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# Berndt-Type Integrals and Series Associated with Ramanujan and Jacobi
Elliptic Functions

Ce Xua, and Jianqiang Zhaob,
a. School of Mathematics and Statistics, Anhui Normal University,
Wuhu 241002, P.R. China
b. Department of Mathematics, The Bishop’s School, La Jolla, CA 92037, USA
Email<EMAIL_ADDRESS><EMAIL_ADDRESS>
Abstract. In this paper, we evaluate in closed forms two families of infinite
integrals containing hyperbolic and trigonometric functions in their
integrands. We call them Berndt-type integrals since he initiated the study of
similar integrals. We first establish explicit evaluations of four classes of
hyperbolic sums by special values of the Gamma function, by two completely
different approaches, which extend those sums considered by Ramanujan and
Zucker previously. We discover the first by refining two results of Ramanujan
concerning some $q$-series. For the second we compare both the Fourier series
expansions and the Maclaurin series expansions of a few Jacobi elliptic
functions. Next, by contour integrations we convert two families of Berndt-
type integrals to the above hyperbolic sums, all of which can be evaluated in
closed forms. We then discover explicit formulas for one of the two families.
Throughout the paper we present many examples which enable us to formulate a
conjectural explicit formula for the other family of the Berndt-type integrals
at the end.

Keywords: Berndt-type integral, $q$-series, hyperbolic and trigonometric
functions, contour integration, Jacobi elliptic functions.

AMS Subject Classifications (2020): 32A27, 42A16, 33E05, 11B68, 05A30.

##  1 Introduction

The primary motivation for this paper arises from several infinite integrals
first considered by Berndt in [3] in which he proved some explicit relations
between hyperbolic series and integrals containing trigonometric and
hyperbolic functions. For example, from [3, Thm. 3.1 and Eq. (5.5)] for
nonnegative integer $a$, Berndt showed that

$\displaystyle(1+(-1)^{a})\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{2a+1}dx}{\cos
x+\cosh
x}=\displaystyle\frac{\pi^{2a+2}i^{a}}{2^{a}}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{2a+1}}{\cosh((2n+1)\pi/2)},$
(1.1)

and

$\displaystyle(1+i^{a+1})\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{a}dx}{\cos
x-\cosh
x}=2i(1+i)^{a-1}\pi^{a+1}\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(-1)^{n+1}n^{a}}{\sinh(n\pi)}\quad(a\geq
2).$ (1.2)

In his second notebook [10] (see [2, p.134, Entry 16(i)-(vi)]), for odd
integers $s<12$ Ramanujan evaluated the hyperbolic sums

$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{s}}{\cosh((2n+1)y/2)}$
(1.3)

in terms of $x$ and $z$. For example, it is proved that

$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{3}}{\cosh((2n+1)y/2)}=\displaystyle\frac{1}{4}z^{4}(1-2x)\sqrt{x(1-x)},$
(1.4)

where the definition of $x,y$ and $z$ are given in (2.14). Zucker [18]
extended these evaluations by replacing $\cosh$ with $\sinh$ in (1.3).

In this paper, we apply some carefully chosen contour integrations related to
hyperbolic and trigonometric functions to build on the previous work on
Berndt-type integrals and hyperbolic infinite sums (see [3, 4, 9, 17, 18]).
First, we will prove the following general results, which extend those of
Yakubovich (see [17, Cor. 3]).

######  Theorem 1.1.

Let $\Gamma(x)$ be the usual gamma function. For any positive integer $m$,

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m-2}}{\sinh^{2}(n\pi)}$
$\displaystyle\,\in\displaystyle\frac{\Gamma^{8m}(1/4)}{\pi^{6m}}\mathbb{Q}+\displaystyle\frac{\delta_{m,1}}{\pi^{2}}\mathbb{Q},$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m-4}}{\sinh^{2}(n\pi)}$
$\displaystyle\,\in\displaystyle\frac{\Gamma^{8m-8}(1/4)}{\pi^{6m-5}}\mathbb{Q}+\displaystyle\frac{\Gamma^{8m-4}(1/4)}{\pi^{6m-3}}\mathbb{Q}+\delta_{m,1}\mathbb{Q},$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m-1}\cosh(n\pi)}{\sinh^{3}(n\pi)}$
$\displaystyle\,\in\displaystyle\frac{\Gamma^{8m}(1/4)}{\pi^{6m+1}}\mathbb{Q}+\displaystyle\frac{\delta_{m,1}}{\pi^{3}}\mathbb{Q},$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m-3}\cosh(n\pi)}{\sinh^{3}(n\pi)}$
$\displaystyle\,\in\displaystyle\frac{\Gamma^{8m}(1/4)}{\pi^{6m}}\mathbb{Q}+\displaystyle\frac{\Gamma^{8m-8}(1/4)}{\pi^{6m-4}}\mathbb{Q},$

where $\delta_{a,b}$ is the Kronecker symbol.

######  Theorem 1.2.

For any integer $m\geq 0$,

$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{4m+2}}{\cosh^{2}((2n+1)\pi/2)}$
$\displaystyle\,\in\displaystyle\frac{\Gamma^{8m+8}}{\pi^{6m+6}}\mathbb{Q},$
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{4m}}{\cosh^{2}((2n+1)\pi/2)}$
$\displaystyle\,\in\displaystyle\frac{\Gamma^{8m}}{\pi^{6m+1}}\mathbb{Q},$
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{4m+1}\sinh((2n+1)\pi/2)}{\cosh^{3}((2n+1)\pi/2))}$
$\displaystyle\,\in\displaystyle\frac{\Gamma^{8m+8}}{\pi^{6m+6}}\mathbb{Q}+\displaystyle\frac{\Gamma^{8m}}{\pi^{6m+2}}\mathbb{Q},$
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{4m+3}\sinh((2n+1)\pi/2)}{\cosh^{3}((2n+1)\pi/2))}$
$\displaystyle\,\in\displaystyle\frac{\Gamma^{8m+8}}{\pi^{6m+7}}\mathbb{Q}.$

A more precise version of the above theorem is given by Theorem 8.10 in which
all the rational constants are given explicitly as a function of $m$ in closed
forms.

In another direction, the Ramanujan’s identity (1.4) was applied by Berndt in
[3] to prove the following beautiful identity:

$\displaystyle\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{dx}{\cos\sqrt{x}+\cosh\sqrt{x}}=\displaystyle\frac{\pi}{4}\displaystyle\frac{\Gamma^{2}(1/4)}{\Gamma^{2}(3/4)}.$
(1.5)

Some recent results on infinite series involving hyperbolic functions may be
found in the works of [7, 11, 12, 13, 15, 16, 17] and the references therein.
Using Theorem 1.1, we will prove the following evaluations (see Theorems 6.3
and 8.12).

######  Theorem 1.3.

For any positive integer $p$,

$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{4p+1}}{(\cos
x-\cosh
x)^{2}}dx=i_{p}\displaystyle\frac{\Gamma^{8p}(1/4)}{\pi^{2p}}+j_{p}\displaystyle\frac{\Gamma^{8p+8}(1/4)}{\pi^{2p+4}},$
(1.6)
$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{4p+1}}{(\cos
x+\cosh
x)^{2}}dx=g_{p}\displaystyle\frac{\Gamma^{8p}(1/4)}{\pi^{2p}}+h_{p}\displaystyle\frac{\Gamma^{8p+8}(1/4)}{\pi^{2p+4}},$
(1.7)

where $g_{p},h_{p}\in\mathbb{Q}$ are given explicitly in closed forms in
Theorem 8.12, and $i_{p},j_{p}\in\mathbb{Q}$ can be explicitly computed for
each fixed $p$.

A natural question is how to evaluate the following more general Berndt-type
integrals?

Question: For integers $a\geq 0$ and $b\geq 1$, can we express the following
integrals in closed forms:

$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{a}dx}{(\cos
x+\cosh x)^{b}}=?$ (1.8)

And if $a\geq 2b$

$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{a}dx}{(\cos
x-\cosh x)^{b}}=?$ (1.9)

##  2 Brief review of complete elliptic integral of the first kind

To fix notation we now give a very brief review of the complete elliptic
integral of the first kind in this section. As usual, $(a)_{n}$ denotes the
ascending Pochhammer symbol defined by

$\displaystyle(a)_{n}:=\displaystyle\frac{\Gamma(a+n)}{\Gamma(a)}=a(a+1)\cdots(a+n-1)\quad{\rm
and}\quad(a)_{0}:=1,$ (2.10)

where $a\in\mathbb{C}$ is any complex number and $n\in\mathbb{N}_{0}$ is a
nonnegative integer. Let $\Gamma\left(z\right)$ be the Gamma function as
usual. If $\Re(z)>0$ then

$\Gamma(z):=\displaystyle\int_{0}^{\infty}e^{-t}t^{z-1}\,dt.$

The Gaussian or ordinary hypergeometric function ${{}_{2}}F_{1}(a,b;c;x)$ is
defined for $|x|<1$ by the power series

$\displaystyle{{}_{2}}F_{1}(a,b;c;x)=\displaystyle\sum\limits_{n=0}^{\infty}\displaystyle\frac{(a)_{n}(b)_{n}}{(c)_{n}}\displaystyle\frac{x^{n}}{n!}\quad(a,b,c\in\mathbb{C}).$
(2.11)

The complete elliptic integral of the first kind is defined by (see [14])

$\displaystyle
K:=K(x):=K(k^{2}):=\displaystyle\int\limits_{0}^{\pi/2}\displaystyle\frac{d\varphi}{\sqrt{1-k^{2}\sin^{2}\varphi}}=\displaystyle\frac{\pi}{2}{{}_{2}}F_{1}\left(\displaystyle\frac{1}{2},\displaystyle\frac{1}{2};1;k^{2}\right).$
(2.12)

Here $x=k^{2}$ and $k\ (0<k<1)$ is the modulus of $K$. The complementary
modulus $k^{\prime}$ is defined by $k^{\prime}=\sqrt{1-k^{2}}$. Furthermore,
we set as usual

$\displaystyle K^{\prime}:=K(k^{\prime
2})=\displaystyle\int\limits_{0}^{\pi/2}\displaystyle\frac{d\varphi}{\sqrt{1-k^{\prime
2}\sin^{2}\varphi}}=\displaystyle\frac{\pi}{2}{{}_{2}}F_{1}\left(\displaystyle\frac{1}{2},\displaystyle\frac{1}{2};1;1-k^{2}\right).$
(2.13)

In order to better state our main results, we shall henceforth adopt the
notations of Ramanujan (see Berndt’s book [2]). Let

$\displaystyle x:=k^{2},\quad
y:=y(x):=\pi\displaystyle\frac{K^{\prime}}{K},\quad q:=q(x):=e^{-y},\quad
z:=z(x):=\displaystyle\frac{2}{\pi}K,\quad
z^{\prime}:=\displaystyle\frac{dz}{dx}.$ (2.14)

Using the identity $(a)_{n+1}=a(a+1)_{n}$, it is easily shown that

$\displaystyle\frac{d}{dx}{{}_{2}}F_{1}(a,b;c;x)=\displaystyle\frac{ab}{c}{{}_{2}}F_{1}(a+1,b+1;c+1;x)$

and more generally,

$\displaystyle\displaystyle\frac{d^{n}}{dx^{n}}{{}_{2}}F_{1}(a,b;c;x)=\displaystyle\frac{(a)_{n}(b)_{n}}{(c)_{n}}{{}_{2}}F_{1}(a+n,b+n;c+n;x)\quad(n\in\mathbb{N}_{0}).$
(2.15)

