#  Joint moments of higher order derivatives of CUE characteristic polynomials
I: asymptotic formulae

Jonathan P. Keating<EMAIL_ADDRESS>Mathematical Institute, University
of Oxford, Oxford OX2 6GG, UK  Fei Wei<EMAIL_ADDRESS>Yau
Mathematical Sciences Center, Tsinghua University

###### Abstract

We derive explicit asymptotic formulae for the joint moments of the $n_{1}$-th
and $n_{2}$-th derivatives of the characteristic polynomials of CUE random
matrices for any non-negative integers $n_{1},n_{2}$. These formulae are
expressed in terms of determinants whose entries involve modified Bessel
functions of the first kind. We also express them in terms of two types of
combinatorial sums. Similar results are obtained for the analogue of Hardy’s
$Z$-function. We use these formulae to formulate general conjectures for the
joint moments of the $n_{1}$-th and $n_{2}$-th derivatives of the Riemann
zeta-function and of Hardy’s $Z$-function. Our conjectures are supported by
comparison with results obtained previously in the number theory literature.

Key words: Joint moments, higher order derivatives, characteristic
polynomials, random unitary matrices, Bessel functions, Riemann zeta function,
Hardy’s $Z$-function.

##  1 Introduction

Deep and remarkable connections exist between random matrix theory and number
theory. These originated in the work of Montgomery [33], who conjectured that
suitably normalised pairs of zeros of the Riemann zeta-function have a
statistical distribution that coincides with Dyson’s [18] description of
spacings between pairs of eigenvalues of random unitary matrices. This
connection was strengthened by the work of Keating and Snaith [30, 29], which
used characteristic polynomials of random unitary matrices to model the
Riemann zeta-function, leading to general conjectures, supported by results
from number theory and by numerical simulations, concerning the value
distribution and moments of the zeta function and other $L$-functions. These
connections and conjectures have been widely studied in the past few years,
e.g., see the survey papers [28, 1, 35].

Let $A\in{\mathbb{U}}(N)$ be taken from the Circular Unitary Ensemble (CUE) of
random matrices, and let $\Lambda_{A}(s)$ be the characteristic polynomial of
$A$, defined by

$\Lambda_{A}(s)=\prod_{n=1}^{N}(1-se^{-\i\theta_{n}}),$

where we set $\i^{2}=-1$ to avoid confusion with the index $i$, and
$e^{\i\theta_{1}},\ldots,e^{\i\theta_{N}}$ are the eigenvalues of $A$. Define

$Z_{A}(s):=e^{-\pi\i
N/2}e^{\i\sum_{n=1}^{N}\theta_{n}/2}s^{-N/2}\Lambda_{A}(s),$

where for $s^{-N/2}$, when $N$ is an odd integer, we use the branch of the
square root that is positive for $s$ real and positive. This is the analogue
of Hardy’s $Z$-function. It satisfies the functional equation
$Z_{A}(s)=(-1)^{N}Z_{A^{*}}(1/s)$, where $A^{*}$ is the conjugate transpose of
$A$. This implies that $Z_{A}(e^{\i\theta})$ is real when $\theta$ is real.

In this paper, we study the joint moments of the $n_{1}$-th and $n_{2}$-th
derivatives of the characteristic polynomial

$\int_{{\mathbb{U}}(N)}|\Lambda_{A}^{(n_{1})}(1)|^{2M}|\Lambda_{A}^{(n_{2})}(1)|^{2k-2M}dA_{N}$
(1)

and

$\int_{{\mathbb{U}}(N)}|Z_{A}^{(n_{1})}(1)|^{2M}|Z_{A}^{(n_{2})}(1)|^{2k-2M}dA_{N},$
(2)

where $n_{1},n_{2}\in\mathbb{N}$ are non-negative integers denoting
derivatives with respect to the variables in question, and $dA_{N}$ is the
Haar measure on ${\mathbb{U}}(N)$.

When $n_{1}=1,n_{2}=0$, the joint moments have been studied extensively. We
here list some results that are most relevant to ours in that case. In [26],
Hughes showed that the limits as $N\to\infty$ of
$(\ref{intro:eq1})/N^{k^{2}+2M}$ and $(\ref{intro:eq2})/N^{k^{2}+2M}$ exist
when $k$ and $M$ are integers, and provided expressions for them that are
analytic in $k$ for ${\rm Re}(k)>M-1/2$. In [16], Conrey, Rubinstein, and
Snaith showed that when $M=k$ (both integers)

$(\ref{intro:eq1})\sim(-1)^{\frac{k(k+1)}{2}}N^{k^{2}+2k}\sum_{h=0}^{k}\binom{k}{h}\left(\frac{d}{dx}\right)^{k+h}\left(e^{-x}x^{-{k^{2}}/{2}}\det_{k\times
k}\Big{(}I_{i+j-1}(2\sqrt{x})\Big{)}\right)\Bigg{|}_{x=0},$

where $I_{\alpha}(x)$ is the modified Bessel function of the first kind, and
the indices $i,j$ in the above determinant are from 1 to $k$ (we will assume
this for all $k\times k$ determinants in this paper). They also gave a similar
asymptotic formula for (2) when $M=k$. In [5], Bailey, et. al. provided
asymptotic ($N\to\infty$) formulae for (1), (2) when $k\geq M\geq 0$ (both
integers). In [20], Forrester and Witte provided alternate asymptotic
expressions for (1), (2) in terms of solutions of $\sigma$-Painlevé III′
equation. In [39], Winn gave an explicit expression for (2) in terms of
certain sums over partitions and also gave an explicit expression for non-
integer $M$ of the form $(2m-1)/2$. In [7], Basor, et. al. established an
exact representation of (2) in terms of a solution of $\sigma$-Painlevé V
equation and showed how this could be used to derive the large-$N$
asymptotics. In [4], Assiotis, Keating and Warren succeeded in proving that
the limits as $N\to\infty$ of $(\ref{intro:eq1})/N^{k^{2}+2M}$ and
$(\ref{intro:eq2})/N^{k^{2}+2M}$ exist for all real $k$ and $M$ in the
appropriate ranges. We refer to [17, 2, 19, 3] for other related results.

For one of $n_{1},n_{2}$ greater than 1, to the best of our knowledge, there
are no previous results for (2). In [17, section 10] Dehaye mentioned that his
methods might be applied to (1) for some $n_{1},n_{2}$, but he did not give
details. In [6, Theorem 4.31], Barhoumi-Andréani obtained an asymptotic
formula for (1) and for the more general moments (11). He obtains the leading
coefficient in the form of a $(k-1)$-fold real integral of a certain function.
In comparison, we provide two types of explicit formulae in the form of
combinatorial sums, see Theorems 2 and 5, which presumably correspond to
evaluation of his integral. Our approach is quite different from Dehaye’s and
Barhoumi-Andréani’s. Our starting point is the Conrey-Farmer-Keating-
Rubinstein-Snaith formula (see Lemma 15), whereas the method mentioned in [17]
is based on expressing (1) in the basis of Schur functions, and that set out
in [6] is based on replacing Lemma 15 by certain Fourier representations (see
(54), (55) and Section 1.5 in [6]) obtained by analysing the reproducing
kernel on the space of all symmetric functions. Barhoumi-Andréani argues that
his equations (54) and (55) can be represented in the form of Hankel
determinants in Section 5.1.3 in [6]. So it is likely that Barhoumi-Andréani’s
result has a link to certain Hankel determinants. In Theorem 1, using our
approach mentioned above, we give an explicit formula for the leading
coefficient in terms of determinants of Hankel matrices whose columns are
shifted by integers, and so which are not Hankel determinants anymore.
Moreover, this shift is important in building the connection between moments
of higher-order derivatives of CUE characteristic polynomials and a solution
of the $\sigma$-Painlevé III′ equation, which we establish in [31].

We remark that when defining
$\widetilde{\Lambda}_{A}(\theta):=\Lambda_{A}(e^{\i\theta})$ and
$\widetilde{Z}_{A}(\theta):=Z_{A}(e^{\i\theta})$, one may consider (1), (2)
with $\Lambda^{(n)}_{A}(1)$ replaced by $\widetilde{\Lambda}^{(n)}_{A}(0)$ and
$Z^{(n)}_{A}(1)$ replaced by $\widetilde{Z}^{(n)}_{A}(0)$ respectively.
However, using Lemma 16 (by taking appropriate orders of derivatives with
respect to the shifts and letting the shifts be zero), it is not hard to prove
that (1), (2) are indeed the main terms in the asymptotics of the
corresponding moments of $\widetilde{\Lambda}^{(n)}_{A}(0)$ and
$\widetilde{Z}^{(n)}_{A}(0)$. Intuitively, $\widetilde{\Lambda}^{(n)}_{A}(0)$
is a linear combination of $\Lambda^{(i)}_{A}(1)$ for $i=1,\ldots,n$, and
$\Lambda^{(i)}_{A}(1)$ contributes $N^{i}$. So the main term in
$|\widetilde{\Lambda}^{(n)}_{A}(0)|^{2h}$ indeed comes from
$|\Lambda^{(n)}_{A}(1)|^{2h}$. Because of this, it suffices to focus on
$\Lambda^{(n)}_{A}(1)$ and $Z^{(n)}_{A}(1)$ in (1), (2).

###  1.1 Main results

The main contribution of this paper is to give asymptotic formulae for (1),
(2) for general integers $n_{1},n_{2}$ and general integers $k\geq M\geq 0$.
More precisely, we show that

$\displaystyle(\ref{intro:eq1})$ $\displaystyle=$ $\displaystyle
a_{k,M}(n_{1},n_{2})N^{k^{2}+2Mn_{1}+2(k-M)n_{2}}+O(N^{k^{2}+2Mn_{1}+2(k-M)n_{2}-1}),$
$\displaystyle(\ref{intro:eq2})$ $\displaystyle=$ $\displaystyle
b_{k,M}(n_{1},n_{2})N^{k^{2}+2Mn_{1}+2(k-M)n_{2}}+O(N^{k^{2}+2Mn_{1}+2(k-M)n_{2}-1})$

for some $a_{k,M}(n_{1},n_{2}),b_{k,M}(n_{1},n_{2})$ depending on
$k,M,n_{1},n_{2}$ with explicit formulae. We then use these asymptotic
formulae to conjecture the joint moments of higher order derivatives of the
Riemann zeta-function on the critical line and of Hardy’s $Z$-function. The
conjectures are supported by results in the number theory literature.

The general results for (1), (2) are given explicitly in Theorems 23-27. The
expressions are long, so here we illustrate them by presenting some non-
trivial special cases. Theorems 1 and 3 are asymptotic formulae for (1), (2)
when $N\to\infty$ for $n_{1}=2,n_{2}=0$. They involve derivatives of
determinants of modified Bessel functions of the first kind shifted by
integers. Theorems 2 and 4 respectively provide an explicit description of the
coefficients appearing in the main terms in Theorems 1 and 3, express them in
terms of partitions.

######  Theorem 1.

Let $k\geq 1$, $0\leq M\leq k$ be integers. Then we have

$\int_{{\mathbb{U}}(N)}|\Lambda_{A}^{\prime\prime}(1)|^{2M}|\Lambda_{A}(1)|^{2k-2M}dA_{N}=a_{k,M}(2,0)N^{k^{2}+4M}+O(N^{k^{2}+4M-1}),$
(3)

where

$\displaystyle a_{k,M}(2,0)$ $\displaystyle=$
$\displaystyle(-1)^{\frac{k(k-1)}{2}}\sum_{l_{1}=0}^{M}\sum_{l_{2}=0}^{M}\sum_{t=0}^{2l_{1}}\sum_{\begin{subarray}{c}h_{1}+\cdots+h_{k}\\\
=\,2M-l_{1}-l_{2}\end{subarray}}\binom{M}{l_{1}}\binom{M}{l_{2}}\binom{2l_{1}}{t}\binom{2M-l_{1}-l_{2}}{h_{1},\ldots,h_{k}}$
(4)
$\displaystyle\times\,\left(\frac{d}{dx}\right)^{2l_{2}+t}\left(e^{-x}x^{-\frac{k^{2}}{2}-2M+l_{1}+l_{2}}\det_{k\times
k}\Big{(}I_{2h_{i}+i+j-1}(2\sqrt{x})\Big{)}\right)\Bigg{|}_{x=0}.$

######  Theorem 2.

$\displaystyle a_{k,M}(2,0)$ $\displaystyle=$
$\displaystyle(-1)^{\frac{k(k-1)}{2}}\sum_{l_{1}=0}^{M}\sum_{l_{2}=0}^{M}\sum_{t=0}^{2l_{1}}\sum_{\begin{subarray}{c}h_{10}+h_{1,1}+\cdots+h_{1k}=2l_{2}+t\\\
h_{21}+\cdots+h_{2k}=2M-(l_{1}+l_{2})\end{subarray}}\binom{M}{l_{1}}\binom{M}{l_{2}}\binom{2l_{1}}{t}\binom{2l_{2}+t}{h_{10},\ldots,h_{1k}}$
$\displaystyle\times\,\binom{2M-(l_{1}+l_{2})}{h_{21},\ldots,h_{2k}}(-1)^{h_{10}}\left(\prod_{i=1}^{k}\frac{1}{(2k+\sum_{s=1}^{2}sh_{si}-i)!}\right)\prod_{1\leq
i<j\leq k}\left(\sum_{s=1}^{2}sh_{sj}-\sum_{s=1}^{2}sh_{si}-j+i\right).$

######  Theorem 3.

Let $k\geq 1$, $0\leq M\leq k$ be integers. Then we have

$\int_{{\mathbb{U}}(N)}|Z_{A}^{\prime\prime}(1)|^{2M}|Z_{A}(1)|^{2k-2M}dA_{N}=b_{k,M}(2,0)N^{k^{2}+4M}+O(N^{k^{2}+4M-1}),$
(5)

where

$b_{k,M}(2,0)=(-1)^{\frac{k(k-1)}{2}}\sum_{\begin{subarray}{c}h+l_{1}+\cdots+l_{k}=2M\\\
h\geq 0,l_{1}\geq 0,\ldots,l_{k}\geq
0\end{subarray}}\binom{2M}{h,l_{1},\ldots,l_{k}}\left(\frac{d}{dx}\right)^{2h}\left(e^{-\frac{x}{2}}x^{h-2M-\frac{k^{2}}{2}}\det_{k\times
k}\left(I_{2l_{i}+i+j-1}(2\sqrt{x})\right)\right)\Bigg{|}_{x=0}.$

######  Theorem 4.

$\displaystyle b_{k,M}(2,0)$ $\displaystyle=$
$\displaystyle(-1)^{\frac{k(k-1)}{2}}\sum_{\begin{subarray}{c}l_{0}+\cdots+l_{k}=2M\\\
l_{0}\geq 0,\ldots,l_{k}\geq
0\end{subarray}}\,\,\,\sum_{\begin{subarray}{c}m_{0}+\cdots+m_{k}=2l_{0}\\\
m_{0}\geq 0,\ldots,m_{k}\geq
0\end{subarray}}(-2)^{-m_{0}}\binom{2M}{l_{0},\ldots,l_{k}}\binom{2l_{0}}{m_{0},\ldots,m_{k}}$
$\displaystyle\times\,\Bigg{(}\prod_{i=1}^{k}\frac{1}{(2k-i+2l_{i}+m_{i})!}\Bigg{)}\Bigg{(}\prod_{1\leq
i<j\leq k}(2l_{j}-2l_{i}+m_{j}-m_{i}-j+i)\Bigg{)}.$

The following two theorems respectively provide alternative forms of the
asymptotic formulae for (1), (2) when $n_{1}=n,n_{2}=0$. Compared to Theorems
23 and 24 with $n_{1}=n$, $n_{2}=0$, they have a significant computational
advantage when $k$ is small and $n$ is large. For more about the differences
in the general case, we refer to Remark 26.

