{\prime}\Big{)}\log(4\pi)$
$\displaystyle\quad-\frac{1-6\theta_{\delta}^{\prime}+6\theta_{\delta}^{\prime