Then,

$\displaystyle\displaystyle\frac{d^{n}z}{dx^{n}}=\displaystyle\frac{\left(1/2\right)^{2}_{n}}{n!}{{}_{2}}F_{1}\left(\displaystyle\frac{1}{2}+n,\displaystyle\frac{1}{2}+n;1+n;x\right).$
(2.16)

Applying the identity (see [1, Theorem 3.5.4(i)])

$\displaystyle{{}_{2}}F_{1}\left(a,b;\displaystyle\frac{a+b+1}{2};\displaystyle\frac{1}{2}\right)=\displaystyle\frac{\Gamma\left(\displaystyle\frac{1}{2}\right)\Gamma\left(\displaystyle\frac{a+b+1}{2}\right)}{\Gamma\left(\displaystyle\frac{a+1}{2}\right)\Gamma\left(\displaystyle\frac{b+1}{2}\right)},$
(2.17)

we can show by an elementary calculation that

$\displaystyle\displaystyle\frac{d^{n}z}{dx^{n}}\bigg{|}_{x=1/2}=\displaystyle\frac{(1/2)^{2}_{n}\sqrt{\pi}}{\Gamma^{2}\left(\displaystyle\frac{n}{2}+\displaystyle\frac{3}{4}\right)}.$
(2.18)

##  3 Refined versions of two results of Ramanujan’s

In [9] Ramanujan studied the $q$-series

$\Phi_{a,m}:=\Phi_{a,m}(q^{2}):=\displaystyle\sum_{n\geq
1}\displaystyle\frac{n^{m}q^{2n}}{(1-q^{2n})^{a+1}}$

for integers $a\geq 0$ and $m\geq 1$. We follow Ramanujan’s notation by
setting

$S_{m}:=\displaystyle\frac{1}{2}\zeta(-m)+\Phi_{0,m}=-\displaystyle\frac{B_{m+1}}{2(m+1)}+\Phi_{0,m}\quad\forall
r\geq 1.$

The three relevant Eisenstein series are defined for $|q|<1$ by

$\displaystyle
P:=1-24\displaystyle\sum\limits_{n=1}^{\infty}\displaystyle\frac{nq^{n}}{1-q^{2n}},\quad
Q:=1+240\displaystyle\sum\limits_{n=1}^{\infty}\displaystyle\frac{n^{3}q^{2n}}{1-q^{2n}},\quad
R:=1-504\displaystyle\sum\limits_{n=1}^{\infty}\displaystyle\frac{n^{5}q^{2n}}{1-q^{2n}}.$

By Entry 13 in Chapter 17 of Ramanujan’s third notebook [2], we have

$\displaystyle P=(1-2x)z^{2}+6x(1-x)zz^{\prime},$ (3.19) $\displaystyle
Q=z^{4}(1-x+x^{2}),$ (3.20) $\displaystyle R=z^{6}(1+x)(1-x/2)(1-2x).$ (3.21)

Ramanujan proved that the $q$-series $\Phi_{a,m}\in\mathbb{Q}[P,Q,R].$ In
order to prove Theorem 1.1 we need to refine this result when $a=0$ and $a=1$.
For two vector spaces $F$ and $G$ we say $f\in F$ modulo $G$ if $f\in F+G$.

######  Proposition 3.1.

Modulo $(2x-1)\mathbb{Q}[x,z]$, we have $S_{1}\in zz^{\prime}\mathbb{Q}[x]$
and for all $m\geq 1$

$\displaystyle S_{4m-1}\in z^{4m}\mathbb{Q}[x],\quad\quad$ $\displaystyle
S_{4m+1}\in\\{0\\},$ (3.22) $\displaystyle S^{\prime}_{4m-1}\in
z^{4m-1}z^{\prime}\mathbb{Q}[x],\quad\quad$ $\displaystyle
S^{\prime}_{4m+1}\in z^{4m+2}\mathbb{Q}[x].$ (3.23)

###### Proof.

First note that $S_{1}=-P/24\in zz^{\prime}\mathbb{Q}[x]$ modulo
$(2x-1)\mathbb{Q}[x,z]$. Now we prove a statement stronger than (3.22): for
all $m\geq 1$

$S_{4m-1}\in Q^{m}\mathbb{Q}+R\,\mathbb{Q}[R,Q]\quad\text{and}\quad
S_{4m+1}\in R\,\mathbb{Q}[R,Q].$

This implies (3.22) since $R\in(2x-1)\mathbb{Q}[x,z]$. Observe that by [9,
(25)]

$S_{1}=-P/24,\quad S_{3}=Q/240,\quad S_{5}=-R/504.$

By [9, (22)], for all $n\geq 1$

$\displaystyle\frac{(2n+5)(n-1)}{12(2n+1)(n+1)}S_{2n+3}=\displaystyle\sum_{k=1}^{n-1}\binom{2n}{2k}S_{2k+1}S_{2n-2k+1}.$
(3.24)

If $n$ is odd, then the sum of the two indices on the right-hand side of the
above is $2n+2$ which is a multiple of 4. Thus one index must have the form
$4m+1$. Therefore $S_{2n+3}\in R\,\mathbb{Q}[R,Q]$ by induction assumption. If
$n$ is even, then by induction assumption either both of $S_{2k+1}$ and
$S_{2n-2k+1}$ have the form $R\,\mathbb{Q}[R,Q]$ (when $k$ is even), or both
have the form $Q^{m_{j}}\mathbb{Q}+R\,\mathbb{Q}[R,Q]$ (when $k$ is odd) with
the sum of the two $Q$-powers given by

$m_{1}+m_{2}=\displaystyle\frac{2k+2}{4}+\displaystyle\frac{2n-2k+2}{4}=\displaystyle\frac{2n+4}{4}.$

This proves (3.22) in the proposition by induction.

To prove (3.23) we differentiate (3.24) with respect to $x$ and get

$\displaystyle\frac{(2n+5)(n-1)}{12(2n+1)(n+1)}S^{\prime}_{2n+3}=\displaystyle\sum_{k=1}^{n-1}\binom{2n}{2k}\big{(}S^{\prime}_{2k+1}S_{2n-2k+1}+S_{2k+1}S^{\prime}_{2n-2k+1}\big{)}.$
(3.25)

Using (3.22) we can now prove (3.23) by induction similarly to the above for
(3.22). We leave the details to the interested reader. ∎

######  Proposition 3.2.

Modulo $(2x-1)\mathbb{Q}[x,z,z^{\prime}]$, we have

$\displaystyle\Phi_{1,2}=\displaystyle\frac{1}{2^{7}}\Big{(}\displaystyle\frac{z^{3}}{3}+z^{2}(z^{\prime})^{2}\Big{)},\quad\Phi^{\prime}_{1,2}=\displaystyle\frac{1}{2^{5}}\Big{(}2z^{3}z^{\prime}-z(z^{\prime})^{3}-z^{2}z^{\prime}z^{\prime\prime}\Big{)},$
(3.26)

and for all $m\geq 1$.

$\displaystyle\Phi_{1,4m+2}\in z^{4m+4}\mathbb{Q}[x],\quad$
$\displaystyle\Phi_{1,4m}\in z^{4m+1}z^{\prime}\mathbb{Q}[x],$ (3.27)
$\displaystyle\Phi^{\prime}_{1,4m+2}\in z^{4m+3}z^{\prime}\mathbb{Q}[x],\quad$
$\displaystyle\Phi^{\prime}_{1,4m}\in
z^{4m+2}\mathbb{Q}[x]+z^{4m}(z^{\prime})^{2}\mathbb{Q}[x]+z^{4m+1}z^{\prime\prime}\mathbb{Q}[x].$
(3.28)

###### Proof.

First we note that $\Phi_{1,2}=(Q-P^{2})/288$ from [9, Table II]. Thus

$288\Phi^{\prime}_{1,2}=4z^{3}z^{\prime}(1-x+x^{2})-12x(1-x)zz^{\prime}\Big{(}-2z^{2}+6x(1-x)\big{(}(z^{\prime})^{2}+zz^{\prime\prime}\big{)}\Big{)}$
(3.29)

modulo $(2x-1)\mathbb{Q}[x,z,z^{\prime}]$ by (3.19) and (3.20). Thus (3.26)
follows.

Now we prove a statement stronger than (3.27): for all $m\geq 1$

$\Phi_{1,4m+2}\in
Q^{m+1}\mathbb{Q}+R\,\mathbb{Q}[R,Q]\quad\text{and}\quad\Phi_{1,4m}\in
PQ^{m}\mathbb{Q}+R\,\mathbb{Q}[R,Q].$

By [9, Table II], we see that

$\Phi_{1,4}=(PQ-R)/720\quad\text{and}\quad\Phi_{1,6}=(Q^{2}-PR)/1008.$

By [9, (28)], for all $n\geq 1$

$\displaystyle\frac{2n+3}{2(2n+1)}S_{2n+1}-\Phi_{1,2n}=\displaystyle\sum_{k=1}^{n}\binom{2n}{2k-1}S_{2k-1}S_{2n-2k+1}.$
(3.30)

Suppose now $n\geq 3$. If $n=2m+1$, then by Prop. 3.1 we see that
$S_{2n+1}=S_{4m+3}\in Q^{m+1}\mathbb{Q}+R\,\mathbb{Q}[R,Q]$ while
$S_{1}S_{2n-1}=S_{1}S_{4m+1}\in R\,\mathbb{Q}[R,Q]$. The other products on the
right-hand side of (3.30) are either of the form
$Q^{m+1}\mathbb{Q}+R\,\mathbb{Q}[R,Q]$ (when $k$ is even) or of the form
$R\,\mathbb{Q}[R,Q]$ (when $k$ is odd). Thus $\Phi_{1,4m+2}\in
Q^{m+1}\mathbb{Q}+R\,\mathbb{Q}[R,Q]$.

If $n=2m$, then by Prop. 3.1 $S_{4m+1}\in R\,\mathbb{Q}[R,Q]$ and
$S_{1}S_{2n-1}=S_{1}S_{4m-1}\in PQ^{m}\mathbb{Q}+R\,\mathbb{Q}[R,Q]$. The
other products must have a factor having the form of $R\,\mathbb{Q}[R,Q]$.
Thus $S_{4m}\in PQ^{m}\mathbb{Q}+R\,\mathbb{Q}[R,Q]$. This completes the proof
of (3.27) of the proposition.