######  Theorem 5.

For $k\geq 1$, $0\leq M\leq k$ and $n\geq 0$ be integers, we have

$\int_{{\mathbb{U}}(N)}|\Lambda_{A}^{(n)}(1)|^{2M}|\Lambda_{A}(1)|^{2k-2M}dA_{N}=a_{k,M}(n,0)N^{k^{2}+2Mn}+O(N^{k^{2}+2Mn-1}),$
(6)

where

$\displaystyle a_{k,M}(n,0)$ $\displaystyle=$
$\displaystyle(-1)^{\frac{k(k-1)}{2}}(n!)^{2M}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s_{l,q}\leq
n\\\ \sum_{l=1}^{k}h_{l,q}=n\\\
q=1,\ldots,M\end{subarray}}(-1)^{\sum_{q=1}^{M}\sum_{l=1}^{k}s_{l,q}}\left(\prod_{q=1}^{M}\frac{1}{(n-\sum_{l=1}^{k}s_{l,q})!}\right)$
$\displaystyle\times\,\left(\prod_{i=1}^{k}\frac{1}{(2k-i+\sum_{q=1}^{M}(s_{i,q}+h_{i,q}))!}\right)\prod_{1\leq
i<j\leq
k}\left(\sum_{q=1}^{M}(s_{j,q}+h_{j,q})-\sum_{q=1}^{M}(s_{i,q}+h_{i,q})-j+i\right).$

######  Theorem 6.

Let $k\geq 1$ and $n\geq 0$ be integers. We have

$\int_{{\mathbb{U}}(N)}|Z_{A}^{(n)}(1)|^{2M}|Z_{A}(1)|^{2k-2M}dA_{N}=b_{k,M}(n,0)N^{k^{2}+2Mn}+O(N^{k^{2}+2Mn-1}),$
(7)

where

$\displaystyle b_{k,M}(n,0)$ $\displaystyle=$
$\displaystyle(-1)^{nM+\frac{k(k-1)}{2}}N^{k^{2}+2Mn}(n!)^{2M}\sum_{\begin{subarray}{c}s_{l,q}\geq
0\\\ \sum_{l=1}^{k}s_{l,q}\leq n\\\
q=1,\ldots,2M\end{subarray}}(-1/2)^{2Mn-\sum_{q=1}^{2M}\sum_{l=1}^{k}s_{l,q}}$
$\displaystyle\times\,\left(\prod_{q=1}^{2M}\frac{1}{(n-\sum_{l=1}^{k}s_{l,q})!}\right)\left(\prod_{i=1}^{k}\frac{1}{(2k-i+\sum_{q=1}^{2M}s_{i,q})!}\right)\prod_{1\leq
i<j\leq k}\left(\sum_{q=1}^{2M}s_{j,q}-\sum_{q=1}^{2M}s_{i,q}-j+i\right).$

Following the philosophy of random matrix theory [30, 21, 11], we make the
following conjecture for the moments of higher order derivatives of the
Riemann zeta function $\zeta(s)$, on the critical line ($\rm{Re}(s)=1/2$), and
of Hardy’s $Z$-function defined as

$Z(t)=\frac{\pi^{-\i t/2}\Gamma(1/4+\i t/2)}{|\Gamma(1/4+\i t/2)|}\zeta(1/2+\i
t).$ (8)

######  Conjecture 7.

For any integers $n_{1},n_{2}\geq 0$ and integers $k\geq 1,0\leq M\leq k$,

$\frac{1}{T}\int_{0}^{T}|\zeta^{(n_{1})}(1/2+\i
t)|^{2M}|\zeta^{(n_{2})}(1/2+\i t)|^{2k-2M}dt\sim
a_{k,M}(n_{1},n_{2})c_{k}(\log T)^{k^{2}+2Mn_{1}+2(k-M)n_{2}},$ (9)

and

$\frac{1}{T}\int_{0}^{T}|Z^{(n_{1})}(t)|^{2M}|Z^{(n_{2})}(t)|^{2k-2M}dt\sim
b_{k,M}(n_{1},n_{2})c_{k}(\log T)^{k^{2}+2Mn_{1}+2(k-M)n_{2}},$ (10)

where $c_{k}$ is the arithmetic factor

$c_{k}=\prod_{p}\left(1-\frac{1}{p}\right)^{k^{2}}\sum_{m=0}^{\infty}\left(\frac{\Gamma(m+k)}{m!\cdot\Gamma(k)}\right)^{2}p^{-m}.$

It follows from [27, Theorem A′′] that the left hand side of (9) is
asymptotically close to $\frac{1}{2n+1}(\log T)^{2n+1}$ when
$k=M=1,n_{1}=n,n_{2}=0$. It follows from [23, Theorem 3] that the left hand
side of (10) is asymptotically close to $\frac{1}{4^{n}(2n+1)}(\log T)^{2n+1}$
when $k=M=1,n_{1}=n,n_{2}=0$. In addition, by [23, Theorems 10 and 11], the
left hand side of (10) is asymptotically close to $\frac{1}{672\pi^{2}}(\log
T)^{8}$ when $k=2,M=1,n_{1}=2,n_{2}=0$, and asymptotically close to
$\frac{19}{604800\pi^{2}}(\log T)^{10}$ when $k=2,M=1,n_{1}=2,n_{2}=1$.
Moreover, from [25, Lemma 1], the left hand side of (10) is asymptotically
close to $\frac{17}{1774080\pi^{2}}(\log T)^{12}$ when $k=M=2$,
$n_{1}=2,n_{2}=0$. All these results verify the conjecture in the respective
special cases due to the fact that $c_{1}=1,c_{2}=6/\pi^{2}$,
$b_{2,1}(2,0)=1/4032,b_{2,1}(2,1)=19/3628800$, $b_{2,2}(2,0)=17/10644480$, and
the following results on $a_{1,1}(n,0)$ and $b_{1,1}(n,0)$ from Theorems 5 and
6.

######  Proposition 8.

For any integer $n\geq 0$,

$a_{1,1}(n,0)=\frac{1}{2n+1},\quad b_{1,1}(n,0)=\frac{1}{(2n+1)4^{n}}.$

Conjecture 7 is closely related to problems concerning large gaps between
zeros of higher order derivatives of the Hardy $Z$-function. Let

$\Lambda^{(m)}=\limsup_{n\rightarrow\infty}\frac{t_{n+1}^{(m)}-t_{n}^{(m)}}{2\pi/\log
t_{n}^{(m)}},$

where $\\{t_{n}^{(m)}\\}$ is the sequence of non-negative zero of
$Z^{(m)}(t)$, counted according to multiplicity. (This is standard notation:
$\Lambda^{(m)}$ is not to be confused with the $m$-th derivative of
$\Lambda_{A}(e^{\i\theta})$.) It is noteworthy that for any fixed $m$, the
quantity $\frac{t_{n+1}^{(m)}-t_{n}^{(m)}}{2\pi/\log t_{n}^{(m)}}$ has average
value 1 under the Riemann hypothesis. This is well known when $m=0$. For
$m\geq 1$, we refer readers to [37, 36, 32, 22, 13, 8, 40] for distribution of
zeros of higher order derivatives of the Riemann zeta function and [25] for
zeros of derivatives of $Z(t)$.

For $m=0$, the quantity $\Lambda^{(0)}$ was introduced by Hall [23]. A related
quantity $\lambda$ may be defined with respect to $\\{\gamma_{n}\\}$, where
$\\{\gamma_{n}\\}$ is the non-decreasing sequence of imaginary parts of the
zeros of the Riemann zeta function in the upper half plane, counted by
multiplicity. Selberg showed that $\lambda>1$ [34]. Conrey, Ghosh and Gonek
showed that $\lambda>2.337$ under the Riemann hypothesis [14] and
$\lambda>2.68$ under the generalized Riemann hypothesis [15]. Note that
$\lambda=\Lambda^{(0)}$ under these assumptions. In [24], Hall proposed a
method, which does not assume the Riemann hypothesis, to use joint moments of
$Z(t)$ and $Z^{\prime}(t)$ to obtain lower bounds of $\Lambda^{(0)}$. He
proved that $\Lambda^{(0)}>4$ times the average gap length assuming (10) for
the case $n_{1}=1,n_{2}=0,k=6$ in our notation, which was predicted previously
by Random Matrix Theory. It is widely believed that $\Lambda^{(0)}=\infty$.
This problem remains open. For $m\geq 1$, we refer to [12, 25] for lower
bounds on $\Lambda^{(1)}$ (which all exceed 1). Additionally, in [9] the
authors also obtained lower bounds of $\Lambda^{(m)}$ for some $m\geq 2$.

We remark that the method in [24] uses a Wirtinger-type inequality, which
involves joint moments of $Z(t)$ and $Z^{\prime}(t)$. It was mentioned in [9]
that this inequality may be generalized to higher order derivatives that
involve joint moments of higher order derivatives of $Z(t)$. Using the results
in this paper, we can give numerical data on $b_{k,M}(n_{1},n_{2})$ for large
$k,M$ and $n_{1},n_{2}$. Under Conjecture 7, we are able to predict the joint
moments of the $n_{1}$-th and $n_{2}$-th derivatives of $Z(t)$ for large
$n_{1},n_{2},k$ and for any $M$ between 0 and $k$. It is possible to use these
numerical values to predict sharper lower bounds on $\Lambda^{(m)}$ through
generalized Wirtinger-type inequalities. We plan to explore this in future
research.

###  1.2 A brief summary of our main methods

Our starting point is [16], which concerns the first order derivative of
moments of characteristic polynomials of CUE matrices. The approach there is
to take the first order derivative of moments of characteristic polynomials
with shifts. In the higher-order case, it is natural to use the same approach
and take the $n$-th derivative. The difficulty here is how to represent
multiple integrals (see formulae (18) and (19)) by expressions in terms of the
determinant involving modified Bessel functions of the first kind, which is
closely related to solutions of the $\sigma$-Painlevé III′ equation.
Furthermore, it is desirable that this expression is computable. Our main idea
here is to introduce $n$ extra variables $t_{1},\ldots,t_{n}$, and to compute
higher order derivatives with respect to them at 0. We find that the higher
order derivative with respect to $t_{1}$ is related to taking the derivative
of a determinant of a matrix whose entries involve the modified Bessel
functions of the first kind, and higher order derivatives with respect to the
other $n-1$ variables contribute combinatorial coefficients and shifts to the
indices of the modified Bessel functions of the first kind, see Propositions
21, 22.

Our method to describe $a_{k,M}(n_{1},n_{2}),b_{k,M}(n_{1},n_{2})$ is
different from that of [16, Theorem 3]. The main tool of [16] is the Laplacian
transform, which converts the complex integral into a computable multiple real
integral. In the higher order derivative case, this multiple integral is very
complicated. Therefore, instead of using the Laplacian transform, we directly
take the higher order derivatives of the determinant involving the modified
Bessel functions of the first kind with shifts that appear in (4). We find
that this procedure can be carried out straightforwardly.

In Theorems 5 and 6, we provide alternate formulae for the joint moments of
higher order derivatives that are more concise than those in Theorems 2 and 4.
Our starting point is similar to that of Theorems 1 and 3. The difference is
that we use a new method (see Proposition 10) to handle the higher order
powers of some expressions produced in taking the $n$-th order derivatives.

We remark that our approaches can be generalised to any type of joint moments
of characteristic polynomials of the form

$\int_{{\mathbb{U}}(N)}|\Lambda_{A}^{(n_{1})}(1)|^{2M_{1}}|\Lambda_{A}^{(n_{2})}(1)|^{2M_{2}}\cdots|\Lambda_{A}^{(n_{d})}(1)|^{2M_{d}}dA_{N}.$
(11)

and to any corresponding joint moments of the analogue of Hardy’s
$Z$-functions.

###  1.3 Notation

We denote by $I_{n}(x)$ the modified Bessel function of the first kind, which
has the following power series expansion

$I_{n}(x)=(\frac{x}{2})^{n}\sum_{j=0}^{\infty}\frac{x^{2j}}{2^{2j}(n+j)!j!}.$

For $x\in\mathbb{C}$, we denote

$\displaystyle z(x)=\frac{1}{1-e^{-x}}$ (12)

The multinomial coefficient is defined as

$\binom{m}{l_{1},\ldots,l_{k}}=\frac{m!}{l_{1}!\cdots l_{k}!},$

where $l_{1}+\cdots+l_{k}=m$. We denote by $S_{k}$ the set of all permutations
of $\\{1,\ldots,k\\}$. For $n$ an integer, when we say $h_{1}+\cdots+h_{k}=n$
or $h_{1}+\cdots+h_{k}\leq n$, we shall always mean $h_{i}\geq 0$ and taking
integer values. For convenience, for any integrable function $F$, we define

$\int_{|w_{i}|=1}F(w_{1},\ldots,w_{k})\prod_{i=1}^{k}dw_{i}:=\int_{|w_{1}|=1}\cdots\int_{|w_{k}|=1}F(w_{1},\ldots,w_{k})\prod_{i=1}^{k}dw_{i}.$

Let $f(t)$ be a differentiable function. The $n$-th derivative of $f$ will be
denoted as $(\frac{d}{dt})^{n}f$ or $f^{(n)}$. For
$w_{1},\ldots,w_{k}\in\mathbb{C}$, the Vandermonde determinant is denoted as

$\Delta(w_{1},\ldots,w_{k})=\det_{k\times k}(w_{i}^{j-1})=\prod_{1\leq i<j\leq
k}(w_{i}-w_{j}).$

For convenience, we will ignore the subscripts and denote it as $\Delta(w)$.
We also allow commutative differential operators as the arguments, such as

$\Delta(\frac{d}{dL}):=\Delta(\frac{d}{dL_{1}},\cdots,\frac{d}{dL_{n}})=\prod_{1\leq
i<j\leq k}\Big{(}\frac{d}{dL_{j}}-\frac{d}{dL_{i}}\Big{)}.$

The square of $\Delta(w)$ will be denoted as $\Delta^{2}(w)$. Similarly, the
composition of the operator $\Delta(\frac{d}{dL})$ will be denoted as
$\Delta^{2}(\frac{d}{dL})$.

###  1.4 Acknowledgement

This research was carried out while F.W. was visiting the University of
Oxford. F.W. would like to thank the Mathematical Institute, University of
Oxford for generous hospitality. We are grateful to Dr. Yacine Barhoumi-
Andréani for helpful feedback on an earlier version of our manuscript. F.W.
was supported by the fellowship of China Postdoctoral Science Foundation
2020M670273 and by Jinxin Xue’s grant NSFC (project No. 20221300605). JPK is
pleased to acknowledge support from ERC Advanced Grant 740900 (LogCorRM).

##  2 Lemmas and propositions

In this section, we prove some lemmas and propositions that will be used in
the proofs of our main results stated in the next section.

The following lemma gives an explicit formula for the higher order derivatives
of the function appearing in the expression of moments of characteristic
polynomials with shifts.

######  Lemma 9.

Let $n\geq 0,k\geq 1$ be integers, then

$\displaystyle\left(\frac{d}{d\alpha}\right)^{n}\frac{e^{-\frac{N}{2}\alpha}}{\prod_{i=1}^{k}(w_{i}-\alpha)}$
$\displaystyle=$
$\displaystyle\frac{e^{-\frac{N}{2}\alpha}}{\prod_{i=1}^{k}(w_{i}-\alpha)}\sum_{m_{1}+2m_{2}+\cdots+nm_{n}=n}\frac{n!}{m_{1}!\cdots
m_{n}!}\left(-\frac{N}{2}+\sum_{l=1}^{k}\frac{1}{w_{l}-\alpha}\right)^{m_{1}}\prod_{j=2}^{n}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{(w_{l}-\alpha)^{j}}\right)^{m_{j}}.$

###### Proof.

It is not hard to check that

$\left(\frac{d}{d\alpha}\right)^{n}\frac{e^{-\frac{N}{2}\alpha}}{\prod_{i=1}^{k}(w_{i}-\alpha)}=\frac{e^{-\frac{N}{2}\alpha}}{\prod_{i=1}^{k}(w_{i}-\alpha)}f_{n}(\alpha),$
(13)

where $f_{n}(\alpha)$ is defined by the following recursive formulae

$\displaystyle f_{1}(\alpha)$ $\displaystyle=$
$\displaystyle-\frac{N}{2}+\sum_{l=1}^{k}\frac{1}{w_{l}-\alpha},$
$\displaystyle f_{l+1}(\alpha)$ $\displaystyle=$ $\displaystyle
f_{l}(\alpha)f_{1}(\alpha)+f_{l}^{\prime}(\alpha),\quad l\geq 1.$

Let $g(\alpha)$ be a function such that $g^{\prime}(\alpha)=f_{1}(\alpha)$.
Consider the function $e^{g(\alpha)}$, then

$\left(\frac{d}{d\alpha}\right)^{n}e^{g(\alpha)}=e^{g(\alpha)}f_{n}(\alpha).$

By Faà di Bruno’s formula

$\displaystyle\left(\frac{d}{d\alpha}\right)^{n}e^{g(\alpha)}=e^{g(\alpha)}\sum_{m_{1}+2m_{2}+\cdots+nm_{n}=n}\frac{n!}{m_{1}!\cdots
m_{n}!}\prod_{j=1}^{n}\left(\frac{f_{1}^{(j-1)}(\alpha)}{j!}\right)^{m_{j}},$

so

$\displaystyle
f_{n}(\alpha)=\sum_{m_{1}+2m_{2}+\cdots+nm_{n}=n}\frac{n!}{m_{1}!\cdots
m_{n}!}\left(-\frac{N}{2}+\sum_{l=1}^{k}\frac{1}{w_{l}-\alpha}\right)^{m_{1}}\prod_{j=2}^{n}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{(w_{l}-\alpha)^{j}}\right)^{m_{j}}.$

By (13), we obtain the claimed result. ∎

The proposition below gives a formula for the powers of the expression
obtained above at $\alpha=0$.

######  Proposition 10.

Let $k\geq 1,r\geq 1$, $n\geq 0$ and $m_{1}\geq 0,\ldots,m_{n}\geq 0$ be
integers. Let $w_{1},\ldots,w_{k}$ be non-zero complex numbers. Then we have

$\displaystyle\Bigg{(}\sum_{m_{1}+2m_{2}+\cdots+nm_{n}=n}\frac{n!}{m_{1}!\cdots
m_{n}!}(-N+\sum_{l=1}^{k}\frac{1}{w_{l}})^{m_{1}}\prod_{j=2}^{n}(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}})^{m_{j}}\Bigg{)}^{r}$
(14) $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s_{lj}\leq n\\\
j=1,\ldots,r\end{subarray}}\,\prod_{j=1}^{r}\left((-N)^{n-\sum_{l=1}^{k}s_{lj}}\binom{n}{\sum_{l=1}^{k}s_{lj}}(\sum_{l=1}^{k}s_{lj})!\right)\frac{1}{\prod_{l=1}^{k}w_{l}^{\sum_{j=1}^{r}s_{lj}}}.$

and

$\displaystyle\Bigg{(}\sum_{m_{1}+2m_{2}+\cdots+nm_{n}=n}\frac{n!}{m_{1}!\cdots
m_{n}!}\prod_{j=1}^{n}(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}})^{m_{j}}\Bigg{)}^{r}=\sum_{\begin{subarray}{c}\sum_{l=1}^{k}h_{lj}=n\\\
j=1,\ldots,r\end{subarray}}\frac{(n!)^{r}}{\prod_{l=1}^{k}w_{l}^{\sum_{j=1}^{r}h_{lj}}}.$
(15)

###### Proof.

Denote

$\displaystyle
a_{n-m_{1}}=\sum_{2m_{2}+\cdots+nm_{n}=n-m_{1}}\frac{1}{m_{2}!\cdots
m_{n}!}\prod_{j=2}^{n}(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}})^{m_{j}}.$

Then $a_{n-m_{1}}$ is the coefficient of $t^{n-m_{1}}$ in the Taylor expansion
of

$\prod_{j=2}^{n}e^{\frac{t^{j}}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}}=\prod_{j=2}^{n}\prod_{l=1}^{k}e^{\frac{1}{j}\frac{t^{j}}{w_{l}^{j}}}.$

So

$\displaystyle a_{n-m_{1}}$ $\displaystyle=$
$\displaystyle\frac{1}{(n-m_{1})!}\frac{d^{(n-m_{1})}}{dt^{(n-m_{1})}}e^{\sum_{j=2}^{n}(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}})t^{j}}\Bigg{|}_{t=0}$
$\displaystyle=$
$\displaystyle\frac{1}{(n-m_{1})!}\frac{d^{(n-m_{1})}}{dt^{(n-m_{1})}}e^{\sum_{j=2}^{\infty}(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}})t^{j}}\Bigg{|}_{t=0}$
$\displaystyle=$
$\displaystyle\frac{1}{(n-m_{1})!}\frac{d^{(n-m_{1})}}{dt^{(n-m_{1})}}\prod_{l=1}^{k}e^{\sum_{j=2}^{\infty}\frac{1}{j}\frac{t^{j}}{w_{l}^{j}}}\Bigg{|}_{t=0}$
$\displaystyle=$
$\displaystyle\frac{1}{(n-m_{1})!}\frac{d^{(n-m_{1})}}{dt^{(n-m_{1})}}\prod_{l=1}^{k}e^{-\ln(1-\frac{t}{w_{l}})-\frac{t}{w_{l}}}\Bigg{|}_{t=0}$
$\displaystyle=$
$\displaystyle\frac{1}{(n-m_{1})!}\frac{d^{(n-m_{1})}}{dt^{(n-m_{1})}}\prod_{l=1}^{k}e^{-\frac{t}{w_{l}}}\frac{1}{1-\frac{t}{w_{l}}}\Bigg{|}_{t=0}.$

Namely, $a_{n-m_{1}}$ is the coefficient of $t^{n-m_{1}}$ in the Taylor
expansion of

$F(t)=\left(\prod_{l=1}^{k}\frac{1}{1-\frac{t}{w_{l}}}\right)e^{-\sum_{l=1}^{k}\frac{t}{w_{l}}}.$

Let $G_{1}(t)=e^{(-N+\sum_{l=1}^{k}\frac{1}{w_{l}})t}$ and let
$G_{2}(t)=e^{(\sum_{l=1}^{k}\frac{1}{w_{l}})t}$, then

$\displaystyle\sum_{m_{1}+2m_{2}+\cdots+nm_{n}=n}\frac{n!}{m_{1}!\cdots
m_{n}!}(-N+\sum_{l=1}^{k}\frac{1}{w_{l}})^{m_{1}}\prod_{j=2}^{n}(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}})^{m_{j}}$
$\displaystyle=$
$\displaystyle\sum_{m_{1}=0}^{n}a_{n-m_{1}}\frac{n!}{m_{1}!}(-N+\sum_{l=1}^{k}\frac{1}{w_{l}})^{m_{1}}$
$\displaystyle=$
$\displaystyle\frac{d^{n}}{dt^{n}}(G_{1}(t)F(t))\Bigg{|}_{t=0}=\frac{d^{n}}{dt^{n}}\left(e^{-Nt}\prod_{l=1}^{k}\frac{1}{1-\frac{t}{w_{l}}}\right)\Bigg{|}_{t=0}$
$\displaystyle=$
$\displaystyle\frac{d^{n}}{dt^{n}}\left(e^{-Nt}\prod_{l=1}^{k}(\sum_{i=0}^{n}\frac{t^{i}}{w_{l}^{i}})\right)\Bigg{|}_{t=0}.$

Similarly,

$\sum_{m_{1}+2m_{2}+\cdots+nm_{n}=n}\frac{n!}{m_{1}!\cdots
m_{n}!}\prod_{j=1}^{n}(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}})^{m_{j}}=\frac{d^{n}}{dt^{n}}\left(\prod_{l=1}^{k}(\sum_{i=0}^{n}\frac{t^{i}}{w_{l}^{i}})\right)\Bigg{|}_{t=0}.$

Note that

$\displaystyle\left(\frac{d^{n}}{dt^{n}}\left(e^{-Nt}\prod_{l=1}^{k}(\sum_{i=0}^{n}\frac{t^{i}}{w_{l}^{i}})\right)\Bigg{|}_{t=0}\right)^{r}$
$\displaystyle=$
$\displaystyle\left(\prod_{j=1}^{r}\frac{d^{n}}{dt_{j}^{n}}\right)\prod_{j=1}^{r}\left(e^{-Nt_{j}}\prod_{l=1}^{k}(\sum_{i=0}^{n}\frac{t_{j}^{i}}{w_{l}^{i}})\right)\Bigg{|}_{t_{1},\ldots,t_{r}=0}$
$\displaystyle=$
$\displaystyle\left(\prod_{j=1}^{r}\frac{d^{n}}{dt_{j}^{n}}\right)e^{-N(t_{1}+\cdots+t_{r})}\prod_{j=1}^{r}\prod_{l=1}^{k}(\sum_{i=0}^{n}\frac{t_{j}^{i}}{w_{l}^{i}})\Bigg{|}_{t_{1},\ldots,t_{r}=0}$
$\displaystyle=$
$\displaystyle\left(\prod_{j=1}^{r}\frac{d^{n}}{dt_{j}^{n}}\right)e^{-N(t_{1}+\cdots+t_{r})}\sum_{\begin{subarray}{c}s_{11},\ldots,s_{1r},\\\
\cdots\cdots\cdots\\\
s_{k1},\ldots,s_{kr}=0\end{subarray}}^{n}\frac{\prod_{j=1}^{r}t_{j}^{\sum_{l=1}^{k}s_{lj}}}{\prod_{l=1}^{k}w_{l}^{\sum_{j=1}^{r}s_{lj}}}\Bigg{|}_{t_{1},\ldots,t_{r}=0}$
$\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}s_{11},\ldots,s_{1r},\\\
\cdots\cdots\cdots\\\
s_{k1},\ldots,s_{kr}=0\end{subarray}}^{n}\prod_{j=1}^{r}\left(\frac{d^{n}}{dt_{j}^{n}}e^{-Nt_{j}}t_{j}^{\sum_{l=1}^{k}s_{lj}}\right)\Bigg{|}_{t_{1},\ldots,t_{r}=0}\frac{1}{\prod_{l=1}^{k}w_{l}^{\sum_{j=1}^{r}s_{lj}}}$
$\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s_{lj}\leq n\\\
j=1,\ldots,r\end{subarray}}\prod_{j=1}^{r}\left((-N)^{n-\sum_{l=1}^{k}s_{lj}}\binom{n}{\sum_{l=1}^{k}s_{lj}}(\sum_{l=1}^{k}s_{lj})!\right)\frac{1}{\prod_{l=1}^{k}w_{l}^{\sum_{j=1}^{r}s_{lj}}}.$

Similarly,

$\left(\frac{d^{n}}{dt^{n}}\left(\prod_{l=1}^{k}(\sum_{i=0}^{n}\frac{t^{i}}{w_{l}^{i}})\right)\Bigg{|}_{t=0}\right)^{r}=\sum_{\begin{subarray}{c}\sum_{l=1}^{k}h_{lj}=n\\\
j=1,\ldots,r\end{subarray}}(n!)^{r}\frac{1}{\prod_{l=1}^{k}w_{l}^{\sum_{j=1}^{r}h_{lj}}}.$

Putting it all together, we obtain the claimed result. ∎

######  Lemma 11 (Equation (2.11) of [16]).

Let $n\geq 0$ be an integer, then

$\frac{1}{2\pi\i}\int_{|w|=1}\frac{e^{Lw+\frac{t}{w}}}{w^{n+1}}dw=\left(\frac{L}{t}\right)^{n/2}I_{n}(2\sqrt{Lt}).$

In our situation, we need to handle the expression (12). We give a formula for
it correlated with the modified Bessel functions of the first kind.

######  Lemma 12.

For integers $n\geq 1,k\geq 1$, we have

$\displaystyle\frac{1}{2\pi\i}\int_{|w|=1}\frac{e^{Lw+\frac{t_{1}}{w}+\cdots+\frac{t_{n}}{w^{n}}}}{w^{2k}}dw$
$\displaystyle=$
$\displaystyle\sum_{m_{2},\ldots,m_{n}=0}^{\infty}\,\left(\prod_{j=2}^{n}\,\frac{t_{j}^{m_{j}}}{m_{j}!}\right)\,I_{\sum_{j=2}^{n}jm_{j}+2k-1}(2\sqrt{Lt_{1}})\left(\frac{L}{t_{1}}\right)^{\frac{\sum_{j=2}^{n}jm_{j}+2k-1}{2}}.$

###### Proof.

We need to compute the coefficient of $w^{2k-1}$ in the Taylor expansion of
$e^{Lw+\frac{t_{1}}{w}+\cdots+\frac{t_{n}}{w^{n}}}$. Let $a_{m}(i)$ be the
coefficient of $w^{i}$ in the Taylor expansion of
$e^{Lw+\frac{t_{1}}{w}+\cdots+\frac{t_{m}}{w^{m}}}$ for $m=1,2,\ldots,n$, then
$a_{n}(2k-1)=\sum_{m_{n}=0}^{\infty}\frac{t_{n}^{m_{n}}}{m_{n}!}a_{n-1}(nm_{n}+2k-1)$.
So by Lemma 11,

$\displaystyle a_{n}(2k-1)$ $\displaystyle=$
$\displaystyle\sum_{m_{2},\ldots,m_{n}=0}^{\infty}\,\left(\prod_{j=2}^{n}\,\frac{t_{j}^{m_{j}}}{m_{j}!}\right)a_{1}(\sum_{j=2}^{n}jm_{j}+2k-1)$
$\displaystyle=$
$\displaystyle\sum_{m_{2},\ldots,m_{n}=0}^{\infty}\,\left(\prod_{j=2}^{n}\,\frac{t_{j}^{m_{j}}}{m_{j}!}\right)\,I_{\sum_{j=2}^{n}jm_{j}+2k-1}(2\sqrt{Lt_{1}})\left(\frac{L}{t_{1}}\right)^{\frac{\sum_{j=2}^{n}jm_{j}+2k-1}{2}}.$

∎

We now state a fact about higher order derivatives of determinants of
functions.

######  Lemma 13.

Let $s\geq 0$, $k\geq 1$ be integers and $a_{i,j}(x)$ be $s$-th differentiable
functions of $x$. Then

$\Big{(}\frac{d}{dx}\Big{)}^{s}\det_{k\times
k}(a_{i,j}(x))=\sum_{\begin{subarray}{c}l_{1}+\cdots+l_{k}=s\\\ l_{1}\geq
0,\ldots,l_{k}\geq 0\end{subarray}}\binom{s}{l_{1},\ldots,l_{k}}\det_{k\times
k}\Big{(}a_{i,j}^{(l_{i})}(x)\Big{)},$

where $a_{i,j}^{(l_{i})}(x)$ means that we take the $l_{i}$-th derivative of
$a_{i,j}(x)$.