To prove (3.28), we differentiate (3.30) to get

$\displaystyle\frac{n+3}{2(n+1)}S^{\prime}_{2n+1}+\Phi^{\prime}_{1,2n}=\displaystyle\sum_{k=1}^{n}\binom{2n}{2k-1}\big{(}S^{\prime}_{2k-1}S_{2n-2k+1}+S_{2k-1}S^{\prime}_{2n-2k+1}\big{)}.$
(3.31)

Then (3.28) follows easily from an induction using Prop. 3.1. We leave the
details to the interested reader. ∎

##  4 Relations between Berndt-type Integrals and Series

In this section, we will establish some explicit relations between the Berndt-
type integrals (1.8), (1.9) with $b=2$ and some hyperbolic summations by using
the contour integrations. To save space, throughout this section we will put

${\tilde{n}}=\displaystyle\frac{2n-1}{2}.$

######  Lemma 4.1.

_(cf.[15])_ Let $n$ be an integer. Then we have

$\displaystyle\displaystyle\frac{\pi}{\sinh(\pi s)}\buildrel s\to
ni\over{=}(-1)^{n}\left(\displaystyle\frac{1}{s-ni}+2\displaystyle\sum_{k=1}^{\infty}(-1)^{k}\bar{\zeta}(2k)(s-ni)^{2k-1}\right),$
(4.32) $\displaystyle\displaystyle\frac{\pi}{\cosh(\pi s)}\buildrel
s\to{\tilde{n}}i\over{=}(-1)^{n}i\left(\displaystyle\frac{1}{s-{\tilde{n}}i}+2\displaystyle\sum_{k=1}^{\infty}(-1)^{k}\bar{\zeta}(2k)(s-{\tilde{n}}i)^{2k-1}\right),$
(4.33)

where ${\bar{\zeta}}(s)$ denotes the alternating Riemann zeta function defined
by

$\displaystyle\bar{\zeta}\left(s\right):=\displaystyle\sum\limits_{n=1}^{\infty}{\displaystyle\frac{{{{\left({-1}\right)}^{n-1}}}}{{{n^{s}}}}}\quad(\Re(s)\geq
1).$

Recall that the Riemann zeta function $\zeta(s)$ is defined by

$\displaystyle\zeta(s):=\displaystyle\sum\limits_{n=1}^{\infty}{\displaystyle\frac{1}{n^{s}}}\quad(\Re(s)>1).$

When $s=2m\ (m\in\mathbb{N})$ is an even number Euler proved the famous
formula

$\displaystyle\zeta(2m)=-\displaystyle\frac{1}{2}\displaystyle\frac{B_{2m}}{(2m)!}(2\pi
i)^{2m},$ (4.34)

where $B_{2m}$ is Bernoulli number which are defined by the generating
function

$\displaystyle\frac{x}{e^{x}-1}=\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{B_{n}}{n!}x^{n}.$

In particular,
$B_{0}=1,B_{1}=-\displaystyle\frac{1}{2},B_{2}=\displaystyle\frac{1}{6},B_{4}=-\displaystyle\frac{1}{30},B_{6}=\displaystyle\frac{1}{42}$
and $B_{8}=-\displaystyle\frac{1}{30}$.

From Lemma 4.1, by direct calculations, we obtain

$\displaystyle\displaystyle\frac{(-1)^{n}}{{\cosh\left(\displaystyle\frac{1+i}{2}z\right)}}\mathop{=}\limits^{z\to{\tilde{n}}\pi(1+i)}2i\left\\{\displaystyle\frac{1}{1+i}\cdot\displaystyle\frac{1}{z-{\tilde{n}}\pi(1+i)}\atop+\displaystyle\sum_{k=1}^{\infty}(-1)^{k}\displaystyle\frac{{\bar{\zeta}}(2k)}{\pi^{2k}}\left(\displaystyle\frac{1+i}{2}\right)^{2k-1}\left(z-{\tilde{n}}\pi(1+i)\right)^{2k-1}\right\\},$
(4.35)
$\displaystyle\displaystyle\frac{(-1)^{n}}{{\cosh\left(\displaystyle\frac{1-i}{2}z\right)}}\mathop{=}\limits^{z\to{\tilde{n}}\pi(i-1)}2i\left\\{\displaystyle\frac{1}{1-i}\cdot\displaystyle\frac{1}{z-{\tilde{n}}\pi(i-1)}\atop+\displaystyle\sum_{k=1}^{\infty}(-1)^{k}\displaystyle\frac{{\bar{\zeta}}(2k)}{\pi^{2k}}\left(\displaystyle\frac{1-i}{2}\right)^{2k-1}\left(z-{\tilde{n}}\pi(i-1)\right)^{2k-1}\right\\},$
(4.36)
$\displaystyle\displaystyle\frac{(-1)^{n}}{{\sinh\left(\displaystyle\frac{1+i}{2}z\right)}}\mathop{=}\limits^{z\to
n\pi(1+i)}2\left\\{\displaystyle\frac{1}{1+i}\cdot\displaystyle\frac{1}{z-n\pi(1+i)}\atop+\displaystyle\sum_{k=1}^{\infty}(-1)^{k}\displaystyle\frac{{\bar{\zeta}}(2k)}{\pi^{2k}}\left(\displaystyle\frac{1+i}{2}\right)^{2k-1}\left(z-n\pi(1+i)\right)^{2k-1}\right\\},$
(4.37)
$\displaystyle\displaystyle\frac{(-1)^{n}}{{\sinh\left(\displaystyle\frac{i-1}{2}z\right)}}\mathop{=}\limits^{z\to
n\pi(1-i)}2\left\\{\displaystyle\frac{1}{i-2}\cdot\displaystyle\frac{1}{z-n\pi(1-i)}\atop+\displaystyle\sum_{k=1}^{\infty}(-1)^{k}\displaystyle\frac{{\bar{\zeta}}(2k)}{\pi^{2k}}\left(\displaystyle\frac{i-1}{2}\right)^{2k-1}\left(z-n\pi(1-i)\right)^{2k-1}\right\\}.$
(4.38)

######  Theorem 4.2.

For any $a\geq 0$, we have

$\displaystyle\displaystyle\frac{(1-i^{a+1})}{\pi^{a}(1+i)^{a-1}}\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{a}dx}{(\cos
x+\cosh x)^{2}}$
$\displaystyle=2\pi\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{{\tilde{n}}^{a}\sinh({\tilde{n}})}{\cosh^{3}({\tilde{n}})}-a\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{{\tilde{n}}^{a-1}}{\cosh^{2}({\tilde{n}})}.$

###### Proof.

Consider

$\displaystyle\displaystyle\lim_{R\rightarrow\infty}\displaystyle\int_{C_{R}}\displaystyle\frac{z^{a}dz}{(\cos
z+\cosh
z)^{2}}=\displaystyle\lim_{R\rightarrow\infty}\displaystyle\int_{C_{R}}F(z)dz,$

where $C_{R}$ denotes the positively oriented quarter-circular contour
consisting of the interval $[0,R]$, the quarter-circle $\Gamma_{R}$ with
$|z|=R$ and $0\leq\arg\leq\pi/2$, and $[iR,0]$ (i.e., the line segment from
$iR$ to $0$ on the imaginary axis). Clearly, there exist poles of order $2$ in
the present case when

$\displaystyle\cos z+\cosh
z=2\cos\left\\{\displaystyle\frac{1}{2}(z+iz)\right\\}\cos\left\\{\displaystyle\frac{1}{2}(z-iz)\right\\}=0.$

Hence, there exist one set of poles at

$\displaystyle z_{n}:=\displaystyle\frac{(2n-1)\pi
i}{1+i}={\tilde{n}}(1+i)\pi,\quad n\in\mathbb{Z},$

and

$\displaystyle s_{n}:=\displaystyle\frac{(2n-1)\pi
i}{1-i}={\tilde{n}}(i-1)\pi,\quad n\in\mathbb{Z}.$

The only poles lying inside $C_{R}$ are $z_{n},\ n\geq 1,\ |z_{n}|<R$. From
(4.35), we have

$\displaystyle\displaystyle\frac{1}{{\cosh\left(\displaystyle\frac{1+i}{2}z\right)}}\mathop{=}\limits^{z\to
z_{n}}(-1)^{n}i\left\\{\displaystyle\frac{2}{1+i}\cdot\displaystyle\frac{1}{z-z_{n}}-\displaystyle\frac{1+i}{12}\left(z-z_{n}\right)+o(z-z_{n})\right\\}.$

Further, one obtains

$\displaystyle\displaystyle\frac{1}{{\cosh^{2}\left(\displaystyle\frac{1+i}{2}z\right)}}\mathop{=}\limits^{z\to
z_{n}}-\displaystyle\frac{4}{(1+i)^{2}}\displaystyle\frac{1}{(z-z_{n})^{2}}+\displaystyle\frac{1}{3}+o(1).$

Hence, if $z\rightarrow z_{n}$ then

$\displaystyle F(z)\mathop{=}\limits^{z\to
z_{n}}\displaystyle\frac{z^{a}}{4{\cosh^{2}\left(\displaystyle\frac{1-i}{2}z\right)}}\left\\{-\displaystyle\frac{4}{(1+i)^{2}}\displaystyle\frac{1}{(z-z_{n})^{2}}+\displaystyle\frac{1}{3}+o(1)\right\\}.$

The residue ${\rm Res}[F(z),z_{n}]$ at each such pole is give by

$\displaystyle{\rm Res}[F(z),z_{n}]$
$\displaystyle=\displaystyle\lim_{z\rightarrow
z_{n}}\displaystyle\frac{d}{dz}\left\\{(z-z_{n})^{2}\displaystyle\frac{z^{a}}{(\cos
z+\cosh z)^{2}}\right\\}$ $\displaystyle=\displaystyle\lim_{z\rightarrow
z_{n}}\displaystyle\frac{d}{dz}\displaystyle\frac{z^{a}}{4{\cosh^{2}\left(\displaystyle\frac{1-i}{2}z\right)}}\left\\{-\displaystyle\frac{4}{(1+i)^{2}}+\displaystyle\frac{1}{3}(z-z_{n})^{2}+o((z-z_{n})^{2})\right\\}$
$\displaystyle=-a(1+i)^{a-3}\displaystyle\frac{\left({\tilde{n}}\pi\right)^{a-1}}{\cosh^{2}\left({\tilde{n}}\pi\right)}+(1-i)(1+i)^{a-2}\left({\tilde{n}}\pi\right)^{a}\displaystyle\frac{\sinh\left({\tilde{n}}\pi\right)}{\cosh^{3}\left({\tilde{n}}\pi\right)}.$
(4.39)

For the integral over $[iR,0]$, we let $z=ir$. Hence,

$\displaystyle\displaystyle\int_{iR}^{0}\displaystyle\frac{z^{a}dz}{(\cos
z+\cosh z)^{2}}=\displaystyle\int_{R}^{0}\displaystyle\frac{i^{a+1}rdr}{(\cos
ir+\cosh
ir)^{2}}=-i^{a+1}\displaystyle\int_{0}^{R}\displaystyle\frac{r^{a}dr}{(\cos
r+\cosh r)^{2}}.$ (4.40)

It is easy to show that, as $R\rightarrow\infty$

$\displaystyle\displaystyle\int_{\Gamma_{R}}\displaystyle\frac{z^{a}dz}{(\cos
z+\cosh z)^{2}}=o(1).$ (4.41)

Applying the residue theorem, letting $R\rightarrow\infty$, and collecting
(4)-(4.41), we conclude that

$\displaystyle(1-i^{a+1})\displaystyle\int_{0}^{\infty}\displaystyle\frac{z^{a}dz}{(\cos
z+\cosh z)^{2}}=2\pi i\displaystyle\sum_{n=1}^{\infty}{\rm Res}[F(z),z_{n}].$
(4.42)

This completes the proof of the theorem. ∎

Letting $a=1,3$ yields

$\displaystyle 2\displaystyle\int_{0}^{\infty}\displaystyle\frac{xdx}{(\cos
x+\cosh
x)^{2}}=-\pi\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{1}{\cosh^{2}\left({\tilde{n}}\pi\right)}+\pi^{2}\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)\sinh\left({\tilde{n}}\pi\right)}{\cosh^{3}\left({\tilde{n}}\pi\right)}$
(4.43)

and

$\displaystyle
0=-3\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{\left({\tilde{n}}\pi\right)^{2}}{\cosh^{2}\left({\tilde{n}}\pi\right)}+2\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{\left({\tilde{n}}\pi\right)^{3}\sinh\left({\tilde{n}}\pi\right)}{\cosh^{3}\left({\tilde{n}}\pi\right)}.$
(4.44)

From [16], we find that

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{2}}{\cosh^{2}\left({\tilde{n}}\pi\right)}=\displaystyle\frac{\Gamma^{8}(1/4)}{192\pi^{6}}.$

Hence, together with (4.44) this gives

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{3}\sinh\left({\tilde{n}}\pi\right)}{\cosh^{3}\left({\tilde{n}}\pi\right)}=\displaystyle\frac{\Gamma^{8}(1/4)}{64\pi^{7}}.$
(4.45)

######  Theorem 4.3.

For any $a\geq 4$, we have

$\displaystyle\displaystyle\frac{(1-i^{a+1})}{\pi^{a}(1+i)^{a-1}}\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{a}dx}{(\cos
x-\cosh
x)^{2}}=a\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{a-1}}{\sinh^{2}(n\pi)}-2\pi\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{a}\cosh(n\pi)}{\sinh^{3}(n\pi)}.$

###### Proof.

Consider

$\displaystyle\displaystyle\lim_{R\rightarrow\infty}\displaystyle\int_{C_{R}}\displaystyle\frac{z^{a}dz}{(\cos
z-\cosh
z)^{2}}=\displaystyle\lim_{R\rightarrow\infty}\displaystyle\int_{C_{R}}G(z)dz,$

where $C_{R}$ denotes the positively oriented quarter-circular contour
consisting of the interval $[0,R]$; the quarter-circle $\Gamma_{R}$ with
$|z|=R$ and $0\leq\arg\leq\pi/2$; and the segment $[iR,0]$ on the imaginary
axis. Clearly, there exist poles of order $2$ in the present case when

$\displaystyle\cos z-\cosh
z=-2\sin\left\\{\displaystyle\frac{1}{2}(z+iz)\right\\}\sin\left\\{\displaystyle\frac{1}{2}(z-iz)\right\\}=0.$

Hence, there exist one set of poles at $t_{n}=n\pi(1-i)$ and
$y_{n}=n\pi(1+i)$, where $n\in\mathbb{N}$. Those lying on the interior of
$C_{R}$ are $y_{n}=n\pi(1+i),n\geq 1,|y_{n}|<R$. Hence, by (4.37), we have

$\displaystyle G(z)\mathop{=}\limits^{z\to
y_{n}}\displaystyle\frac{z^{a}}{4{\sinh^{2}\left(\displaystyle\frac{i-1}{2}z\right)}}\left\\{\displaystyle\frac{4}{(1+i)^{2}}\displaystyle\frac{1}{(z-y_{n})^{2}}-\displaystyle\frac{1}{3}+o(1)\right\\}.$

Further, the residue is

$\displaystyle{\rm Res}[G(z),y_{n}]$
$\displaystyle=\displaystyle\lim_{z\rightarrow
y_{n}}\displaystyle\frac{d}{dz}\left\\{(z-y_{n})^{2}\displaystyle\frac{z^{a}}{(\cos
z-\cosh z)^{2}}\right\\}$ $\displaystyle=\displaystyle\lim_{z\rightarrow
y_{n}}\displaystyle\frac{d}{dz}\displaystyle\frac{z^{a}}{4{\sinh^{2}\left(\displaystyle\frac{i-1}{2}z\right)}}\left\\{\displaystyle\frac{4}{(1+i)^{2}}-\displaystyle\frac{1}{3}(z-y_{n})^{2}+o((z-y_{n})^{2})\right\\}$
$\displaystyle=a(1+i)^{a-3}\displaystyle\frac{(n\pi)^{a-1}}{\sinh^{2}(n\pi)}+(i-1)(1+i)^{a-2}\displaystyle\frac{(n\pi)^{a}\cosh(n\pi)}{\sinh^{3}(n\pi)}.$
(4.46)

The integral over $\Gamma_{R}$ tends to $0$ as $R\rightarrow\infty$. Hence, by
(4) and the residue theorem, we find that

$\displaystyle(1-i^{a+1})\displaystyle\int_{0}^{\infty}\displaystyle\frac{z^{a}dz}{(\cos
z-\cosh z)^{2}}=2\pi i\displaystyle\sum_{n=1}^{\infty}{\rm Res}[G(z),y_{n}].$
(4.47)

Thus, we have finished the proof of the theorem. ∎

Setting $a=7$ yields

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{7}\cosh(n\pi)}{\sinh^{3}(n\pi)}=\displaystyle\frac{7}{2\pi}\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{6}}{\sinh^{2}(n\pi)}.$
(4.48)

From [16], we find that

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{6}}{\sinh^{2}(n\pi)}=\displaystyle\frac{\Gamma^{16}(1/4)}{7\cdot
2^{14}\pi^{12}}.$

Hence,

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{7}\cosh(n\pi)}{\sinh^{3}(n\pi)}=\displaystyle\frac{\Gamma^{16}(1/4)}{2^{15}\pi^{13}}.$
(4.49)

In general, setting $a=4p+1$ and $4p-1$ in Theorems 4.2 and 4.3, we get the
following corollaries.

######  Corollary 4.4.

For any integer $p\geq 0$,

$\displaystyle\displaystyle\frac{(-1)^{p}2^{2p+1}}{\pi^{4p+1}}\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{4p+1}dx}{(\cos
x+\cosh x)^{2}}$
$\displaystyle=\pi\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{4p+1}\sinh\left({\tilde{n}}\pi\right)}{\cosh^{3}\left({\tilde{n}}\pi\right)}-(4p+1)\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n-1)^{4p}}{\cosh^{2}\left({\tilde{n}}\pi\right)}.$

######  Corollary 4.5.

For any integer $p\geq 1$,

$\displaystyle\displaystyle\frac{(-1)^{p}}{2^{2p-1}\pi^{4p+1}}\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{4p+1}dx}{(\cos
x-\cosh x)^{2}}$
$\displaystyle=(4p+1)\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4p}}{\sinh^{2}(n\pi)}-2\pi\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4p+1}\cosh(n\pi)}{\sinh^{3}(n\pi)}.$

######  Corollary 4.6.

For any positive integer $p$,

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{4p-1}\sinh\left({\tilde{n}}\pi\right)}{\cosh^{3}\left({\tilde{n}}\pi\right)}=\displaystyle\frac{4p-1}{\pi}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n-1)^{4p-2}}{\cosh^{2}\left({\tilde{n}}\pi\right)}.$
(4.50)

######  Corollary 4.7.

For any positive integer $p\geq 2$,

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4p-1}\cosh(n\pi)}{\sinh^{3}(n\pi)}=\displaystyle\frac{4p-1}{2\pi}\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4p-2}}{\sinh^{2}(n\pi)}.$
(4.51)

######  Remark 4.8.

It is possible to establish the explicit relations between the Berndt-type
integrals with $b\geq 3$ in (1.8), (1.9) and some hyperbolic summations by
using the method of contour integration.

##  5 Berndt-type integrals

In this section, we first provide a weaker version of Theorem 1.3 whose proof
is much shorter.

######  Theorem 5.1.

For any positive integer $p$, the integrals

$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{4p+1}}{(\cos
x+\cosh
x)^{2}}dx\quad\text{and}\quad\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{4p+1}}{(\cos
x-\cosh x)^{2}}dx$ (5.52)

can be evaluated by $\Gamma(1/4)$ and $\pi$.

###### Proof.

From [16, Thm. 3.6], we know that the series

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{2p}}{\sinh^{2}(ny)}\quad\text{and}\quad\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{2p}}{\cosh^{2}((2n-1)y/2)}$

can be expressed in terms of $x,z$ and $z^{\prime}$. Observing that

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{2p+1}\cosh(ny)}{\sinh^{3}(ny)}=-\displaystyle\frac{1}{2}\displaystyle\frac{dx}{dy}\displaystyle\frac{d}{dx}\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{2p}}{\sinh^{2}(ny)}$

and

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{2p+1}\sinh((2n-1)y/2)}{\cosh^{3}((2n-1)y/2))}=-\displaystyle\frac{dx}{dy}\displaystyle\frac{d}{dx}\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{2p}}{\cosh^{2}((2n-1)y/2)}$

with

$\displaystyle dx/dy=-x(1-x)z^{2},$ (5.53)

(see [2, pp. 120, Entry 9(i)]) we conclude that the two series

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{2p+1}\cosh(ny)}{\sinh^{3}(ny)}\quad\text{and}\quad\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{2p+1}\sinh((2n-1)y/2)}{\cosh^{3}((2n-1)y/2))}$

can be expressed in terms of a linear combinations of $x,z,z^{\prime}$ and
$z^{\prime\prime}$. Further, using (2.14),(2.18) and taking $x=1/2,n=0,1,2$,
we obtain that

$\displaystyle
y=\pi,z\Big{(}\displaystyle\frac{1}{2}\Big{)}=\displaystyle\frac{\Gamma^{2}\Big{(}\displaystyle\frac{1}{4}\Big{)}}{2\pi^{3/2}},\
z^{\prime}\Big{(}\displaystyle\frac{1}{2}\Big{)}=4\displaystyle\frac{\sqrt{\pi}}{\Gamma^{2}\Big{(}\displaystyle\frac{1}{4}\Big{)}}\quad{\rm
and}\quad
z^{\prime\prime}\Big{(}\displaystyle\frac{1}{2}\Big{)}=\displaystyle\frac{\Gamma^{2}\Big{(}\displaystyle\frac{1}{4}\Big{)}}{2\pi^{3/2}},$
(5.54)

where we have used the two classical relations

$\Gamma(x+1)=x\Gamma(x)\quad{\rm
and}\quad\Gamma(x)\Gamma(1-x)=\displaystyle\frac{\pi}{\sin(\pi x)}.$

Hence, all of the four series

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{2p}}{\sinh^{2}(n\pi)},$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{2p}}{\cosh^{2}((2n-1)\pi/2)},$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{2p+1}\cosh(n\pi)}{\sinh^{3}(n\pi)},\quad$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{2p+1}\sinh((2n-1)\pi/2)}{\cosh^{3}((2n-1)\pi/2))}$

can be evaluated by $\Gamma(1/4)$ and $\pi$. Finally, using Corollaries 4.4
and 4.5 with the help of the above, we obtain the desired description. ∎

From [16, Cor. 3.7 and Cor. 3.8], we can calculate some explicit formulas. For
example, from [16, Eqs. (3.73),(3.74) and (3.83)], we have