###### Proof.

Expand the determinant as a sum over all permutations $\sigma$ of the numbers
$1,2,\ldots,k$:

$\det_{k\times k}(a_{i,j}(x))=\sum_{\sigma\in
S_{k}}\text{sign}(\sigma)\prod_{i=1}^{k}a_{i,\sigma(i)}(x).$

Then

$\displaystyle\Big{(}\frac{d}{dx}\Big{)}^{s}\det_{k\times k}(a_{i,j}(x))$
$\displaystyle=$ $\displaystyle\sum_{\sigma\in
S_{k}}\text{sign}(\sigma)\Big{(}\frac{d}{dx}\Big{)}^{s}\Big{(}\prod_{i=1}^{k}a_{i,\sigma(i)}(x)\Big{)}$
$\displaystyle=$ $\displaystyle\sum_{\sigma\in
S_{k}}\text{sign}(\sigma)\sum_{\begin{subarray}{c}l_{1}+\cdots+l_{k}=s\\\
l_{1}\geq 0,\ldots,l_{k}\geq
0\end{subarray}}\binom{s}{l_{1},\ldots,l_{k}}\prod_{i=1}^{k}(\frac{d}{dx})^{l_{i}}(a_{i,\sigma(i)}(x))$
$\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}l_{1}+\cdots+l_{k}=s\\\ l_{1}\geq
0,\ldots,l_{k}\geq
0\end{subarray}}\binom{s}{l_{1},\ldots,l_{k}}\sum_{\sigma\in
S_{k}}\text{sign}(\sigma)\prod_{i=1}^{k}(\frac{d}{dx})^{l_{i}}(a_{i,\sigma(i)}(x))$
$\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}l_{1}+\cdots+l_{k}=s\\\ l_{1}\geq
0,\ldots,l_{k}\geq 0\end{subarray}}\binom{s}{l_{1},\ldots,l_{k}}\det_{k\times
k}\Big{(}(\frac{d}{dx})^{l_{i}}(a_{i,j}(x))\Big{)}.$

∎

The following are two identities concerning $\Lambda_{A}^{(n)}(1)$ and
$Z_{A}^{(n)}(1)$. They ensure that the derivatives with respect to the shifts
at 0 give $|\Lambda_{A}^{(n)}(1)|^{2}$ and $|Z_{A}^{(n)}(1)|^{2}$.

######  Lemma 14.

For $n\geq 0$,

$\Lambda_{A^{*}}^{(n)}(1)=\overline{\Lambda_{A}^{(n)}(1)}$

and

$Z_{A^{*}}^{(n)}(1)=(-1)^{N}\overline{Z_{A}^{(n)}(1)}.$

###### Proof.

By the definition of $\Lambda_{A}(s)$, it is not hard to check that
$\Lambda_{A^{*}}^{(n)}(1)=\overline{\Lambda_{A}^{(n)}(1)}$. Since

$\displaystyle Z_{A}(s)$ $\displaystyle=$ $\displaystyle e^{-\pi\i
N/2}e^{\i\sum_{n=1}^{N}\theta_{n}/2}s^{-N/2}\Lambda_{A}(s),$ $\displaystyle
Z_{A^{*}}(s)$ $\displaystyle=$ $\displaystyle(-1)^{N}e^{\pi\i
N/2}e^{-\i\sum_{n=1}^{N}\theta_{n}/2}s^{-N/2}\Lambda_{A^{*}}(s).$

Combining with $\Lambda_{A^{*}}^{(n)}(1)=\overline{\Lambda_{A}^{(n)}(1)}$, we
have $Z_{A^{*}}^{(n)}(1)=(-1)^{N}\overline{Z_{A}^{(n)}(1)}$. ∎

We use $\Xi$ to denote the subset of permutations $\sigma\in S_{2k}$ of
$\\{1,2,\ldots,2k\\}$ for which
$\sigma(1)<\sigma(2)<\cdot\cdot\cdot<\sigma(k)$ and
$\sigma(k+1)<\sigma(k+2)<\cdot\cdot\cdot<\sigma(2k)$. The following lemma was
proved in section 2 of [10].

######  Lemma 15.

Assume that $\alpha_{1},\ldots,\alpha_{k}$ are distinct complex numbers. We
have

$\int_{{\mathbb{U}}(N)}\prod_{j=1}^{k}\Lambda_{A}(e^{-\alpha_{j}})\Lambda_{A^{*}}(e^{\alpha_{j+k}})dA_{N}=\sum_{\sigma\in\Xi}e^{N\sum_{j=1}^{k}(\alpha_{\sigma(j)}-\alpha_{j})}\prod_{1\leq
i,j\leq k}z(\alpha_{\sigma(j)}-\alpha_{\sigma(k+i)}),$

where $z(x)$ is defined in (12).

By the above lemma and the definition of $Z_{A}(s)$,

$\displaystyle\int_{{\mathbb{U}}(N)}\prod_{j=1}^{k}Z_{A}(e^{-\alpha_{j}})Z_{A^{*}}(e^{\alpha_{j+k}})dA_{N}$
$\displaystyle=$
$\displaystyle(-1)^{Nk}e^{-\frac{N}{2}\sum_{j=1}^{2k}\alpha_{j}}\sum_{\sigma\in\Xi}e^{N\sum_{j=1}^{k}\alpha_{\sigma(j)}}\prod_{1\leq
i,j\leq k}Z(\alpha_{\sigma(j)}-\alpha_{\sigma(k+i)})$

By the residue theorem, the summations in Lemma 15 and the above formula can
be expressed as integrals in the following way.

######  Lemma 16 (Lemma 3 of [16]).

$\displaystyle\int_{{\mathbb{U}}(N)}\prod_{j=1}^{k}\Lambda_{A}(e^{-\alpha_{j}})\Lambda_{A^{*}}(e^{\alpha_{j+k}})dA_{N}$
$\displaystyle=$
$\displaystyle\frac{1}{k!(2\pi\i)^{k}}\int_{|w_{i}|=1}e^{N\sum_{j=1}^{k}(w_{j}-\alpha_{j})}\prod_{\begin{subarray}{c}1\leq
i\leq k\\\ 1\leq j\leq 2k\end{subarray}}z(w_{i}-\alpha_{j})\prod_{i\neq
j}z(w_{i}-w_{j})^{-1}\prod_{j=1}^{k}dw_{j},$

and

$\displaystyle\int_{{\mathbb{U}}(N)}\prod_{j=1}^{k}Z_{A}(e^{-\alpha_{j}})Z_{A^{*}}(e^{\alpha_{j+k}})dA_{N}$
$\displaystyle=$
$\displaystyle(-1)^{Nk}\frac{e^{-\frac{N}{2}\sum_{j=1}^{2k}\alpha_{j}}}{k!(2\pi\i)^{k}}\int_{|w_{i}|=1}e^{N\sum_{j=1}^{k}w_{j}}\prod_{\begin{subarray}{c}1\leq
i\leq k\\\ 1\leq j\leq 2k\end{subarray}}z(w_{i}-\alpha_{j})\prod_{i\neq
j}z(w_{i}-w_{j})^{-1}\prod_{j=1}^{k}dw_{j}.$

######  Remark 17.

The integral contours $\\{w_{i}:|w_{i}|=1\\}$ can be replaced by any contours
enclosing the variables $\alpha_{i}$, by the residue theorem.

The following lemma is a non-symmetric version of [16, Lemma 5].

######  Lemma 18.

Let $f_{1},\ldots,f_{k}$ be $2k-2$ times differentiable functions. Let
$\Delta(\frac{d}{dL})$ be the differential operator

$\Delta(\frac{d}{dL}):=\prod_{1\leq i<j\leq
k}\Big{(}\frac{d}{dL_{j}}-\frac{d}{dL_{i}}\Big{)}=\det_{k\times
k}\Big{(}\frac{d^{j-1}}{dL_{i}^{j-1}}\Big{)}.$

Then

$\Delta^{2}(\frac{d}{dL})\prod_{i=1}^{k}f_{i}(L_{i})\Bigg{|}_{L_{i}=X}\\\
=\sum_{\sigma\in S_{k}}\det_{k\times k}(f_{\sigma(i)}^{(i+j-2)}(X)).$

In particular, when $f_{1}=\cdots=f_{k}=f$, we have

$\Delta^{2}(\frac{d}{dL})(\prod_{i=1}^{k}f(L_{i}))\Big{|}_{L_{i}=X}=k!\det_{k\times
k}(f^{(i+j-2)}(X)).$

###### Proof.

Firstly,

$\displaystyle\Delta(\frac{d}{dL})\prod_{i=1}^{k}f_{i}(L_{i})$
$\displaystyle=$ $\displaystyle\det_{k\times
k}\left(\frac{d^{j-1}}{dL_{i}^{j-1}}\right)\prod_{i=1}^{k}f_{i}(L_{i})$
$\displaystyle=$ $\displaystyle\sum_{\sigma\in S_{k}}{\rm
sign}(\sigma)\left(\frac{d}{dL_{1}}\right)^{\sigma(1)-1}\left(\frac{d}{dL_{2}}\right)^{\sigma(2)-1}\cdots\left(\frac{d}{dL_{k}}\right)^{\sigma(k)-1}\prod_{i=1}^{k}f_{i}(L_{i})$
$\displaystyle=$ $\displaystyle\sum_{\sigma\in S_{k}}{\rm
sign}(\sigma)\prod_{i=1}^{k}f_{i}^{(\sigma(i)-1)}(L_{i})$ $\displaystyle=$
$\displaystyle\det_{k\times k}(f_{i}^{(j-1)}(L_{i})).$

Secondly,

$\displaystyle\Delta^{2}(\frac{d}{dL})\prod_{i=1}^{k}f_{i}(L_{i})\Bigg{|}_{L_{i}=X}$
$\displaystyle=$ $\displaystyle\Delta(\frac{d}{dL})\det_{k\times
k}(f_{i}^{(j-1)}(L_{i}))\Bigg{|}_{L_{i}=X}$ $\displaystyle=$
$\displaystyle\Delta(\frac{d}{dL})\sum_{\sigma}{\rm
sign}(\sigma)\prod_{i=1}^{k}f_{i}^{(\sigma(i)-1)}(L_{i})\Bigg{|}_{L_{i}=X}$
$\displaystyle=$ $\displaystyle\sum_{\tau\in S_{k}}\sum_{\sigma\in S_{k}}{\rm
sign}(\sigma){\rm
sign}(\tau)\prod_{i=1}^{k}f_{i}^{(\sigma(i)-1+\tau(i)-1)}(L_{i})\Bigg{|}_{L_{i}=X}$
$\displaystyle=$ $\displaystyle\sum_{\sigma\in S_{k}}{\rm
sign}(\sigma)\det_{k\times k}(f_{i}^{(\sigma(i)+j-2)}(X))$ $\displaystyle=$
$\displaystyle\sum_{\sigma\in S_{k}}\det_{k\times
k}(f_{\sigma(i)}^{(i+j-2)}(X)).$

∎

The following proposition will be used in the proofs of our main theorems 25
and 27.

######  Proposition 19.

Let $k\geq 1$, $Q_{l}\geq 0$ be integers for $l=1,\dots,k$. Then we have

$\displaystyle\frac{1}{(2\pi\i)^{k}}\int_{|w_{i}|=1}\frac{e^{\sum_{j=1}^{k}Nw_{j}}\Delta^{2}(w)}{\prod_{l=1}^{k}w_{l}^{2k+Q_{l}}}\prod_{j=1}^{k}dw_{j}=\sum_{\sigma\in
S_{k}}\det_{k\times
k}\left(\frac{N^{2k+Q_{\sigma(i)}-1-(i+j-2)}}{(2k+Q_{\sigma(i)}-1-(i+j-2))!}\right).$

###### Proof.

Note that

$\displaystyle\frac{1}{(2\pi\i)^{k}}\int_{|w_{i}|=1}\frac{e^{\sum_{j=1}^{k}Nw_{j}}\Delta^{2}(w)}{\prod_{l=1}^{k}w_{l}^{2k+Q_{l}}}\prod_{j=1}^{k}dw_{j}=\frac{1}{(2\pi\i)^{k}}\Delta^{2}(\frac{d}{dL})\int_{|w_{i}|=1}\frac{e^{\sum_{j=1}^{k}w_{j}L_{j}}}{\prod_{l=1}^{k}w_{l}^{2k+Q_{l}}}\prod_{j=1}^{k}dw_{j}\Bigg{|}_{L_{j}=N}.$
(17)

Let

$f_{j}(L_{j}):=\frac{1}{2\pi\i}\int_{|w|=1}\frac{e^{L_{j}w}}{w^{2k+Q_{j}}}dw.$

Then

$(\ref{eq6})=\Delta^{2}(\frac{d}{dL})\prod_{j=1}^{k}f_{j}(L_{j})\Bigg{|}_{L_{j}=N}.$

By Lemma 18,

$\displaystyle(\ref{eq6})=\sum_{\sigma\in S_{k}}\det_{k\times
k}(f_{\sigma(i)}^{(i+j-2)}(N)).$

Note that $f_{j}(L_{j})$ is the coefficient of $w^{2k+Q_{j}-1}$ in the Taylor
expansion of $e^{L_{j}w}$, namely

$f_{i}(L_{i})=\frac{(L_{i})^{2k+Q_{i}-1}}{(2k+Q_{i}-1)!}.$

So

$\displaystyle(\ref{eq6})=\sum_{\sigma\in S_{k}}\det_{k\times
k}\left(\frac{N^{2k+Q_{\sigma(i)}-1-(i+j-2)}}{(2k+Q_{\sigma(i)}-1-(i+j-2))!}\right).$

∎

The following lemma follows from the proof of [16, Theorem 3, see
(4.39)-(4.41)].

######  Lemma 20.

$\det_{k\times
k}\Big{(}\frac{1}{(2k+1+m_{i}-i-j)!}\Big{)}=\prod_{i=1}^{k}\frac{1}{(2k-i+m_{i})!}\prod_{1\leq
i<j\leq k}(m_{j}-m_{i}-j+i).$

##  3 Main results

In the following, we first calculate the integrals

$\displaystyle\frac{1}{(2\pi\i)^{k}k!}\int_{|w_{i}|=1}\frac{e^{N\sum_{i=1}^{k}w_{i}}}{\prod_{i=1}^{k}w_{i}^{2k}}\Delta^{2}(w)\left(\sum_{l=1}^{k}\frac{1}{w_{l}}-\frac{N}{2}\right)^{m_{1}}\prod_{j=2}^{n}\left(\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{m_{j}}\prod_{i=1}^{k}dw_{i}$
(18)

and

$\frac{1}{(2\pi\i)^{k}k!}\int_{|w_{i}|=1}\frac{e^{N\sum_{i=1}^{k}w_{i}}}{\prod_{i=1}^{k}w_{i}^{2k}}\Delta^{2}(w)\left(\sum_{l=1}^{k}\frac{1}{w_{l}}-N\right)^{m_{1}}\prod_{j=2}^{n}\left(\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{m_{j}}\prod_{i=1}^{k}dw_{i}$
(19)

######  Proposition 21.

Let $(m_{1},\ldots,m_{n})$ be a tuple of non-negative integers, then

$\displaystyle(\ref{eq3})$ $\displaystyle=$ $\displaystyle
N^{k^{2}+\sum_{s=1}^{n}sm_{s}}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}m_{sl}=m_{s}\\\
s=2,\ldots,n\end{subarray}}\left(\,\prod_{s=2}^{n}\binom{m_{s}}{m_{s1},\ldots,m_{sk}}\right)\frac{d^{m_{1}}}{dx^{m_{1}}}\Bigg{(}e^{-\frac{x}{2}}x^{-\frac{k^{2}}{2}-\frac{1}{2}\sum_{s=2}^{n}sm_{s}}$
$\displaystyle\times\,\det_{k\times
k}\left(I_{\sum_{s=2}^{n}sm_{si}+i+j-1}(2\sqrt{x})\right)\Bigg{)}\Bigg{|}_{x=0}.$

More explicitly,

$\displaystyle(\ref{eq3})$ $\displaystyle=$ $\displaystyle
N^{k^{2}+\sum_{s=1}^{n}sm_{s}}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}m_{sl}=m_{s}\\\
s=2,\ldots,n\\\
\sum_{l=0}^{k}m_{1l}=m_{1}\end{subarray}}(-\frac{1}{2})^{m_{10}}\left(\,\prod_{s=2}^{n}\binom{m_{s}}{m_{s1},\ldots,m_{sk}}\right)\binom{m_{1}}{m_{10},\ldots,m_{1k}}$
$\displaystyle\times\,\left(\prod_{i=1}^{k}\frac{1}{(2k+\sum_{s=1}^{n}sm_{si}-i)!}\right)\prod_{1\leq
i<j\leq k}\left(\sum_{s=1}^{n}sm_{sj}-\sum_{s=1}^{n}sm_{si}-j+i\right).$

###### Proof.

In the following, we write $t=(t_{1},\ldots,t_{n})$ and when we say $t=0$ we
mean $t_{1}=\cdots=t_{n}=0$. We rewrite

$\displaystyle(\ref{eq3})=\frac{1}{(2\pi\i)^{k}k!}\prod_{i=1}^{n}\left(\frac{d}{dt_{i}}\right)^{m_{i}}e^{-\frac{N}{2}t_{1}}\Delta^{2}(\frac{d}{dL})\int_{|w_{i}|=1}\frac{e^{\sum_{i=1}^{k}(L_{i}w_{i}+\frac{t_{1}}{w_{i}}+\frac{t_{2}}{w_{i}^{2}}+\cdots+\frac{t_{n}}{w_{i}^{n}})}}{\prod_{i=1}^{k}w_{i}^{2k}}\prod_{i=1}^{k}dw_{i}\Bigg{|}_{L_{i}=N,t=0}.$

Let

$f(x)=\frac{1}{2\pi\i}\int_{|w|=1}\frac{e^{xw+\sum_{j=1}^{n}\frac{t_{j}}{w^{j}}}}{w^{2k}}dw.$

By Lemma 18, we have

$\displaystyle(\ref{eq3})=\prod_{i=1}^{n}\left(\frac{d}{dt_{i}}\right)^{m_{i}}e^{-\frac{N}{2}t_{1}}\det_{k\times
k}(f^{(i+j-2)}(N))\Bigg{|}_{t=0}.$

By Lemma 12,

$f(L)=\sum_{l_{2},\ldots,l_{n}=0}^{\infty}\,\left(\prod_{j=2}^{n}\,\frac{t_{j}^{l_{j}}}{l_{j}!}\right)\,\sum_{l_{1}=0}^{\infty}\frac{L^{l_{1}}t_{1}^{l_{1}}L^{\sum_{j=2}^{n}jl_{j}+2k-1}}{l_{1}!(\sum_{j=2}^{n}jl_{j}+2k-1+l_{1})!}.$

Since $i+j-2\leq 2k-2$, we have

$f^{(i+j-2)}(N)=\sum_{l_{2},\ldots,l_{n}=0}^{\infty}\,\left(\prod_{j=2}^{n}\,\frac{t_{j}^{l_{j}}}{l_{j}!}\right)\,\sum_{l_{1}=0}^{\infty}\frac{N^{l_{1}}t_{1}^{l_{1}}N^{\sum_{j=2}^{n}jl_{j}+2k-1-(i+j-2)}}{l_{1}!(\sum_{j=2}^{n}jl_{j}+2k-1+l_{1}-(i+j-2))!}.$

Since $\det(a_{i,j})=\det(a_{k+1-i,k+1-j})$,

$\displaystyle(\ref{eq3})=\prod_{l=1}^{n}\left(\frac{d}{dt_{l}}\right)^{m_{l}}e^{-\frac{N}{2}t_{1}}\det_{k\times
k}\left(\sum_{l_{2},\ldots,l_{n}=0}^{\infty}\,\left(\prod_{s=2}^{n}\,\frac{t_{s}^{l_{s}}}{l_{s}!}\right)I_{\sum_{s=2}^{n}sl_{s}+i+j-1}(2\sqrt{Nt_{1}})\left(\frac{N}{t_{1}}\right)^{\frac{\sum_{s=2}^{n}sl_{s}+i+j-1}{2}}\right)\Bigg{|}_{t=0}.$

By Lemma 13, we further obtain

$\displaystyle(\ref{eq3})=$ $\displaystyle\sum_{\begin{subarray}{c}m_{sl}\geq
0\\\ \sum_{l=1}^{k}m_{sl}=m_{s}\\\
s=2,\ldots,n\end{subarray}}\left(\prod_{s=2}^{n}\binom{m_{s}}{m_{s1},\ldots,m_{sk}}\right)\frac{d^{m_{1}}}{dt_{1}^{m_{1}}}\Bigg{(}e^{-\frac{N}{2}t_{1}}$
$\displaystyle\times\,\det_{k\times
k}\left(I_{\sum_{s=2}^{n}sm_{si}+i+j-1}(2\sqrt{Nt_{1}})\left(\frac{N}{t_{1}}\right)^{\frac{\sum_{s=2}^{n}sm_{si}+i+j-1}{2}}\right)\Bigg{)}\Bigg{|}_{t_{1}=0}.$

Let $Nt_{1}=x$, then

$\displaystyle(\ref{eq3})$ $\displaystyle=$ $\displaystyle
N^{k^{2}+\sum_{s=1}^{n}sm_{s}}\sum_{\begin{subarray}{c}m_{sl}\geq 0\\\
\sum_{l=1}^{k}m_{sl}=m_{s}\\\
s=2,\ldots,n\end{subarray}}\left(\prod_{s=2}^{n}\binom{m_{s}}{m_{s1},\ldots,m_{sk}}\right)\frac{d^{m_{1}}}{dx^{m_{1}}}\Bigg{(}e^{-\frac{x}{2}}x^{-\frac{k^{2}}{2}-\frac{1}{2}\sum_{s=2}^{n}sm_{s}}$
$\displaystyle\times\,\det_{k\times
k}\left(I_{\sum_{s=2}^{n}sm_{si}+i+j-1}(2\sqrt{x})\right)\Bigg{)}\Bigg{|}_{x=0}.$

Concerning the explicit formula of (18), note that

$\displaystyle\frac{d^{m_{1}}}{dx^{m_{1}}}\Bigg{(}e^{-\frac{x}{2}}x^{-\frac{k^{2}}{2}-\frac{1}{2}\sum_{s=2}^{n}sm_{s}}\det_{k\times
k}\left(I_{\sum_{s=2}^{n}sm_{si}+i+j-1}(2\sqrt{x})\right)\Bigg{)}\Bigg{|}_{x=0}$
$\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}m_{10},\ldots,m_{1k}\geq 0\\\
m_{10}+\ldots+m_{1k}=m_{1}\end{subarray}}\binom{m_{1}}{m_{10},\ldots,m_{1k}}(-\frac{1}{2})^{m_{10}}\det_{k\times
k}\left(\left(\sum_{l=0}^{\infty}\frac{x^{l}}{l!(\sum_{s=2}^{n}sm_{si}+i+j-1+l)!}\right)^{(m_{1i})}\right)\Bigg{|}_{x=0}$
$\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}m_{10},\ldots,m_{1k}\geq 0\\\
m_{10}+\ldots+m_{1k}=m_{1}\end{subarray}}\binom{m_{1}}{m_{10},\ldots,m_{1k}}(-\frac{1}{2})^{m_{10}}\det_{k\times
k}\left(\frac{1}{(\sum_{s=1}^{n}sm_{si}+i+j-1)!}\right).$

We next use Lemma 20 to finish this proof. To use Lemma 20, we first change
the variables $i\rightarrow k+1-i$, $j\rightarrow k+1-j$ so $m_{si}$ becomes
$\tilde{m}_{si}:=m_{s(k+1-i)}$. The summation above over $m_{si}$ has a
symmetric property in the sense that
$\sum_{i=1}^{k}m_{si}=\sum_{i=1}^{k}\tilde{m}_{si}$. By this observation and
Lemma 20, we obtain the claim in this proposition. ∎

A similar argument to that of Proposition 21 leads to the following result for
(19).

######  Proposition 22.

For any tuple $(m_{1},\ldots,m_{n})$ of non-negative integers, we have

$\displaystyle(\ref{519eq4})$ $\displaystyle=$ $\displaystyle
N^{k^{2}+\sum_{s=1}^{n}sm_{s}}\sum_{\begin{subarray}{c}m_{sl}\geq 0\\\
\sum_{l=1}^{k}m_{sl}=m_{s}\\\
s=2,\ldots,n\end{subarray}}\prod_{s=2}^{n}\binom{m_{s}}{m_{s1},\ldots,m_{sk}}\left(\frac{d}{dx}\right)^{m_{1}}\Bigg{(}e^{-x}x^{-\frac{k^{2}}{2}-\frac{1}{2}\sum_{s=2}^{n}sm_{s}}$
$\displaystyle\times\,\det_{k\times
k}\left(I_{\sum_{s=2}^{n}sm_{si}+i+j-1}(2\sqrt{x})\right)\Bigg{)}\Bigg{|}_{x=0}.$

More explicitly,

$\displaystyle(\ref{519eq4})$ $\displaystyle=$ $\displaystyle
N^{k^{2}+\sum_{s=1}^{n}sm_{s}}\sum_{\begin{subarray}{c}s=2,\ldots,n\\\
\sum_{l=1}^{k}m_{sl}=m_{s}\\\
\sum_{l=0}^{k}m_{1l}=m_{1}\end{subarray}}(-1)^{m_{10}}\left(\,\prod_{s=2}^{n}\binom{m_{s}}{m_{s1},\ldots,m_{sk}}\right)\binom{m_{1}}{m_{10},\ldots,m_{1k}}$
$\displaystyle\times\,\left(\prod_{i=1}^{k}\frac{1}{(2k+\sum_{s=1}^{n}sm_{si}-i)!}\right)\prod_{1\leq
i<j\leq k}\left(\sum_{s=1}^{n}sm_{sj}-\sum_{s=1}^{n}sm_{si}-j+i\right).$

The following is our first main theorem. It concerns joint moments of the
$n_{1}$-th and $n_{2}$-th derivatives of characteristic polynomials of CUE
matrices. Theorems 1 and 2 follow directly by setting $n_{1}=2,n_{2}=0$ and
switching the roles of $k-M$ and $M$.

######  Theorem 23.

For $k\geq 1$, $0\leq M\leq k$ and $n_{1}\geq n_{2}\geq 0$ be integers. Let
$P=\\#\\{{\bf
a}_{i}=(a_{i1},\ldots,a_{in_{1}}):\sum_{j=1}^{n_{1}}ja_{ij}=n_{1},a_{ij}\geq
0\\}$ and $Q=\\#\\{{\bf
b}_{i}=(b_{i1},\ldots,b_{in_{2}}):\sum_{j=1}^{n_{2}}jb_{ij}=n_{2},b_{ij}\geq
0\\}$. Denote by ${\bf a}_{i}!=\prod_{j=1}^{n_{1}}a_{ij}!$ and ${\bf
b}_{i}!=\prod_{j=1}^{n_{2}}b_{ij}!$. Then we have

$\displaystyle\int_{{\mathbb{U}}(N)}|\Lambda_{A}^{(n_{1})}(1)|^{2k-2M}|\Lambda_{A}^{(n_{2})}(1)|^{2M}dA_{N}$
$\displaystyle=$ $\displaystyle
N^{k^{2}+2(k-M)n_{1}+2Mn_{2}}(-1)^{(k-M)n_{1}+Mn_{2}+\frac{k(k-1)}{2}}$
$\displaystyle\times\,\sum_{\begin{subarray}{c}l_{1}+\cdots+l_{P}=k-M\\\
l^{\prime}_{1}+\cdots+l^{\prime}_{P}=k-M\end{subarray}}\sum_{t_{1}=0}^{\sum_{i=1}^{P}l^{\prime}_{i}a_{i1}}\binom{k-M}{l_{1},\ldots,l_{P}}\binom{k-M}{l^{\prime}_{1},\ldots,l^{\prime}_{P}}\binom{\sum_{i=1}^{P}l^{\prime}_{i}a_{i1}}{t_{1}}\frac{(n_{1}!)^{2k-2M}}{\prod_{i=1}^{P}({\bf
a}_{i}!\prod_{j=1}^{n_{1}}j^{a_{ij}})^{l_{i}+l^{\prime}_{i}}}$
$\displaystyle\times\,\sum_{\begin{subarray}{c}u_{1}+\cdots+u_{Q}=M\\\
u^{\prime}_{1}+\cdots+u^{\prime}_{Q}=M\end{subarray}}\sum_{t_{2}=0}^{\sum_{i=1}^{Q}u^{\prime}_{i}b_{i1}}\binom{M}{u_{1},\ldots,u_{Q}}\binom{M}{u^{\prime}_{1},\ldots,u^{\prime}_{Q}}\binom{\sum_{i=1}^{Q}u^{\prime}_{i}b_{i1}}{t_{2}}\frac{(n_{2}!)^{2M}}{\prod_{i=1}^{Q}({\bf
b}_{i}!\prod_{j=1}^{n_{2}}j^{b_{ij}})^{u_{i}+u^{\prime}_{i}}}$
$\displaystyle\times\sum_{\begin{subarray}{c}s=2,\ldots,n_{1}\\\
\sum_{i=1}^{k}h_{si}=V_{s}\end{subarray}}\prod_{s=2}^{n_{1}}\binom{V_{s}}{h_{s1},\ldots,h_{sk}}\left(\frac{d}{dx}\right)^{V_{1}}\Bigg{(}e^{-x}x^{-\frac{k^{2}}{2}-(k-M)n_{1}-Mn_{2}+\frac{1}{2}\sum_{i=1}^{P}(l_{i}+l^{\prime}_{i})a_{i1}+\frac{1}{2}\sum_{i=1}^{Q}(u_{i}+u^{\prime}_{i})b_{i1}}$
$\displaystyle\times\,\det_{k\times
k}\Big{(}I_{\sum_{s=2}^{n_{1}}sh_{si}+i+j-1}(2\sqrt{x})\Big{)}\Bigg{)}\Bigg{|}_{x=0}+O(N^{k^{2}+2(k-M)n_{1}+2Mn_{2}-1}),$

where
$V_{1}=\sum_{i=1}^{P}(l_{i}+l_{i}^{\prime})a_{i1}+\sum_{i=1}^{Q}(u_{i}+u^{\prime}_{i})b_{i1}-t_{1}-t_{2}$,
$V_{j}=\sum_{i=1}^{P}(l_{i}+l^{\prime}_{i})a_{ij}+\sum_{i=1}^{Q}(u_{i}+u^{\prime}_{i})b_{ij}$
for $j=2,\ldots,n_{2}$, and $V_{j}=\sum_{i=1}^{P}(l_{i}+l^{\prime}_{i})a_{ij}$
for $j=\max(2,n_{2}+1),\ldots,n_{1}$.