$\displaystyle
S_{2,2}(y)=\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{2}}{\sinh^{2}(ny)}=\displaystyle\frac{x(1-x)z^{2}}{24}\big{[}z^{2}-4(1-2x)zz^{\prime}-12x(1-x)(z^{\prime})^{2}\big{]},$
$\displaystyle
S_{4,2}(y)=\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4}}{\sinh^{2}(ny)}=\displaystyle\frac{x(1-x)z^{5}}{120}\big{[}4(1-x+x^{2})z^{\prime}+(2x-1)z\big{]},$
$\displaystyle
C^{\prime}_{2,2}(y)=\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{2}}{\cosh^{2}((2n-1)y/2)}=\displaystyle\frac{x(1-x)z^{3}}{3}\big{[}z-(1-2x)z^{\prime}\big{]}.$

Hence,

$\displaystyle\begin{aligned}
&\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{3}\cosh(ny)}{\sinh^{3}(ny)}=-\displaystyle\frac{1}{2y^{\prime}}\displaystyle\frac{d}{dx}\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{2}}{\sinh^{2}(ny)}\\\
&\qquad=\displaystyle\frac{(x^{2}-x)z^{3}}{48}\left\\{\begin{array}[]{l}24(x^{2}-x)^{2}(z^{\prime})^{3}+4z^{2}\left[(x-3x^{2}+2x^{3})z^{\prime\prime}+(7x^{2}-7x+1)z^{\prime}\right]\\\
+12(x^{2}-x)zz^{\prime}\left[2(x^{2}-x)z^{\prime\prime}+(6x-3)z^{\prime}\right]+(2x-1)z^{3}\end{array}\right\\},\end{aligned}$
$\displaystyle\begin{aligned}
&\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{5}\cosh(ny)}{\sinh^{3}(ny)}=-\displaystyle\frac{1}{2y^{\prime}}\displaystyle\frac{d}{dx}\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4}}{\sinh^{2}(ny)}\\\
&\qquad=\displaystyle\frac{(x-x^{2})z^{6}}{240}\left\\{\begin{array}[]{l}4(x-2x^{2}+2x^{3}-x^{4})zz^{\prime\prime}+2(2-11x+21x^{2}-14x^{3})zz^{\prime}\\\
+20(x-2x^{2}+2x^{3}-x^{4})z^{\prime}+(6x-6x^{2}-1)z^{2}\end{array}\right\\},\end{aligned}$
$\displaystyle\begin{aligned}
&\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{3}\sinh((2n-1)y/2)}{\cosh^{3}((2n-1)y/2))}=-\displaystyle\frac{1}{y^{\prime}}\displaystyle\frac{d}{dx}\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{2}}{\cosh^{2}((2n-1)y/2)}\\\
&\qquad=\displaystyle\frac{(x-x^{2})z^{4}}{3}\left\\{\begin{array}[]{l}3(3x^{2}-x-2x^{3})(z^{\prime})^{2}+(1-2x)z^{2}\\\
+z\left[(3x^{2}-2x^{3}-x)z^{\prime\prime}+(10x-10x^{2}-1)z^{\prime}\right]\end{array}\right\\}.\end{aligned}$

Setting $x=1/2$ yields

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{3}\cosh(n\pi)}{\sinh^{3}(n\pi)}=\displaystyle\frac{\Gamma^{8}(1/4)}{2^{10}\pi^{7}}-\displaystyle\frac{1}{16\pi^{3}},$
(5.55)
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{5}\cosh(n\pi)}{\sinh^{3}(n\pi)}=\displaystyle\frac{\Gamma^{16}(1/4)}{3\cdot
2^{16}\pi^{12}}+\displaystyle\frac{\Gamma^{8}(1/4)}{2^{10}\pi^{8}},$ (5.56)
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{3}\sinh((2n-1)\pi/2)}{\cosh^{3}((2n-1)\pi/2))}=\displaystyle\frac{\Gamma^{8}(1/4)}{64\pi^{7}}.$

We can also obtain more similar evaluations with higher powers of $n$ or
$2n-1$ on the numerators.

##  6 Proof of Theorem 1.1 and first part of Theorem 1.3

We now apply the results from previous sections to prove Theorem 1.1. The
crucial ones are the properties of the $q$-series $\Phi_{1,m}$ listed in Prop.
3.2.

######  Theorem 6.1.

For any positive integer $m$,

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m-2}}{\sinh^{2}(n\pi)}$
$\displaystyle\,\in\displaystyle\frac{\Gamma^{8m}(1/4)}{\pi^{6m}}\mathbb{Q}+\displaystyle\frac{\delta_{m,1}}{\pi^{2}}\mathbb{Q},$
(6.57)
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m-4}}{\sinh^{2}(n\pi)}$
$\displaystyle\,\in\displaystyle\frac{\Gamma^{8m-8}(1/4)}{\pi^{6m-5}}\mathbb{Q}+\delta_{m,1}\mathbb{Q},$
(6.58)
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m-1}\cosh(n\pi)}{\sinh^{3}(n\pi)}$
$\displaystyle\,\in\displaystyle\frac{\Gamma^{8m}(1/4)}{\pi^{6m+1}}\mathbb{Q}+\displaystyle\frac{\delta_{m,1}}{\pi^{3}}\mathbb{Q},$
(6.59)
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m-3}\cosh(n\pi)}{\sinh^{3}(n\pi)}$
$\displaystyle\,\in\displaystyle\frac{\Gamma^{8m}(1/4)}{\pi^{6m}}\mathbb{Q}+\displaystyle\frac{\Gamma^{8m-8}(1/4)}{\pi^{6m-4}}\mathbb{Q}.$
(6.60)

###### Proof.

From the last equation of [18, Table 1(i)], setting $s=1$ and replacing $\pi
c$ by $y$ we get

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{1}{\sinh^{2}(ny)}=4\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n}{e^{2ny}-1}.$
(6.61)

Setting $y=\pi$ yields the well-known identity (see [18, Eq. (20)])

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{1}{\sinh^{2}(n\pi)}=\displaystyle\frac{1}{6}-\displaystyle\frac{1}{2\pi}.$
(6.62)

This settles the $m=1$ case of (6.58). The $m=1$ case of (6.57) is given by
the first entry in [16, Table 4.1] (denoted by $S_{2,2}(\pi)$ in [16,
Conjecture 4.1]). We will also calculate it by (6.64) in Examples 6.2. The
special case of (6.59) when $m=1$ is given by (5.55). This can also be proved
by using (3.29) as is done in Examples 6.2.

As the last special case we also need to prove (6.60) when $m=1$.
Differentiating (6.61) with respect to $y$ we have

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n\cosh(ny)}{\sinh^{3}(ny)}=4\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{2}e^{2ny}}{(e^{2ny}-1)^{2}}=\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{2}}{\sinh^{2}(ny)}.$
(6.63)

Hence, setting $y=\pi$ and using the $m=1$ case of (6.57) just shown above we
get

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n\cosh(n\pi)}{\sinh^{3}(n\pi)}=\displaystyle\frac{\Gamma^{8}(1/4)}{3\cdot
2^{9}\pi^{6}}-\displaystyle\frac{1}{8\pi^{2}}.$

For general cases, by Prop. 3.2 we see that for all $m\geq 1$

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m+2}}{\sinh^{2}(ny)}$
$\displaystyle\,=4\Phi_{1,4m+2}\in z^{4m+4}\mathbb{Q}[x],$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m}}{\sinh^{2}(ny)}$
$\displaystyle\,=4\Phi_{1,4m}\in z^{4m+1}z^{\prime}\mathbb{Q}[x],$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m+3}\cosh(ny)}{\sinh^{3}(n\pi)}$
$\displaystyle\,=2x(1-x)z^{2}\Phi^{\prime}_{1,4m+2}\in
z^{4m+5}z^{\prime}\mathbb{Q}[x],$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m+1}\cosh(ny)}{\sinh^{3}(ny)}$
$\displaystyle\,=2x(1-x)z^{2}\Phi^{\prime}_{1,4m}\in
z^{4m+4}\mathbb{Q}[x]+z^{4m+2}(z^{\prime})^{2}\mathbb{Q}[x]+z^{4m+3}z^{\prime\prime}\mathbb{Q}[x],$

by (5.53). Thus the theorem follows immediately by taking $x=1/2$ and using
(5.54). ∎

######  Example 6.2.

According to the theorem for all $m\geq 1$ we can put

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m-2}}{\sinh^{2}(n\pi)}=$
$\displaystyle\,\alpha_{4m-2}\displaystyle\frac{\Gamma^{8m}(1/4)}{\pi^{6m}}-\displaystyle\frac{\delta_{m,1}}{8\pi^{2}},$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m}}{\sinh^{2}(n\pi)}=$
$\displaystyle\,\alpha_{4m}\displaystyle\frac{\Gamma^{8m}(1/4)}{\pi^{6m+1}},$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m-1}\cosh(n\pi)}{\sinh^{3}(n\pi)}=$
$\displaystyle\,\beta_{4m-1}\displaystyle\frac{\Gamma^{8m}(1/4)}{\pi^{6m+1}}+\displaystyle\frac{\delta_{m,1}\gamma_{3}}{\pi^{3}},$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4m+1}\cosh(n\pi)}{\sinh^{3}(n\pi)}=$
$\displaystyle\,\beta_{4m+1}\displaystyle\frac{\Gamma^{8m+8}(1/4)}{\pi^{6m+6}}+\gamma_{4m+1}\displaystyle\frac{\Gamma^{8m}(1/4)}{\pi^{6m+2}}.$

The following table provides the explicit expression of $\Phi_{1,2m}$ modulo
$R\mathbb{Q}[P,Q,R]$ for small values of $m$, extending those contained in [9,
Table II]. Using the values of $P,Q$ in (3.19) and (3.20) and setting $x=1/2$
we get the third and last row.