Moreover,

$\displaystyle\sum_{\begin{subarray}{c}s=2,\ldots,n_{1}\\\
\sum_{i=1}^{k}h_{si}=V_{s}\end{subarray}}\prod_{s=2}^{n_{1}}\binom{V_{s}}{h_{s1},\ldots,h_{sk}}\left(\frac{d}{dx}\right)^{V_{1}}\Bigg{(}e^{-x}x^{-\frac{k^{2}}{2}-(k-M)n_{1}-Mn_{2}+\frac{1}{2}\sum_{i=1}^{P}(l_{i}+l^{\prime}_{i})a_{i1}+\frac{1}{2}\sum_{i=1}^{Q}(u_{i}+u^{\prime}_{i})b_{i1}}$
$\displaystyle\times\,\det_{k\times
k}\Big{(}I_{\sum_{s=2}^{n_{1}}sh_{si}+i+j-1}(2\sqrt{x})\Big{)}\Bigg{)}\Bigg{|}_{x=0}$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}s=2,\ldots,n_{1}\\\
\sum_{i=1}^{k}h_{si}=V_{s}\\\
\sum_{i=0}^{k}h_{1i}=V_{1}\end{subarray}}(-1)^{h_{10}}\Bigg{(}\prod_{s=2}^{n_{1}}\binom{V_{s}}{h_{s1},\ldots,h_{sk}}\Bigg{)}\binom{V_{1}}{h_{10},\ldots,h_{1k}}\left(\prod_{i=1}^{k}\frac{1}{(2k+\sum_{s=1}^{n_{1}}sh_{si}-i)!}\right)$
$\displaystyle\times\,\prod_{1\leq i<j\leq
k}\left(\sum_{s=1}^{n_{1}}sh_{sj}-\sum_{s=1}^{n_{1}}sh_{si}-j+i\right).$

###### Proof.

Let $\alpha=(\alpha_{1},\ldots,\alpha_{2k})$. By Lemmas 14 and 16,

$\displaystyle\int_{{\mathbb{U}}(N)}|\Lambda_{A}^{(n_{1})}(1)|^{2k-2M}|\Lambda_{A}^{(n_{2})}(1)|^{2M}dA_{N}$
$\displaystyle=$
$\displaystyle(-1)^{n_{1}(k-M)+Mn_{2}}\frac{1}{k!(2\pi\i)^{k}}\prod_{j=1}^{k-M}(\frac{d}{d\alpha_{j}}\frac{d}{d\alpha_{k+j}})^{n_{1}}\prod_{j=k-M+1}^{k}(\frac{d}{d\alpha_{j}}\frac{d}{d\alpha_{k+j}})^{n_{2}}$
$\displaystyle\times\,\int_{|w_{i}|=1}e^{N\sum_{j=1}^{k}(w_{j}-\alpha_{j})}\prod_{i\neq
j}z(w_{i}-w_{j})^{-1}\prod_{\begin{subarray}{c}1\leq i\leq k\\\ 1\leq j\leq
2k\end{subarray}}z(w_{i}-\alpha_{j})\prod_{j=1}^{k}dw_{j}\Bigg{|}_{\alpha=0}$
$\displaystyle+\,O(N^{k^{2}+(2k-2M)n_{1}+2Mn_{2}-1}).$

Suppose that $\alpha_{j}=\alpha_{j}(N)$ and $|\alpha_{j}|\leq 1/N$ for
$j=1,2,\ldots,2k$ above, we have furthermore that the last expression is

$\displaystyle=$
$\displaystyle(-1)^{n_{1}(k-M)+n_{2}M+\frac{k(k-1)}{2}}\frac{1}{k!(2\pi\i)^{k}}\prod_{j=1}^{k-M}\left(\frac{d}{d\alpha_{j}}\frac{d}{d\alpha_{k+j}}\right)^{n_{1}}\prod_{j=k-M+1}^{k}\left(\frac{d}{d\alpha_{j}}\frac{d}{d\alpha_{k+j}}\right)^{n_{2}}$
$\displaystyle\times\,\int_{|w_{i}|=1}\frac{e^{N\sum_{j=1}^{k}(w_{j}-\alpha_{j})}\Delta^{2}(w)}{\prod_{j=1}^{2k}\prod_{i=1}^{k}(w_{i}-\alpha_{j})}\prod_{j=1}^{k}dw_{j}\Bigg{|}_{\alpha=0}+\,O(N^{k^{2}+(2k-2M)n_{1}+2Mn_{2}-1}).$

By Lemma 9,

$\displaystyle\prod_{j=1}^{k-M}\left(\frac{d}{d\alpha_{j}}\frac{d}{d\alpha_{k+j}}\right)^{n_{1}}\prod_{j=k-M+1}^{k}(\frac{d}{d\alpha_{j}}\frac{d}{d\alpha_{k+j}})^{n_{2}}\int_{|w_{i}|=1}\frac{e^{N\sum_{j=1}^{k}w_{j}}\Delta^{2}(w)}{\prod_{j=1}^{2k}\prod_{i=1}^{k}(w_{i}-\alpha_{j})}e^{-N\sum_{j=1}^{k}\alpha_{j}}\prod_{j=1}^{k}dw_{j}\Bigg{|}_{\alpha=0}$
$\displaystyle=$
$\displaystyle\int_{|w_{i}|=1}\frac{e^{N\sum_{j=1}^{k}w_{j}}\Delta^{2}(w)}{\prod_{i=1}^{k}w_{i}^{2k}}\Bigg{(}\sum_{m_{1}+2m_{2}+\cdots+n_{1}m_{n_{1}}=n_{1}}\frac{n_{1}!}{m_{1}!\cdots
m_{n_{1}}!}\left(-N+\sum_{l=1}^{k}\frac{1}{w_{l}}\right)^{m_{1}}$ (20)
$\displaystyle\times\,\prod_{j=2}^{n_{1}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{m_{j}}\Bigg{)}^{k-M}\Bigg{(}\sum_{r_{1}+2r_{2}+\cdots+n_{2}r_{n_{2}}=n_{2}}\frac{n_{2}!}{r_{1}!\cdots
r_{n_{2}}!}\left(-N+\sum_{l=1}^{k}\frac{1}{w_{l}}\right)^{r_{1}}$
$\displaystyle\times\,\prod_{j=2}^{n_{2}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{r_{j}}\Bigg{)}^{M}\Bigg{(}\sum_{m_{1}+2m_{2}+\cdots+n_{1}m_{n_{1}}=n_{1}}\frac{n_{1}!}{m_{1}!\cdots
m_{n_{1}}!}\prod_{j=1}^{n_{1}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{m_{j}}\Bigg{)}^{k-M}$
$\displaystyle\times\,\Bigg{(}\sum_{r_{1}+2r_{2}+\cdots+n_{2}r_{n_{2}}=n_{2}}\frac{n_{2}!}{r_{1}!\cdots
r_{n_{2}}!}\prod_{j=1}^{n_{2}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{r_{j}}\Bigg{)}^{M}\prod_{j=1}^{k}dw_{j}.$

Expanding all the powers involved, we obtain

$\displaystyle=$
$\displaystyle\int_{|w_{i}|=1}\frac{e^{N\sum_{j=1}^{k}w_{j}}\Delta^{2}(w)}{\prod_{i=1}^{k}w_{i}^{2k}}\Bigg{(}\sum_{l_{1}+\cdots+l_{P}=k-M}\binom{k-M}{l_{1},\ldots,l_{P}}\frac{(n_{1}!)^{k-M}}{\prod_{i=1}^{P}({\bf
a}_{i}!)^{l_{i}}}\left(-N+\sum_{l=1}^{k}\frac{1}{w_{l}}\right)^{\sum_{i=1}^{P}l_{i}a_{i1}}$
(21)
$\displaystyle\times\,\prod_{j=2}^{n_{1}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{\sum_{i=1}^{P}l_{i}a_{ij}}\Bigg{)}\Bigg{(}\sum_{u_{1}+\cdots+u_{Q}=M}\binom{M}{u_{1},\ldots,u_{Q}}\frac{(n_{2}!)^{M}}{\prod_{i=1}^{Q}({\bf
b}_{i}!)^{u_{i}}}\left(-N+\sum_{l=1}^{k}\frac{1}{w_{l}}\right)^{\sum_{i=1}^{Q}u_{i}b_{i1}}$
$\displaystyle\times\,\prod_{j=2}^{n_{2}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{\sum_{i=1}^{Q}u_{i}b_{ij}}\Bigg{)}\Bigg{(}\sum_{l^{\prime}_{1}+\cdots+l^{\prime}_{P}=k-M}\binom{k-M}{l^{\prime}_{1},\ldots,l^{\prime}_{P}}\frac{(n_{1}!)^{k-M}}{\prod_{i=1}^{P}({\bf
a}_{i}!)^{l^{\prime}_{i}}}\prod_{j=1}^{n_{1}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{\sum_{i=1}^{P}l^{\prime}_{i}a_{ij}}\Bigg{)}$
$\displaystyle\times\,\Bigg{(}\sum_{u^{\prime}_{1}+\cdots+u^{\prime}_{Q}=M}\binom{M}{u^{\prime}_{1},\ldots,u^{\prime}_{Q}}\frac{(n_{2}!)^{M}}{\prod_{i=1}^{Q}({\bf
b}_{i}!)^{u^{\prime}_{i}}}\prod_{j=1}^{n_{2}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{\sum_{i=1}^{Q}u^{\prime}_{i}b_{ij}}\Bigg{)}\prod_{j=1}^{k}dw_{j}.$

Next, we expand
$\left(\sum_{l=1}^{k}\frac{1}{w_{l}}\right)^{\sum_{i=1}^{P}l^{\prime}_{i}a_{i1}}$
and
$\left(\sum_{l=1}^{k}\frac{1}{w_{l}}\right)^{\sum_{i=1}^{Q}u^{\prime}_{i}b_{i1}}$
by viewing $\sum_{l=1}^{k}\frac{1}{w_{l}}-N$ as a whole. We then simplify the
expression by viewing $\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}$ as a whole for
$j=2,\ldots,n_{1}$, and obtain

$\displaystyle(\ref{516night1})$ $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}l_{1}+\cdots+l_{P}=k-M\\\
l^{\prime}_{1}+\cdots+l^{\prime}_{P}=k-M\end{subarray}}\binom{k-M}{l_{1},\ldots,l_{P}}\binom{k-M}{l^{\prime}_{1},\ldots,l^{\prime}_{P}}\frac{(n_{1}!)^{2k-2M}}{\prod_{i=1}^{P}(\prod_{j=1}^{n_{1}}a_{ij}!j^{a_{ij}})^{l_{i}+l^{\prime}_{i}}}\sum_{t_{1}=0}^{\sum_{i=1}^{P}l^{\prime}_{i}a_{i1}}\binom{\sum_{i=1}^{P}l^{\prime}_{i}a_{i1}}{t_{1}}N^{t_{1}}$
$\displaystyle\times\,\sum_{\begin{subarray}{c}u_{1}+\cdots+u_{Q}=M\\\
u^{\prime}_{1}+\cdots+u^{\prime}_{Q}=M\end{subarray}}\binom{M}{u_{1},\ldots,u_{Q}}\binom{M}{u^{\prime}_{1},\ldots,u^{\prime}_{Q}}\frac{(n_{2}!)^{2M}}{\prod_{i=1}^{Q}(\prod_{j=1}^{n_{2}}b_{ij}!j^{b_{ij}})^{u_{i}+u^{\prime}_{i}}}\sum_{t_{2}=0}^{\sum_{i=1}^{Q}u^{\prime}_{i}b_{i1}}\binom{\sum_{i=1}^{Q}u^{\prime}_{i}b_{i1}}{t_{2}}N^{t_{2}}$
$\displaystyle\times\,\int_{|w_{i}|=1}\frac{e^{N\sum_{j=1}^{k}w_{j}}\Delta^{2}(w)}{\prod_{i=1}^{k}w_{i}^{2k}}(\sum_{l=1}^{k}\frac{1}{w_{l}}-N)^{V_{1}}\prod_{j=2}^{n_{1}}\left(\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{V_{j}}\prod_{j=1}^{k}dw_{j},$

where the quantities $V_{j}$ are defined in the theorem. Finally, we apply
Proposition 22 to the last integral with $m_{j}=V_{j}$ for $j=1,\ldots,n_{1}$.
In this process, we can compute that
$\sum_{s=2}^{n_{1}}sV_{s}=\sum_{s=1}^{n_{1}}sV_{s}-V_{1}=2(k-M)n_{1}+2Mn_{2}-\sum_{i=1}^{P}(l_{i}+l^{\prime}_{i})a_{i1}-\sum_{i=1}^{Q}(u_{i}+u^{\prime}_{i})b_{i1}$.
Hence, we obtain the claimed result. ∎

We present below our main theorem on joint moments of the $n_{1}$-th and
$n_{2}$-th derivatives of the analogue of Hardy’s $Z$-function. Theorems 3 and
4 are obtained by choosing $n_{1}=2,n_{2}=0$ and switching $n-M$ and $M$.

######  Theorem 24.

For $k\geq 1$, $0\leq M\leq k$ and $n_{1}\geq n_{2}\geq 0$ be integers. Let
$P=\\#\\{{\bf
a}_{i}=(a_{i1},\ldots,a_{in_{1}}):\sum_{j=1}^{n_{1}}ja_{ij}=n_{1},a_{ij}\geq
0\\}$ and $Q=\\#\\{{\bf
b}_{i}=(b_{i1},\ldots,b_{in_{2}}):\sum_{j=1}^{n_{2}}jb_{ij}=n_{2},b_{ij}\geq
0\\}$. Denote by ${\bf a}_{i}!=\prod_{j=1}^{n_{1}}a_{ij}!$ and ${\bf
b}_{i}!=\prod_{j=1}^{n_{2}}b_{ij}!$. Then we have

$\displaystyle\int_{{\mathbb{U}}(N)}|Z_{A}^{(n_{1})}(1)|^{2k-2M}|Z_{A}^{(n_{2})}(1)|^{2M}dA_{N}$
$\displaystyle=$
$\displaystyle(-1)^{n_{1}(k-M)+n_{2}M+\frac{k(k-1)}{2}}N^{k^{2}+2n_{1}(k-M)+2n_{2}M}$
$\displaystyle\times\,\sum_{\begin{subarray}{c}l_{1}+\cdots+l_{P}=2k-2M\\\
u_{1}+\cdots+u_{Q}=2M\end{subarray}}\binom{2k-2M}{l_{1},\ldots,l_{P}}\binom{2M}{u_{1},\ldots,u_{Q}}\frac{(n_{1}!)^{2k-2M}}{(\prod_{i=1}^{P}({\bf
a}_{i}!)^{l_{i}})(\prod_{j=1}^{n_{1}}j^{\sum_{i=1}^{P}l_{i}a_{ij}})}$
$\displaystyle\times\,\frac{(n_{2}!)^{2M}}{(\prod_{i=1}^{Q}({\bf
b}_{i}!)^{u_{i}})(\prod_{j=1}^{n_{2}}j^{\sum_{i=1}^{Q}u_{i}b_{ij}})}\sum_{\begin{subarray}{c}s=2,\ldots,n_{1}\\\
\sum_{i=1}^{k}h_{si}=V_{s}\end{subarray}}\prod_{s=2}^{n_{1}}\binom{V_{s}}{h_{s1},\ldots,h_{sk}}$
$\displaystyle\times\,\left(\frac{d}{dx}\right)^{V_{1}}\left(e^{-\frac{x}{2}}x^{-\frac{k^{2}}{2}-(k-M)n_{1}-Mn_{2}+\frac{1}{2}V_{1}}\det_{k\times
k}\Big{(}I_{\sum_{s=2}^{n_{1}}sh_{si}+i+j-1}(2\sqrt{x})\Big{)}\right)\Bigg{|}_{x=0}$
$\displaystyle+\,O(N^{k^{2}+2n_{1}(k-M)+2n_{2}M-1}),$

where $V_{j}=\sum_{i=1}^{P}l_{i}a_{i,j}+\sum_{i=1}^{Q}u_{i}b_{i,j}$ for
$j=1,\ldots,n_{2}$ and $V_{j}=\sum_{i=1}^{P}l_{i}a_{i,j}$ for
$j=n_{2}+1,\ldots,n_{1}$.