$k$ | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20
---|---|---|---|---|---|---|---|---|---|---
$\phantom{\displaystyle\frac{1}{1}}\hskip-4.0pt4\Phi_{1,k}$ | $\tfrac{Q-P^{2}}{72}$ | $\tfrac{PQ}{180}$ | $\tfrac{Q^{2}}{252}$ | $\tfrac{PQ^{2}}{180}$ | $\tfrac{Q^{3}}{132}$ | $\tfrac{7PQ^{3}}{260}$ | $\tfrac{Q^{4}}{12}$ | $\tfrac{539PQ^{4}}{1020}$ | $\tfrac{203Q^{5}}{76}$ | $\tfrac{539PQ^{5}}{20}$
$\phantom{\displaystyle\frac{1}{1}}\hskip-4.0ptx=\tfrac{1}{2}$ | $\tfrac{z^{4}-3z^{2}(z^{\prime})^{2}}{96}$ | $\tfrac{z^{5}z^{\prime}}{160}$ | $\tfrac{z^{8}}{448}$ | $\tfrac{3z^{9}z^{\prime}}{640}$ | $\tfrac{9z^{12}}{2816}$ | $\tfrac{567z^{13}z^{\prime}}{33280}$ | $\tfrac{27z^{16}}{1024}$ | $\tfrac{43659z^{17}z^{\prime}}{174080}$ | $\tfrac{49329z^{20}}{77824}$ | $\tfrac{392931z^{21}z^{\prime}}{40960}$
$\phantom{\displaystyle\frac{1}{1}}\hskip-4.0pt\alpha_{k}$ | $\tfrac{1}{3\cdot 2^{9}}$ | $\tfrac{1}{5\cdot 2^{8}}$ | $\tfrac{1}{7\cdot 2^{14}}$ | $\tfrac{3}{5\cdot 2^{14}}$ | $\tfrac{3^{2}}{11\cdot 2^{20}}$ | $\tfrac{7\cdot 3^{4}}{5\cdot 13\cdot 2^{20}}$ | $\tfrac{3^{3}}{2^{26}}$ | $\tfrac{11\cdot 3^{4}\cdot 7^{2}}{5\cdot 17\cdot 2^{26}}$ | $\tfrac{7\cdot 29\cdot 3^{5}}{19\cdot 2^{32}}$ | $\tfrac{11\cdot 3^{6}\cdot 7^{2}}{5\cdot 2^{32}}$

In particular, if $k=2$ then by setting $x=1/2$ we see that

$4\Phi_{1,2}=\displaystyle\frac{\Gamma^{8}(1/4)}{3\cdot
2^{9}}-\displaystyle\frac{1}{8\pi^{2}}.$ (6.64)

This proves the $m=1$ case of (6.57). We may write down the $k=4,6,8,10$ cases
explicitly below:

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{4}}{\sinh^{2}(n\pi)}=$
$\displaystyle\,\displaystyle\frac{\Gamma^{8}(1/4)}{5\cdot
2^{8}\pi^{7}},\quad$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{6}}{\sinh^{2}(n\pi)}=$
$\displaystyle\,\displaystyle\frac{\Gamma^{16}(1/4)}{7\cdot 2^{14}\pi^{12}},$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{8}}{\sinh^{2}(n\pi)}=$
$\displaystyle\,\displaystyle\frac{3\Gamma^{16}(1/4)}{5\cdot
2^{14}\pi^{13}},\quad$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{10}}{\sinh^{2}(n\pi)}=$
$\displaystyle\,\displaystyle\frac{9\Gamma^{24}(1/4)}{11\cdot
2^{20}\pi^{18}}.$

Similarly, applying the relations (3.31) to compute $\Phi^{\prime}_{1,2m}$’s
and then setting $x=1/2$, we arrive at the following table.

$k$ | 3 | 5 | 7 | 9 | 11 | 13 | 15 | 17 | 19 | 21 | 23
---|---|---|---|---|---|---|---|---|---|---|---
$\phantom{\displaystyle\frac{1}{1}}\hskip-4.0pt\beta_{k}$ | $\tfrac{1}{2^{10}}$ | $\tfrac{1}{3\cdot 2^{16}}$ | $\tfrac{1}{2^{15}}$ | $\tfrac{1}{2^{22}}$ | $\tfrac{3^{2}}{2^{21}}$ | $\tfrac{19\cdot 3^{2}}{7\cdot 2^{28}}$ | $\tfrac{5\cdot 3^{4}}{2^{27}}$ | $\tfrac{67\cdot 3^{3}}{2^{34}}$ | $\tfrac{7\cdot 3^{5}\cdot 29}{2^{33}}$ | $\tfrac{15629\cdot 3^{5}}{11\cdot 2^{40}}$ | $\tfrac{389\cdot 3^{6}\cdot 7^{2}}{2^{39}}$
$\phantom{\displaystyle\frac{1}{1}}\hskip-4.0pt\gamma_{k}$ | $-\tfrac{1}{2^{4}}$ | $\tfrac{1}{2^{10}}$ | $0$ | $\tfrac{3^{3}}{5\cdot 2^{16}}$ | $0$ | $\tfrac{7\cdot 3^{4}}{5\cdot 2^{22}}$ | $0$ | $\tfrac{11\cdot 7^{2}\cdot 3^{4}}{5\cdot 2^{28}}$ | $0$ | $\tfrac{11\cdot 7^{3}\cdot 3^{7}}{5\cdot 2^{34}}$ | $0$

In particular, $k=3$ implies the special case $m=1$ in (6.59). We may write
down the $k=5,7,9,11$ cases explicitly below.

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{5}\cosh(n\pi)}{\sinh^{3}(n\pi)}=\displaystyle\frac{\Gamma^{16}(1/4)}{3\cdot
2^{16}\pi^{12}}+\displaystyle\frac{\Gamma^{8}(1/4)}{2^{10}\pi^{8}},\quad$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{7}\cosh(n\pi)}{\sinh^{3}(n\pi)}=\displaystyle\frac{\Gamma^{16}(1/4)}{2^{15}\pi^{13}},$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{9}\cosh(n\pi)}{\sinh^{3}(n\pi)}=\displaystyle\frac{\Gamma^{24}(1/4)}{2^{22}\pi^{18}}+\displaystyle\frac{27\Gamma^{16}(1/4)}{5\cdot
2^{16}\pi^{14}},\quad$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{n^{11}\cosh(n\pi)}{\sinh^{3}(n\pi)}=\displaystyle\frac{9\Gamma^{24}(1/4)}{2^{21}\pi^{19}}.$

Note that the $k=5$ case agrees with (5.56).

We now prove the evaluation of the first integral in Theorem 1.3.

######  Theorem 6.3.

For any positive integer $p$,

$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{4p+1}}{(\cos
x-\cosh
x)^{2}}dx=i_{p}\displaystyle\frac{\Gamma^{8p}(1/4)}{\pi^{2p}}+j_{p}\displaystyle\frac{\Gamma^{8p+8}(1/4)}{\pi^{2p+4}}$
(6.65)

for some $i_{p},j_{p}\in\mathbb{Q}$.

###### Proof.

This follows from Cor. 4.5 and Theorem 6.1 easily. ∎

######  Example 6.4.

By computing more cases in Theorem 6.1 we can obtain the following
evaluations:

$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{5}}{(\cos
x-\cosh
x)^{2}}dx=-\displaystyle\frac{\Gamma^{8}(1/4)}{2^{8}\pi^{2}}+\displaystyle\frac{\Gamma^{16}(1/4)}{3\cdot
2^{14}\pi^{6}},$
$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{9}}{(\cos
x-\cosh x)^{2}}dx=\displaystyle\frac{27\Gamma^{16}(1/4)}{5\cdot
2^{12}\pi^{4}}-\displaystyle\frac{\Gamma^{24}(1/4)}{2^{18}\pi^{8}},$
$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{13}}{(\cos
x-\cosh x)^{2}}dx=-\displaystyle\frac{567\Gamma^{24}(1/4)}{5\cdot
2^{16}\pi^{6}}+\displaystyle\frac{171\Gamma^{32}(1/4)}{7\cdot
2^{22}\pi^{10}},$
$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{17}}{(\cos
x-\cosh x)^{2}}dx=\displaystyle\frac{43659\Gamma^{32}(1/4)}{5\cdot
2^{20}\pi^{8}}-\displaystyle\frac{1809\Gamma^{40}(1/4)}{2^{26}\pi^{12}},$
$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{21}}{(\cos
x-\cosh x)^{2}}dx=-\displaystyle\frac{8251551\Gamma^{40}(1/4)}{5\cdot
2^{24}\pi^{10}}+\displaystyle\frac{3797847\Gamma^{48}(1/4)}{11\cdot
2^{30}\pi^{14}},$
$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{25}}{(\cos
x-\cosh x)^{2}}dx=\displaystyle\frac{8622870795\Gamma^{48}(1/4)}{13\cdot
2^{28}\pi^{12}}-\displaystyle\frac{138429081\Gamma^{56}(1/4)}{2^{34}\pi^{16}},$
$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{29}}{(\cos
x-\cosh x)^{2}}dx=-\displaystyle\frac{2498907956391\Gamma^{56}(1/4)}{5\cdot
2^{32}\pi^{14}}+\displaystyle\frac{104367224493\Gamma^{64}(1/4)}{2^{38}\pi^{18}}.$

##  7 Some preliminary results of Jacobi elliptic functions

In order to prove Theorem 1.2 and the second part of Theorem 1.3 we need to
switch gears and turn to Jacobi elliptic functions. This large family of
functions have a very long history and many books have been devoted to its
study. We will only need some of them in this paper.

The Jacobi elliptic function ${\rm sn}(u)$ is defined via the inversion of the
elliptic integral

$\displaystyle
u=\displaystyle\int_{0}^{\varphi}\displaystyle\frac{dt}{\sqrt{1-k^{2}\sin^{2}t}}\quad(0<k^{2}<1),$
(7.66)

namely, ${\rm sn}(u):=\sin\varphi$, where $k=\mod u$ is referred to as the
elliptic modulus, and $\varphi={\rm am}(u,k)={\rm am}(u)$ is referred to as
the Jacobi amplitude. The Jacobi elliptic functions ${\rm cn}u$ and ${\rm
dn}u$ may be defined as follows:

$\displaystyle{\rm cn}(u):=\sqrt{1-{\rm sn}^{2}(u)},\quad{\rm
dn}(u):=\sqrt{1-k^{2}{\rm sn}^{2}(u)}.$

Using the Jacobi elliptic functions ${\rm sn}(u)$, ${\rm cn}(u)$ and ${\rm
dn}(u)$, we give the definitions of Jacobi elliptic functions ${\rm nc}(u)$,
${\rm dc}(u)$ and ${\rm ds}(u)$, as follows:

$\displaystyle{\rm nc}(u):=\displaystyle\frac{1}{{\rm cn}(u)},\quad{\rm
dc}(u):=\displaystyle\frac{{\rm dn}(u)}{{\rm cn}(u)},\quad{\rm
ds}(u):=\displaystyle\frac{{\rm dn}(u)}{{\rm sn}(u)}.$

We will need to use the Maclaurin (or Laurent) series of these functions.
First, we consider their higher derivatives.

######  Lemma 7.1.

For any integer $m\geq 0$ there is a polynomial
$f_{m}(X,Y,Z)\in\mathbb{Z}[X,Y,Z]$ such that

$\displaystyle\frac{d^{m}}{du^{m}}{\rm
nc}(u)=f_{m}(\sin{\varphi},\cos{\varphi},k^{2})(1-\sin^{2}{\varphi})^{-m-1/2}\cdot\left\\{\begin{array}[]{ll}(1-k^{2}\sin^{2}{\varphi})^{1/2},&\hbox{if
$m$ is odd;}\\\ 1,&\hbox{if $m$ is even.}\end{array}\right.$

###### Proof.

We proceed by induction on $m$. By definition

${\rm nc}(u)=(1-\sin^{2}{\varphi})^{-1/2}$

so we get $f_{0}(X,Y,Z)=1$. By the chain rule

$\displaystyle\frac{d}{du}{\rm
nc}(u)=\sin{\varphi}\cos{\varphi}(1-\sin^{2}{\varphi})^{-3/2}\displaystyle\frac{d{\varphi}}{du}=\sin{\varphi}\cos{\varphi}(1-\sin^{2}{\varphi})^{-3/2}(1-k^{2}\sin^{2}{\varphi})^{1/2}.$

Thus $f_{1}(X,Y,Z)=XY$. Suppose the lemma holds for $m$, namely,
$f_{m}(X,Y,Z)\in\mathbb{Z}[X,Y,Z]$ for some $m\geq 1$. If $m+1>1$ is even then

$\displaystyle\displaystyle\frac{d^{m+1}}{du^{m+1}}{\rm nc}(u)=$
$\displaystyle\,\displaystyle\frac{d}{d{\varphi}}\Big{(}f_{m}(\sin{\varphi},\cos{\varphi},k^{2})(1-\sin^{2}{\varphi})^{-m-1/2}(1-k^{2}\sin^{2}{\varphi})^{1/2}\Big{)}\cdot(1-k^{2}\sin^{2}{\varphi})^{1/2}$
$\displaystyle=$
$\displaystyle\,\displaystyle\frac{d}{d{\varphi}}\Big{(}f_{m}(\sin{\varphi},\cos{\varphi},k^{2})\Big{)}(1-\sin^{2}{\varphi})^{-m-1/2}(1-k^{2}\sin^{2}{\varphi})$
$\displaystyle\,+(2m+1)f_{m}(\sin{\varphi},\cos{\varphi},k^{2})\sin{\varphi}\cos{\varphi}(1-\sin^{2}{\varphi})^{-m-3/2}(1-k^{2}\sin^{2}{\varphi})$
$\displaystyle\,-f_{m}(\sin{\varphi},\cos{\varphi},k^{2})(1-\sin^{2}{\varphi})^{-m-1/2}(k^{2}\sin{\varphi}\cos{\varphi}).$

Hence

$\displaystyle f_{m+1}(X,Y,Z)=$
$\displaystyle\,\left(Y\displaystyle\frac{\partial}{\partial
X}f_{m}(X,Y,Z)-X\displaystyle\frac{\partial}{\partial
Y}f_{m}(X,Y,Z)\right)(1-X^{2})(1-X^{2}Z)$
$\displaystyle\,+XY\Big{(}(2m+1)(1-X^{2}Z)-Z(1-X^{2})\Big{)}f_{m}(X,Y,Z)\in\mathbb{Z}[X,Y,Z]$

by induction assumption. If $m+1>1$ is odd then

$\displaystyle\displaystyle\frac{d^{m+1}}{du^{m+1}}{\rm nc}(u)=$
$\displaystyle\,\displaystyle\frac{d}{d{\varphi}}\Big{(}f_{m}(\sin{\varphi},\cos{\varphi},k^{2})(1-\sin^{2}{\varphi})^{-m-1/2}\Big{)}\cdot(1-k^{2}\sin^{2}{\varphi})^{1/2}$
$\displaystyle=$
$\displaystyle\,\left\\{\displaystyle\frac{d}{d{\varphi}}\Big{(}f_{m}(\sin{\varphi},\cos{\varphi},k^{2})\Big{)}(1-\sin^{2}{\varphi})^{-m-1/2}\right.$
$\displaystyle\,\left.+(2m+1)f_{m}(\sin{\varphi},\cos{\varphi},k^{2})\sin{\varphi}\cos{\varphi}(1-\sin^{2}{\varphi})^{-m-3/2}\right\\}(1-k^{2}\sin^{2}{\varphi})$

Hence

$\displaystyle f_{m+1}(X,Y,Z)=$
$\displaystyle\,\left(Y\displaystyle\frac{\partial}{\partial
X}f_{m}(X,Y,Z)-X\displaystyle\frac{\partial}{\partial
Y}f_{m}(X,Y,Z)\right)(1-X^{2})(1-X^{2}Z)$
$\displaystyle\,+(2m+1)XY(1-X^{2}Z)f_{m}(X,Y,Z)\in\mathbb{Z}[X,Y,Z]$

by induction assumption. This completes the proof of the lemma. ∎

By Lemma 7.1 we obtain the following corollary at once.

######  Corollary 7.2.

The Maclaurin series expansion of ${\rm nc}(u)$ has the form

$\displaystyle\sum_{m\geq
0}\displaystyle\frac{f_{m}(x)}{m!}u^{m},\quad\text{where
}f_{m}(x)\in\mathbb{Z}[x]\ \forall m\geq 0.$ (7.67)

The proof of the following lemma is completely similar to the above.

######  Lemma 7.3.

For any integer $m\geq 0$ there is a polynomial
$g_{m}(X,Y,Z),h_{m}(X,Y,Z),l_{m}(X,Y,Z)\in\mathbb{Z}[X,Y,Z]$ such that

$\displaystyle\displaystyle\frac{d^{m}}{du^{m}}{\rm sn}(u)=$
$\displaystyle\,g_{m}(\sin{\varphi},\cos{\varphi},k^{2})\cdot\left\\{\begin{array}[]{ll}(1-k^{2}\sin^{2}{\varphi})^{1/2},&\hbox{if
$m$ is odd;}\\\ 1,&\hbox{if $m$ is even,}\end{array}\right.$
$\displaystyle\displaystyle\frac{d^{m}}{du^{m}}{\rm cn}(u)=$
$\displaystyle\,h_{m}(\sin{\varphi},\cos{\varphi},k^{2})(1-\sin^{2}{\varphi})^{-m+1/2}\cdot\left\\{\begin{array}[]{ll}1,&\hbox{if
$m$ is odd;}\\\ (1-k^{2}\sin^{2}{\varphi})^{1/2},&\hbox{if $m$ is
even.}\end{array}\right.$ $\displaystyle\displaystyle\frac{d^{m}}{du^{m}}{\rm
dn}(u)=$
$\displaystyle\,l_{m}(\sin{\varphi},\cos{\varphi},k^{2})\cdot\left\\{\begin{array}[]{ll}1,&\hbox{if
$m$ is odd;}\\\ (1-k^{2}\sin^{2}{\varphi})^{1/2},&\hbox{if $m$ is
even.}\end{array}\right.$

Using the fact the the Maclaurin series of ${\rm sn}(u)$ starts with
$u+O(u^{2})$ we can prove the following results immediately.

######  Corollary 7.4.

There are polynomials where $p_{m}(x),q_{m}(x)\in\mathbb{Q}[x]$ such that the
Maclaurin series of ${\rm dc}(u)$ has the form

$\displaystyle\sum_{m\geq 0}\displaystyle\frac{p_{m}(x)}{m!}u^{m}$ (7.68)

and the Maclaurin series of $u\cdot{\rm ds}(u)$ has the form

$\displaystyle\sum_{m\geq 0}\displaystyle\frac{q_{m}(x)}{m!}u^{m}.$ (7.69)

######  Remark 7.5.

See [8] and the references therein for recent results on Jacobian elliptic
functions.

##  8 Evaluations of hyperbolic summations via Jacobi functions

In this section, we will apply the Fourier series expansions and power series
expansions of Jacobi elliptic functions ${\rm dc},{\rm nc}$ and ${\rm ds}$ to
establish some explicit evaluations of hyperbolic summations.

To begin with, we record the Fourier series expansions of Jacobi elliptic
functions ${\rm nc}(u)$, ${\rm dc}(u)$ and ${\rm ds}(u)$

$\displaystyle{\rm
dc}(u)=\displaystyle\frac{\pi}{2K}\sec\Big{(}\displaystyle\frac{\pi
u}{2K}\Big{)}+\displaystyle\frac{2\pi}{K}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}q^{2n+1}\cos\Big{(}(2n+1)\displaystyle\frac{\pi
u}{2K}\Big{)}}{1-q^{2n+1}},$ (8.70) $\displaystyle{\rm
nc}(u)=\displaystyle\frac{\pi}{2K\sqrt{1-k^{2}}}\sec\Big{(}\displaystyle\frac{\pi
u}{2K}\Big{)}-\displaystyle\frac{2\pi}{K\sqrt{1-k^{2}}}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}q^{2n+1}\cos\Big{(}(2n+1)\displaystyle\frac{\pi
u}{2K}\Big{)}}{1+q^{2n+1}},$ (8.71) $\displaystyle{\rm
ds}(u)=\displaystyle\frac{\pi}{2K}\csc\Big{(}\displaystyle\frac{\pi
u}{2K}\Big{)}-\displaystyle\frac{2\pi}{K}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{q^{2n+1}\sin\Big{(}(2n+1)\displaystyle\frac{\pi
u}{2K}\Big{)}}{1+q^{2n+1}},$ (8.72)

given in [5] and in the classical text [14, pp. 511-512].

Recall that the Euler numbers $E_{m}$ are defined by the generating function

$\sec(x)=\displaystyle\sum_{m=0}^{\infty}\displaystyle\frac{E_{m}}{(2m)!}x^{2m}.$

In particular, $E_{0}=1,E_{1}=1,E_{2}=5,E_{3}=61,E_{4}=1385,\ldots$.

######  Theorem 8.1.

Let $x,y$ and $z$ be defined by (2.14). For any integer $m\geq 0$,

$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{2m}}{e^{(2n+1)y}-1}=\displaystyle\frac{(-1)^{m}}{4}\Big{(}z^{2m+1}p_{2m}(x)-E_{m}\Big{)}\in\mathbb{Q}[x,z],$
(8.73)

where $p_{n}(x)$ appears in the coefficients of the Maclaurin series of ${\rm
dc}(u)$ in (7.68).

###### Proof.

On the one hand, using the notations of (2.14), the (8.70) can be rewritten as

$\displaystyle{\rm dc}(u)$
$\displaystyle=\displaystyle\frac{1}{z}\sec\Big{(}\displaystyle\frac{u}{z}\Big{)}+\displaystyle\frac{4}{z}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}e^{-(2n+1)y}\cos\Big{(}(2n+1)\displaystyle\frac{u}{z}\Big{)}}{1-e^{-(2n+1)y}}$
$\displaystyle=\displaystyle\frac{1}{z}\displaystyle\sum_{m=0}^{\infty}\displaystyle\frac{E_{m}}{(2m)!}\Big{(}\displaystyle\frac{u}{z}\Big{)}^{2m}+\displaystyle\frac{4}{z}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}}{e^{(2n+1)y}-1}\displaystyle\sum_{m=0}^{\infty}\displaystyle\frac{(-1)^{m}}{(2m!)}\Big{(}(2n+1)\displaystyle\frac{u}{z}\Big{)}^{2m}$
$\displaystyle=\displaystyle\sum_{m=0}^{\infty}\left\\{E_{m}+4(-1)^{m}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{2m}}{e^{(2n+1)y}-1}\right\\}\displaystyle\frac{u^{2m}}{(2m)!z^{2m+1}}.$
(8.74)

Thus, comparing the coefficients of $u^{2m}$ in (8) and (7.68) yields (8.73)
quickly. ∎

######  Example 8.2.

With the help of _Mathematica_ , we calculate the power series expansion of
Jacobi elliptic function ${\rm dc}(u)$ as

$\displaystyle{\rm dc}(u)$
$\displaystyle=1+\displaystyle\frac{1}{2}(1-x)u^{2}+\displaystyle\frac{1}{4!}(5-6x+x^{2})u^{4}+\displaystyle\frac{1}{6!}(61-107x+47x^{2}-x^{3})u^{6}$
$\displaystyle\quad+\displaystyle\frac{1}{8!}(1385-3116x+2142x^{2}-412x^{3}+x^{4})u^{8}$
$\displaystyle\quad+\displaystyle\frac{1}{10!}(50521-138933x+130250x^{2}-45530x^{3}+3693x^{4}-x^{5})u^{10}+\cdots.$

Comparing the coefficients of $u^{0},u^{2},u^{4}$ and $u^{6}$, we see that

$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}}{e^{(2n+1)y}-1}=\displaystyle\frac{z}{4}-\displaystyle\frac{1}{4},$
(8.75)
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{2}}{e^{(2n+1)y}-1}=\displaystyle\frac{1}{4}-\displaystyle\frac{1}{4}z^{3}(1-x),$
(8.76)
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{4}}{e^{(2n+1)y}-1}=-\displaystyle\frac{5}{4}+\displaystyle\frac{1}{4}z^{5}(5-6x+x^{2}),$
(8.77)
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{6}}{e^{(2n+1)y}-1}=\displaystyle\frac{61}{4}-\displaystyle\frac{1}{4}z^{7}(61-107x+47x^{2}-x^{3}).$
(8.78)

######  Remark 8.3.

In his recent work [4] J.M. Campbell also applied the method based on Fourier
series expansions related to Jacobi elliptic functions ${\rm dc},{\rm nc}$ to
discover some new identities evaluating hyperbolic infinite sums in terms of
the complete elliptic integrals $K$ and $E$.