Moreover,

$\displaystyle\sum_{\begin{subarray}{c}s=2,\ldots,n_{1}\\\
\sum_{i=1}^{k}h_{si}=V_{s}\end{subarray}}\left(\prod_{s=2}^{n_{1}}\binom{V_{s}}{h_{s1},\ldots,h_{sk}}\right)\left(\frac{d}{dx}\right)^{V_{1}}\Bigg{(}e^{-\frac{x}{2}}x^{-\frac{k^{2}}{2}-(k-M)n_{1}-Mn_{2}+\frac{1}{2}V_{1}}$
$\displaystyle\times\,\det_{k\times
k}\Big{(}I_{\sum_{s=2}^{n_{1}}sh_{si}+i+j-1}(2\sqrt{x})\Big{)}\Bigg{)}\Bigg{|}_{x=0}$
$\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}\sum_{i=1}^{k}h_{si}=V_{s}\\\
s=2,\ldots,n_{1}\\\
\sum_{i=0}^{k}h_{1i}=V_{1}\end{subarray}}(-\frac{1}{2})^{h_{10}}\left(\prod_{s=2}^{n_{1}}\binom{V_{s}}{h_{s1},\ldots,h_{sk}}\right)\binom{V_{1}}{h_{10},\ldots,h_{1k}}$
$\displaystyle\times\,\left(\prod_{i=1}^{k}\frac{1}{(2k+\sum_{s=1}^{n_{1}}sh_{si}-i)!}\right)\prod_{1\leq
i<j\leq
k}\left(\sum_{s=1}^{n_{1}}sh_{sj}-\sum_{s=1}^{n_{1}}sh_{si}-j+i\right).$

###### Proof.

Let $\alpha=(\alpha_{1},\ldots,\alpha_{2k})$. By Lemmas 14 and 16 , we have

$\displaystyle\int_{{\mathbb{U}}(N)}|Z_{A}^{(n_{1})}(1)|^{2k-2M}|Z_{A}^{(n_{2})}(1)|^{2M}dA_{N}$
$\displaystyle=$
$\displaystyle(-1)^{n_{1}(k-M)+Mn_{2}}\frac{1}{k!(2\pi\i)^{k}}\prod_{j=1}^{k-M}\left(\frac{d}{d\alpha_{j}}\frac{d}{d\alpha_{k+j}}\right)^{n_{1}}\prod_{j=k-M+1}^{k}\left(\frac{d}{d\alpha_{j}}\frac{d}{d\alpha_{k+j}}\right)^{n_{2}}$
$\displaystyle\times\,\int_{|w_{i}|=1}e^{N\sum_{j=1}^{k}w_{j}}\prod_{i\neq
j}z(w_{i}-w_{j})^{-1}\prod_{\begin{subarray}{c}1\leq i\leq k\\\ 1\leq j\leq
2k\end{subarray}}z(w_{i}-\alpha_{j})$
$\displaystyle\times\,e^{-\frac{N}{2}\sum_{j=1}^{2k}\alpha_{j}}\prod_{j=1}^{k}dw_{j}\Bigg{|}_{\alpha=0}+O(N^{k^{2}+(2k-2M)n_{1}+2Mn_{2}-1}).$

Suppose that $\alpha_{j}=\alpha_{j}(N)$ and $|\alpha_{j}|\leq 1/N$ for
$j=1,2,\ldots,2k$ above, the right-hand side is

$\displaystyle=$
$\displaystyle(-1)^{n_{1}(k-M)+n_{2}M+\frac{k(k-1)}{2}}\frac{1}{k!(2\pi\i)^{k}}\prod_{j=1}^{k-M}\left(\frac{d}{d\alpha_{j}}\frac{d}{d\alpha_{k+j}}\right)^{n_{1}}\prod_{j=k-M+1}^{k}\left(\frac{d}{d\alpha_{j}}\frac{d}{d\alpha_{k+j}}\right)^{n_{2}}$
$\displaystyle\times\,\int_{|w_{i}|=1}\frac{e^{N\sum_{j=1}^{k}w_{j}}\Delta^{2}(w)}{\prod_{j=1}^{2k}\prod_{i=1}^{k}(w_{i}-\alpha_{j})}e^{-\frac{N}{2}\sum_{j=1}^{2k}\alpha_{j}}\prod_{j=1}^{k}dw_{j}\Bigg{|}_{\alpha=0}+\,O(N^{k^{2}+(2k-2M)n_{1}+2Mn_{2}-1}).$

Using Lemma 9 to the derivatives with respect to $\alpha_{j}$ in the above
formula, we have

$\displaystyle=$
$\displaystyle\frac{1}{k!(2\pi\i)^{k}}(-1)^{n_{1}(k-M)+n_{2}M+\frac{k(k-1)}{2}}\int_{|w_{i}|=1}\frac{e^{N\sum_{j=1}^{k}w_{j}}\Delta^{2}(w)}{\prod_{i=1}^{k}w_{i}^{2k}}\Bigg{(}\sum_{m_{1}+2m_{2}+\cdots+n_{1}m_{n_{1}}=n_{1}}\frac{n_{1}!}{m_{1}!\cdots
m_{n_{1}}!}$ (22)
$\displaystyle\times\,\left(-\frac{N}{2}+\sum_{l=1}^{k}\frac{1}{w_{l}}\right)^{m_{1}}\prod_{j=2}^{n_{1}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{m_{j}}\Bigg{)}^{2k-2M}\Bigg{(}\sum_{r_{1}+2r_{2}+\cdots+n_{2}r_{n_{2}}=n_{2}}\frac{n_{2}!}{r_{1}!\cdots
r_{n_{2}}!}$
$\displaystyle\times\,\left(-\frac{N}{2}+\sum_{l=1}^{k}\frac{1}{w_{l}}\right)^{r_{1}}\prod_{j=2}^{n_{2}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{r_{j}}\Bigg{)}^{2M}\prod_{j=1}^{k}dw_{j}+O(N^{k^{2}+(2k-2M)n_{1}+2Mn_{2}-1}).$

Expanding the powers, we write

$\displaystyle\Bigg{(}\sum_{m_{1}+2m_{2}+\cdots+n_{1}m_{n_{1}}=n_{1}}\frac{n_{1}!}{m_{1}!\cdots
m_{n_{1}}!}\left(-\frac{N}{2}+\sum_{l=1}^{k}\frac{1}{w_{l}}\right)^{m_{1}}\prod_{j=2}^{n_{1}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{m_{j}}\Bigg{)}^{2k-2M}$
$\displaystyle=$
$\displaystyle\sum_{l_{1}+\cdots+l_{P}=2k-2M}\binom{2k-2M}{l_{1},\ldots,l_{P}}\frac{(n_{1}!)^{2k-2M}}{\prod_{i=1}^{P}({\bf
a}_{i}!)^{l_{i}}}\left(-\frac{N}{2}+\sum_{l=1}^{k}\frac{1}{w_{l}}\right)^{\sum_{i=1}^{P}l_{i}a_{i1}}\prod_{j=2}^{n_{1}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{\sum_{i=1}^{P}l_{i}a_{ij}},$

and

$\displaystyle\Bigg{(}\sum_{r_{1}+2r_{2}+\cdots+n_{2}r_{n_{2}}=n_{2}}\frac{n_{2}!}{r_{1}!\cdots
r_{n_{2}}!}\left(-\frac{N}{2}+\sum_{l=1}^{k}\frac{1}{w_{l}}\right)^{r_{1}}\prod_{j=2}^{n_{2}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{r_{j}}\Bigg{)}^{2M}$
$\displaystyle=$
$\displaystyle\sum_{u_{1}+\cdots+u_{Q}=2M}\binom{2M}{u_{1},\ldots,u_{Q}}\frac{(n_{2}!)^{2M}}{\prod_{i=1}^{Q}({\bf
b}_{i}!)^{u_{i}}}\left(-\frac{N}{2}+\sum_{l=1}^{k}\frac{1}{w_{l}}\right)^{\sum_{i=1}^{Q}u_{i}b_{i1}}\prod_{j=2}^{n_{2}}\left(\frac{1}{j}\sum_{l=1}^{k}\frac{1}{w_{l}^{j}}\right)^{\sum_{i=1}^{Q}u_{i}b_{ij}}.$

The claim in this theorem now follows from Proposition 21 with
$m_{j}=\sum_{i=1}^{P}l_{i}a_{i,j}+\sum_{i=1}^{Q}u_{i}b_{i,j}$ for
$j=1,\ldots,n_{2}$ and $m_{j}=\sum_{i=1}^{P}l_{i}a_{i,j}$ for
$j=n_{2}+1,\ldots,n_{1}$. Here we can compute that
$\sum_{s=2}^{n_{1}}sm_{s}=\sum_{s=1}^{n_{1}}sm_{s}-m_{1}=2(k-M)n_{1}+2Mn_{2}-m_{1}$.
∎

The following is our third main result. It provides an alternative formula for
Theorem 23. Compared with Theorem 23, this theorem provides an effective
approach to compute the joint moments when $k$ is small and $n_{1},n_{2}$ are
large. We will explain this in detail after the proof of the theorem. Theorem
5 is obtained by setting $n_{1}=n,n_{2}=0$ and switching $k-M$ and $M$.

######  Theorem 25.

For $k\geq 1$, $0\leq M\leq k$ and $n_{1},n_{2}\geq 0$ be integers, then we
have

$\displaystyle\int_{{\mathbb{U}}(N)}|\Lambda_{A}^{(n_{1})}(1)|^{2k-2M}|\Lambda_{A}^{(n_{2})}(1)|^{2M}dA_{N}$
$\displaystyle=$
$\displaystyle(-1)^{\frac{k(k-1)}{2}}N^{k^{2}+(2k-2M)n_{1}+2Mn_{2}}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s_{l,q_{1}}\leq
n_{1}\\\ \sum_{l=1}^{k}h_{l,q_{1}}=n_{1}\\\
q_{1}=1,\ldots,k-M\end{subarray}}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s^{\prime}_{l,q_{2}}\leq
n_{2}\\\ \sum_{l=1}^{k}h^{\prime}_{l,q_{2}}=n_{2}\\\
q_{2}=1,\ldots,M\end{subarray}}(-1)^{\sum_{q_{1}=1}^{k-M}\sum_{l=1}^{k}s_{l,q_{1}}+\sum_{q_{2}=1}^{M}\sum_{l=1}^{k}s^{\prime}_{l,q_{2}}}$
$\displaystyle\times\,(n_{1}!)^{2k-2M}(n_{2}!)^{2M}\left(\prod_{q_{1}=1}^{k-M}\frac{1}{(n_{1}-\sum_{l=1}^{k}s_{l,q_{1}})!}\right)\left(\prod_{q_{2}=1}^{M}\frac{1}{(n_{2}-\sum_{l=1}^{k}s^{\prime}_{l,q_{2}})!}\right)$
$\displaystyle\times\,\left(\prod_{i=1}^{k}\frac{1}{(2k-i+W_{i})!}\right)\prod_{1\leq
i<j\leq k}(W_{j}-W_{i}-j+i)+O(N^{k^{2}+(2k-2M)n_{1}+2Mn_{2}-1}),$

where for $i=1,\ldots,k$,

$\displaystyle
W_{i}=\sum_{q_{1}=1}^{k-M}(s_{i,q_{1}}+h_{i,q_{1}})+\sum_{q_{2}=1}^{M}(s^{\prime}_{i,q_{2}}+h^{\prime}_{i,q_{2}}).$
(23)

###### Proof.

Following the proof of Theorem 23, we have (20). Instead of expanding the
powers, here we apply (14) and (15) in Proposition 10 to tackle the powers. We
then obtain

$\displaystyle(\ref{518mixmomentofLamda:eq})$ $\displaystyle=$
$\displaystyle\int_{|w_{i}|=1}\frac{e^{N\sum_{j=1}^{k}w_{j}}\Delta^{2}(w)}{\prod_{i=1}^{k}w_{i}^{2k}}\Bigg{(}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s_{l,q}\leq
n_{1}\\\
q=1,\ldots,k-M\end{subarray}}\prod_{q=1}^{k-M}\Big{(}(-N)^{n_{1}-\sum_{l=1}^{k}s_{l,q}}\binom{n_{1}}{\sum_{l=1}^{k}s_{l,q}}(\sum_{l=1}^{k}s_{l,q})!\Big{)}$
(24)
$\displaystyle\times\,\frac{1}{\prod_{l=1}^{k}w_{l}^{\sum_{q=1}^{k-M}s_{l,q}}}\Bigg{)}\Bigg{(}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s^{\prime}_{l,q}\leq
n_{2}\\\
q=1,\ldots,M\end{subarray}}\prod_{q=1}^{M}\Big{(}(-N)^{n_{2}-\sum_{l=1}^{k}s^{\prime}_{l,q}}\binom{n_{2}}{\sum_{l=1}^{k}s^{\prime}_{l,q}}(\sum_{l=1}^{k}s^{\prime}_{l,q})!\Big{)}$
$\displaystyle\times\,\frac{1}{\prod_{l=1}^{k}w_{l}^{\sum_{q=1}^{M}s^{\prime}_{l,q}}}\Bigg{)}\Bigg{(}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}h_{l,q}=n_{1}\\\
q=1,\ldots,k-M\end{subarray}}\frac{(n_{1}!)^{k-M}}{\prod_{l=1}^{k}w_{l}^{\sum_{q=1}^{k-M}h_{l,q}}}\Bigg{)}$
$\displaystyle\times\,\Bigg{(}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}h^{\prime}_{l,q}=n_{2}\\\
q=1,\ldots,M\end{subarray}}\frac{(n_{2}!)^{M}}{\prod_{l=1}^{k}w_{l}^{\sum_{q=1}^{M}h^{\prime}_{l,q}}}\Bigg{)}\prod_{j=1}^{k}dw_{j}.$

Combining Proposition 19,

$\displaystyle(\ref{516night})$ $\displaystyle=$
$\displaystyle(2\pi\i)^{k}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s_{l,q_{1}}\leq
n_{1}\\\ \sum_{l=1}^{k}h_{l,q_{1}}=n_{1}\\\
q_{1}=1,\ldots,k-M\end{subarray}}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s^{\prime}_{l,q_{2}}\leq
n_{2}\\\ \sum_{l=1}^{k}h^{\prime}_{l,q_{2}}=n_{2}\\\
q_{2}=1,\ldots,M\end{subarray}}(n_{1}!)^{k-M}(n_{2}!)^{M}$
$\displaystyle\times\,\prod_{q_{1}=1}^{k-M}\left((-N)^{n_{1}-\sum_{l=1}^{k}s_{l,q_{1}}}\binom{n_{1}}{\sum_{l=1}^{k}s_{l,q_{1}}}(\sum_{l=1}^{k}s_{l,q_{1}})!\right)$
$\displaystyle\times\,\prod_{q_{2}=1}^{M}\left((-N)^{n_{2}-\sum_{l=1}^{k}s^{\prime}_{l,q_{2}}}\binom{n_{2}}{\sum_{l=1}^{k}s^{\prime}_{l,q_{2}}}(\sum_{l=1}^{k}s^{\prime}_{l,q_{2}})!\right)$
$\displaystyle\times\,\sum_{\sigma\in S_{k}}\det_{k\times
k}\Bigg{(}\frac{N^{2k+W_{\sigma(i)}-1-(i+j-2)}}{(2k+W_{\sigma(i)}-1-(i+j-2))!}\Bigg{)}.$

In the determinant, we firstly extract the factor $N^{W_{\sigma(i)}+k+1-i}$
from the $i$-th row and secondly extract the factor $N^{k-j}$ from the $j$-th
column, then

$\displaystyle(\ref{516night})$ $\displaystyle=$
$\displaystyle(2\pi\i)^{k}N^{k^{2}+2(k-M)n_{1}+2Mn_{2}}\sum_{\sigma\in
S_{k}}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s_{l,q_{1}}\leq n_{1}\\\
\sum_{l=1}^{k}h_{l,q_{1}}=n_{1}\\\
q_{1}=1,\ldots,k-M\end{subarray}}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s^{\prime}_{l,q_{2}}\leq
n_{2}\\\ \sum_{l=1}^{k}h^{\prime}_{l,q_{2}}=n_{2}\\\
q_{2}=1,\ldots,M\end{subarray}}(n_{1}!)^{2k-2M}(n_{2}!)^{2M}(-1)^{(k-M)n_{1}+Mn_{2}}$
$\displaystyle\times\,(-1)^{\sum_{q_{1}=1}^{k-M}\sum_{l=1}^{k}s_{l,q_{1}}+\sum_{q_{2}=1}^{M}\sum_{l=1}^{k}s^{\prime}_{l,q_{2}}}\left(\prod_{q_{1}=1}^{k-M}\frac{1}{(n_{1}-\sum_{l=1}^{k}s_{l,q_{1}})!}\right)\left(\prod_{q_{2}=1}^{M}\frac{1}{(n_{2}-\sum_{l=1}^{k}s^{\prime}_{l,q_{2}})!}\right)$
$\displaystyle\times\,\det_{k\times
k}\Bigg{(}\frac{1}{(2k+1+W_{\sigma(i)}-i-j)!}\Bigg{)}.$

It is not hard to check that the summations over
$s_{i,q_{1}},h_{i,q_{1}},s^{\prime}_{i,q_{2}},h^{\prime}_{i,q_{2}}$ are
invariant under any permutation $\sigma$ of $i$ on the set $\\{1,\ldots,k\\}$.
Indeed, for a fixed $\sigma$, we change the variables
$s_{\sigma(i),q_{1}}=\tilde{s}_{i,q_{1}},h_{\sigma(i),q_{1}}=\tilde{h}_{i,q_{1}}$,
$s^{\prime}_{\sigma(i),q_{2}}=\tilde{s^{\prime}}_{i,q_{2}},h^{\prime}_{\sigma(i),q_{2}}=\tilde{h^{\prime}}_{i,q_{2}}$,
then $\sum_{l=1}^{k}s_{l,q_{1}}=\sum_{l=1}^{k}\tilde{s}_{l,q_{1}}$,
$\sum_{l=1}^{k}s^{\prime}_{l,q_{2}}=\sum_{l=1}^{k}\tilde{s}^{\prime}_{l,q_{2}}$,
$\sum_{l=1}^{k}h_{l,q_{1}}=\sum_{l=1}^{k}\tilde{h}_{l,q_{1}}$,
$\sum_{l=1}^{k}h^{\prime}_{l,q_{2}}=\sum_{l=1}^{k}\tilde{h}^{\prime}_{l,q_{2}}$.
We also have
$W_{\sigma(i)}=\sum_{q_{1}=1}^{k-M}(s_{\sigma(i),q_{1}}+h_{\sigma(i),q_{1}})+\sum_{q_{2}=1}^{M}(s^{\prime}_{\sigma(i),q_{2}}+h^{\prime}_{\sigma(i),q_{2}})=\sum_{q_{1}=1}^{k-M}(\tilde{s}_{i,q_{1}}+\tilde{h}_{i,q_{1}})+\sum_{q_{2}=1}^{M}(\tilde{s}^{\prime}_{i,q_{2}}+\tilde{h}^{\prime}_{i,q_{2}}).$
So the whole expression is independent of $\sigma$ and we can set
$\sigma=\text{id}$. Hence, $\sum_{\sigma\in S_{k}}$ can be replaced $k!$.

By Lemma 20,

$\displaystyle(\ref{516night})$ $\displaystyle=$ $\displaystyle
k!(2\pi\i)^{k}N^{k^{2}+2(k-M)n_{1}+2Mn_{2}}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s_{l,q_{1}}\leq
n_{1}\\\ \sum_{l=1}^{k}h_{l,q_{1}}=n_{1}\\\
q_{1}=1,\ldots,k-M\end{subarray}}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s^{\prime}_{l,q_{2}}\leq
n_{2}\\\ \sum_{l=1}^{k}h^{\prime}_{l,q_{2}}=n_{2}\\\
q_{2}=1,\ldots,M\end{subarray}}(n_{1}!)^{2k-2M}(n_{2}!)^{2M}(-1)^{(k-M)n_{1}+Mn_{2}}$
$\displaystyle\times\,(-1)^{\sum_{q_{1}=1}^{k-M}\sum_{l=1}^{k}s_{l,q_{1}}+\sum_{q_{2}=1}^{M}\sum_{l=1}^{k}s^{\prime}_{l,q_{2}}}\prod_{q_{1}=1}^{k-M}\frac{1}{(n_{1}-\sum_{l=1}^{k}s_{l,q_{1}})!}\prod_{q_{2}=1}^{M}\frac{1}{(n_{2}-\sum_{l=1}^{k}s^{\prime}_{l,q_{2}})!}$
$\displaystyle\times\,\prod_{i=1}^{k}\frac{1}{(2k-i+W_{i})!}\prod_{1\leq
i<j\leq k}(W_{j}-W_{i}-j+i),$

where $W_{i}$ is given in (23). Hence, we obtain the claimed result in this
theorem. ∎

######  Remark 26.

We now explain the difference between Theorem 23 and Theorem 25. When using
Theorem 23, it is required to list all tuples of the sets for the two
combinatorial objects $P,Q$. This may not be an easy task when $n_{1},n_{2}$
are large. In comparison, the summation in Theorem 25 is easier when $k$ is
small. So Theorem 25 is computationally effective. However, the formula in
Theorem 23 has a nice structure depending on determinants of matrices whose
entries involve the modified Bessel functions of the first kind, which can be
used to build connections between moments and Painlevé equations. This is
studied in detail in our second paper [31].

The following is our fourth main result. It provides an alternative formula
for Theorem 24. Theorem 6 is obtained from it by choosing $n_{1}=n,n_{2}=0$
and switching the positions of $k-M$ and $M$.

######  Theorem 27.

For $k\geq 1$, $0\leq M\leq k$ and $n_{1},n_{2}\geq 0$ be integers, we have

$\displaystyle\int_{{\mathbb{U}}(N)}|Z_{A}^{(n_{1})}(1)|^{2k-2M}|Z_{A}^{(n_{2})}(1)|^{2M}dA_{N}$
$\displaystyle=$
$\displaystyle(-1)^{n_{1}(k-M)+n_{2}M+\frac{k(k-1)}{2}}N^{k^{2}+(2k-2M)n_{1}+2Mn_{2}}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s_{l,q_{1}}\leq
n_{1}\\\ \sum_{l=1}^{k}h_{l,q_{2}}\leq n_{2}\\\ q_{1}=1,\ldots,2k-2M\\\
q_{2}=1,\ldots,2M\end{subarray}}(n_{1}!)^{2k-2M}(n_{2}!)^{2M}$
$\displaystyle\times\,(-1/2)^{2kn_{1}-2Mn_{1}-\sum_{q_{1}=1}^{2k-2M}\sum_{l=1}^{k}s_{l,q_{1}}}(-1/2)^{2Mn_{2}-\sum_{q_{2}=1}^{2M}\sum_{l=1}^{k}h_{l,q_{2}}}\left(\prod_{q_{1}=1}^{2k-2M}\frac{1}{(n_{1}-\sum_{l=1}^{k}s_{l,q_{1}})!}\right)$
$\displaystyle\times\,\left(\prod_{q_{2}=1}^{2M}\frac{1}{(n_{2}-\sum_{l=1}^{k}h_{l,q_{2}})!}\right)\left(\prod_{i=1}^{k}\frac{1}{(2k-i+\sum_{q_{1}=1}^{2k-2M}s_{i,q_{1}}+\sum_{q_{2}=1}^{2M}h_{i,q_{2}})!}\right)$
$\displaystyle\times\,\prod_{1\leq i<j\leq
k}\left(\Big{(}\sum_{q_{1}=1}^{2k-2M}s_{j,q_{1}}+\sum_{q_{2}=1}^{2M}h_{j,q_{2}}\Big{)}-\Big{(}\sum_{q_{1}=1}^{2k-2M}s_{i,q_{1}}+\sum_{q_{2}=1}^{2M}h_{i,q_{2}}\Big{)}-j+i\right)$
$\displaystyle+\,O(N^{k^{2}+(2k-2M)n_{1}+2Mn_{2}-1}).$

###### Proof.

Following the proof of Theorem 24, we have (22). We now apply (14) in
Proposition 10 to tackle the powers and obtain

$\displaystyle(\ref{generalization of mixed moment:eq})$ $\displaystyle=$
$\displaystyle\frac{1}{k!(2\pi\i)^{k}}(-1)^{n_{1}(k-M)+n_{2}M+\frac{k(k-1)}{2}}\int_{|w_{i}|=1}\frac{e^{N\sum_{j=1}^{k}w_{j}}\Delta^{2}(w)}{\prod_{i=1}^{k}w_{i}^{2k}}$
$\displaystyle\times\,\Bigg{(}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}s_{l,q}\leq
n_{1}\\\
q=1,\ldots,2k-2M\end{subarray}}\prod_{q=1}^{2k-2M}\left((-N/2)^{n_{1}-\sum_{l=1}^{k}s_{l,q}}\binom{n_{1}}{\sum_{l=1}^{k}s_{l,q}}(\sum_{l=1}^{k}s_{l,q})!\right)\frac{1}{\prod_{l=1}^{k}w_{l}^{\sum_{q=1}^{2k-2M}s_{l,q}}}\Bigg{)}$
$\displaystyle\times\,\Bigg{(}\sum_{\begin{subarray}{c}\sum_{l=1}^{k}h_{l,q}\leq
n_{2}\\\
q=1,\ldots,2M\end{subarray}}\prod_{q=1}^{2M}\left((-N/2)^{n_{2}-\sum_{l=1}^{k}h_{l,q}}\binom{n_{2}}{\sum_{l=1}^{k}h_{l,q}}(\sum_{l=1}^{k}h_{l,q})!\right)\frac{1}{\prod_{l=1}^{k}w_{l}^{\sum_{q=1}^{2M}h_{l,q}}}\Bigg{)}$
$\displaystyle\times\,\prod_{j=1}^{k}dw_{j}+O(N^{k^{2}+(2k-2M)n_{1}+2Mn_{2}-1}).$

Using a similar argument to the analysis of (24), we obtain the claimed result
in this theorem. ∎

To finish, we prove Proposition 8.

######  Lemma 28 (Equation (6) of [38]).

For all $n,m,k_{1},\ldots,k_{m}\in\mathbb{N}$, with $k_{1}+\cdots+k_{m}=n$,
$n\geq 1$ and $m\geq 2$, we have

$\binom{n}{k_{1},\ldots,k_{m}}=\sum_{i=1}^{m}\binom{n-1}{k_{1},\ldots,k_{i}-1,\ldots,k_{m}}.$

###### Proof of Proposition 8.

We first show $a_{1,1}(n,0)$. Based on Theorem 5 with $k=1$, it suffices to
prove

$\displaystyle(n!)^{2}\sum_{s=0}^{n}(-1)^{s}\frac{1}{(n-s)!}\frac{1}{(n+1+s)!}=\frac{1}{2n+1}.$

Equivalently,

$\displaystyle\frac{(n!)^{2}}{(2n+1)!}\sum_{s=0}^{n}(-1)^{n-s}\binom{2n+1}{s}=\frac{1}{2n+1}.$

Using Lemma 28 with $m=2$, that is,
$\binom{2n+1}{s}=\binom{2n}{s}+\binom{2n}{s-1}$, it is easy to see the above
equality holds.

In the following, we show $b_{1,1}(n,0)$. Let $k=M=1$ in Theorem 6, we then
have

$\displaystyle\int_{{\mathbb{U}}(N)}|Z_{A}^{(n)}(1)|^{2}dA_{N}$
$\displaystyle=$
$\displaystyle(-1)^{n}N^{2n+1}(n!)^{2}\sum_{i,j=0}^{n}(-1/2)^{2n-i-j}\frac{1}{(n-i)!(n-j)!(i+j+1)!}N^{2n+1}+O(N^{2n})$
$\displaystyle=$
$\displaystyle(-1)^{n}N^{2n+1}\frac{(n!)^{2}}{(2n+1)!}\sum_{i,j=0}^{n}(-1/2)^{i+j}\binom{2n+1}{i,j,2n+1-i-j}N^{2n+1}+O(N^{2n}).$

Applying Lemma 28 with $m=3$,

$\displaystyle\sum_{i,j=0}^{n}(-1/2)^{i+j}\binom{2n+1}{i,j,2n+1-i-j}$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{n}\sum_{j=0}^{n}(-1/2)^{i+j}\binom{2n}{i-1,j,2n+1-i-j}+\sum_{i=0}^{n}\sum_{j=1}^{n}(-1/2)^{i+j}\binom{2n}{i,j-1,2n+1-i-j}$
$\displaystyle+\,\sum_{i=0}^{n}\sum_{j=0}^{n}(-1/2)^{i+j}\binom{2n}{i,j,2n-i-j}$
$\displaystyle=$ $\displaystyle
2\sum_{i=0}^{n}\sum_{j=1}^{n}(-1/2)^{i+j}\binom{2n}{i,j-1,2n+1-i-j}+\sum_{i=0}^{n}\sum_{j=0}^{n-1}(-1/2)^{i+j}\binom{2n}{i,j,2n-i-j}$
$\displaystyle+\,\sum_{i=0}^{n}(-1/2)^{i+n}\frac{(2n)!}{n!i!(n-i)!}$
$\displaystyle=$
$\displaystyle\sum_{i=0}^{n}(-1/2)^{i+n}\frac{(2n)!}{n!i!(n-i)!}=\frac{(-1)^{n}(2n)!}{(n!)^{2}2^{n}}\sum_{i=0}^{n}(-1/2)^{i}\binom{n}{i}=\frac{(-1)^{n}(2n)!}{(n!)^{2}4^{n}}.$

Putting this all together, we obtain the claimed result in this proposition. ∎

##  4 Numerical data

The following are $b_{k,k}(2,0)$ for $k=1,\ldots,6$:

$\displaystyle\displaystyle\frac{1}{2^{4}\cdot 5}$
$\displaystyle\displaystyle\frac{17}{2^{10}\cdot 3^{3}\cdot 5\cdot 7\cdot 11}$
$\displaystyle\displaystyle\frac{11593}{2^{18}\cdot 3^{7}\cdot 5^{2}\cdot
7^{3}\cdot 11\cdot 13\cdot 17}$ $\displaystyle\displaystyle\frac{103\cdot
413129}{2^{28}\cdot 3^{12}\cdot 5^{5}\cdot 7^{3}\cdot 11^{2}\cdot 13^{2}\cdot
17\cdot 19\cdot 23}$ $\displaystyle\displaystyle\frac{2616269\cdot
322433}{2^{40}\cdot 3^{17}\cdot 5^{8}\cdot 7^{5}\cdot 11^{4}\cdot 13^{3}\cdot
17^{2}\cdot 19\cdot 23\cdot 29}$ $\displaystyle\displaystyle\frac{53\cdot
5830411\cdot 94098709}{2^{54}\cdot 3^{24}\cdot 5^{13}\cdot 7^{8}\cdot
11^{4}\cdot 13^{4}\cdot 17^{3}\cdot 19^{2}\cdot 23\cdot 29\cdot 31}$

$a_{2,1}(0,0)=1/12$.

$a_{2,1}(1,n_{2})$ for $n_{2}=0,1$:

${\frac{1}{45}},{\frac{61}{10080}}.$

$a_{2,1}(2,n_{2})$ for $n_{2}=0,1,2$:

${\frac{1}{112}},{\frac{1133}{453600}},{\frac{449}{415800}}.$

$a_{2,1}(3,n_{2})$ for $n_{2}=0,1,2,3$:

${\frac{1}{225}},{\frac{529}{415800}},{\frac{3943}{6879600}},{\frac{48953}{155232000}}.$

$b_{2,1}(0,0)=1/12$.

$b_{2,1}(1,n_{2})$ for $n_{2}=0,1$:

${\frac{1}{720}},{\frac{1}{6720}}.$

$b_{2,1}(2,n_{2})$ for $n_{2}=0,1,2$:

${\frac{1}{4032}},{\frac{19}{3628800}},{\frac{17}{10644480}}.$

$b_{2,1}(3,n_{2})$ for $n_{2}=0,1,2,3$:

${\frac{1}{57600}},{\frac{19}{10644480}},{\frac{127}{1761177600}},{\frac{41}{1419264000}}.$

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