######  Theorem 8.4.

Let $x,y$ and $z$ be defined by (2.14). For any positive integer
$m\in\mathbb{N}$,

$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{2m}}{e^{(2n+1)y}+1}=\displaystyle\frac{(-1)^{m}}{4}\Big{(}E_{m}-\sqrt{1-x}z^{2m+1}f_{2m}(x)\Big{)}\in\mathbb{Q}[\sqrt{1-x},x,z],$
(8.79)

where $f_{m}(x)$ appears in the coefficients of the Maclaurin series of ${\rm
nc}(u)$ in (7.2).

###### Proof.

The proof is completely similar to that of Theorem 8.1. On the one hand,
applying (2.14), we obtain

$\displaystyle{\rm
nc}(u)=\displaystyle\frac{1}{\sqrt{1-x}}\displaystyle\sum_{m=0}^{\infty}\left\\{E_{m}-4(-1)^{m}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{2m}}{e^{(2n+1)y}+1}\right\\}\displaystyle\frac{u^{2m}}{(2m)!z^{2m+1}}.$
(8.80)

Comparing the coefficients of $u^{2m}$ in (8.80) and (7.67), we can derive
(8.79) easily. ∎

######  Example 8.5.

Using _Mathematica_ , we can find that

$\displaystyle{\rm nc}(u)$
$\displaystyle=1+\displaystyle\frac{1}{2}u^{2}+\displaystyle\frac{1}{4!}(5-4x)u^{4}+\displaystyle\frac{1}{6!}(61-76x+16x^{2})u^{6}$
$\displaystyle\quad+\displaystyle\frac{1}{8!}(1385-2424x+1104x^{2}-64x^{3})u^{8}$
$\displaystyle\quad+\displaystyle\frac{1}{10!}(50521-113672x+79728x^{2}-16832x^{3}+256x^{4})u^{10}+\cdots.$

Comparing the coefficients of $u^{0},u^{2},u^{4}$ and $u^{6}$, we deduce that

$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}}{e^{(2n+1)y}+1}=\displaystyle\frac{1}{4}-\displaystyle\frac{1}{4}\sqrt{1-x}z,$
(8.81)
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{2}}{e^{(2n+1)y}+1}=\displaystyle\frac{1}{4}\sqrt{1-x}z^{3}-\displaystyle\frac{1}{4},$
(8.82)
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{4}}{e^{(2n+1)y}+1}=\displaystyle\frac{5}{4}-\displaystyle\frac{1}{4}\sqrt{1-x}(5-4x)z^{5},$
(8.83)
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{6}}{e^{(2n+1)y}+1}=-\displaystyle\frac{61}{4}+\displaystyle\frac{1}{4}\sqrt{1-x}(61-76x+16x^{2})z^{7},.$
(8.84)

######  Remark 8.6.

The evaluations (8.75)-(8.78) can be found in [2, Entry 17(vi)-(ix)]. Those in
(8.75), (8.76) and (8.81) can also be found in [4, Eq. (15)], [4, Thm. 1] and
[4, Eq. (35)], respectively.

Utilizing (5.53) and the fact that ($m\in\mathbb{N}$)

$\displaystyle\displaystyle\frac{d}{dx}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{2m}}{e^{(2n+1)y}-1}=-\displaystyle\frac{1}{4}\displaystyle\frac{dy}{dx}\cdot\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{2m+1}}{\sinh^{2}\Big{(}\displaystyle\frac{2n+1}{2}y\Big{)}},$
$\displaystyle\displaystyle\frac{d}{dx}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{2m}}{e^{(2n+1)y}+1}=-\displaystyle\frac{1}{4}\displaystyle\frac{dy}{dx}\cdot\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{2m+1}}{\cosh^{2}\Big{(}\displaystyle\frac{2n+1}{2}y\Big{)}},$

we obtain the following corollary immediately from Theorems 8.1 and 8.4.

######  Corollary 8.7.

For any positive integer $m$,

$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{2m+1}}{\sinh^{2}\Big{(}\displaystyle\frac{2n+1}{2}y\Big{)}}=(-1)^{m}x(1-x)z^{2}\displaystyle\frac{d}{dx}\Big{(}z^{2m+1}p_{2m}(x)\Big{)}\in\mathbb{Q}[x,z,z^{\prime}],$
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{2m+1}}{\cosh^{2}\Big{(}\displaystyle\frac{2n+1}{2}y\Big{)}}=-(-1)^{m}x(1-x)z^{2}\displaystyle\frac{d}{dx}\Big{(}\sqrt{1-x}z^{2m+1}f_{2m}(x)\Big{)}\in\mathbb{Q}[\sqrt{1-x},x,z,z^{\prime}].$

For example, by direct calculations, we have

$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{3}}{\sinh^{2}\Big{(}\displaystyle\frac{2n+1}{2}y\Big{)}}=z^{4}x(1-x)\big{(}z-3z^{\prime}(1-x)\big{)},$
(8.85)
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}(2n+1)^{3}}{\cosh^{2}\Big{(}\displaystyle\frac{2n+1}{2}y\Big{)}}=-\displaystyle\frac{1}{2}\sqrt{1-x}xz^{4}\big{(}6(x-1)z^{\prime}+z\big{)}.$
(8.86)

The identities (8.85) and (8.86) are the same as [16, Eqs. (3.35) and (3.45)].

We now turn to the similar sums as those in Theorem 8.4 but with even power of
$2n+1$ replaced by the odd ones. These sums are closely related to the
Bernoulli number.

######  Theorem 8.8.

For any positive integer $m\geq 1$,

$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{2m-1}}{e^{(2n+1)y}+1}=\displaystyle\frac{1}{8m}\Big{(}(-1)^{m}q_{2m}(x)z^{2m}+2(2^{2m-1}-1)B_{2m}\Big{)}\in\mathbb{Q}[x,z],$
(8.87)

where $q_{n}(x)$ appears in the coefficients of the Maclaurin series of
$u\,{\rm ds}(u)$ in (7.69).

###### Proof.

The proof is completely similar to the proof of Theorem 8.1. Applying (8.72)
gives

$\displaystyle u{\rm ds}(u)$
$\displaystyle=1-\displaystyle\sum_{m=1}^{\infty}\displaystyle\frac{(-1)^{m}2(2^{2m-1}-1)B_{2m}}{(2m)!z^{2m}}u^{2m}$
$\displaystyle\quad+4\displaystyle\sum_{m=1}^{\infty}\displaystyle\frac{(-1)^{m}}{(2m-1)!z^{2m}}\left\\{\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{2m-1}}{e^{(2n+1)y}+1}\right\\}u^{2m},$
(8.88)

where we used the Laurent series expansions of $\csc(x)$ and power series
expansions of $\sin(x)$

$\displaystyle\csc(x)=\displaystyle\sum_{m=0}^{\infty}(-1)^{m-1}\displaystyle\frac{2(2^{2m-1}-1)B_{2m}}{(2m)!}x^{2m-1}\quad\text{and}\quad\sin(x)=-\displaystyle\sum_{m=1}^{\infty}\displaystyle\frac{(-1)^{m}}{(2m-1)!}x^{2m-1}.$

One the other hand, by Corollary 7.4 and comparing the coefficients of
$u^{2m}$ of (8) and (7.69) we arrive at the conclusion. ∎

######  Corollary 8.9.

For any integer $m\geq 0$ we have

$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{2m}}{\cosh^{2}((2n+1)y/2)}$
$\displaystyle\,=\displaystyle\frac{(-1)^{m}}{2m}x(1-x)z^{2m+1}\Big{(}zq^{\prime}_{2m}(x)+2mz^{\prime}q_{2m}(x)\Big{)},$
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{2m+1}\sinh((2n+1)y/2)}{\cosh^{3}((2n+1)y/2))}$
$\displaystyle\,=\displaystyle\frac{(-1)^{m}}{2m}x(1-x)z^{2}\displaystyle\frac{d}{dx}\bigg{\\{}x(1-x)z^{2m+1}\Big{(}zq^{\prime}_{2m}(x)+2mz^{\prime}q_{2m}(x)\Big{)}\bigg{\\}},$

where $q^{\prime}_{m}(x):=\displaystyle\frac{dq_{m}(x)}{dx}$.

###### Proof.

This follows quickly from Theorem 8.8 and the identities

$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{2m}}{\cosh^{2}((2n+1)y/2)}$
$\displaystyle\,=4x(1-x)z^{2}\displaystyle\frac{d}{dx}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{2m-1}}{e^{(2n+1)y}+1},$
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{2m+1}\sinh((2n+1)y/2)}{\cosh^{3}((2n+1)y/2))}$
$\displaystyle\,=x(1-x)z^{2}\displaystyle\frac{d}{dx}\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n+1)^{2m}}{\cosh^{2}((2n+1)y/2)}.$

∎

We can now prove a more precise version of Theorem 1.2.

######  Theorem 8.10.

Set $\Gamma=\Gamma(1/4)$. For any integer $m\geq 0$ we have

$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{4m+2}}{\cosh^{2}((2n+1)\pi/2)}$
$\displaystyle\,=-\displaystyle\frac{q^{\prime}_{4m+2}(1/2)}{2^{4m+7}(2m+1)}\cdot\displaystyle\frac{\Gamma^{8m+8}}{\pi^{6m+6}},$
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{4m}}{\cosh^{2}((2n+1)\pi/2)}$
$\displaystyle\,=\displaystyle\frac{q_{4m}(1/2)}{2^{4m+1}}\cdot\displaystyle\frac{\Gamma^{8m}}{\pi^{6m+1}},$
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{4m+1}\sinh((2n+1)\pi/2)}{\cosh^{3}((2n+1)\pi/2))}$
$\displaystyle\,=\displaystyle\frac{1}{4^{2m+5}m}\left\\{\begin{array}[]{l}\Gamma^{8}\big{(}4m\,q_{4m}(1/2)+q_{4m}^{\prime\prime}(1/2)\big{)}\\\
+2^{8}m(4m+1)q_{4m}(1/2)\pi^{4}\phantom{\displaystyle\frac{1}{1}}\end{array}\right\\}\displaystyle\frac{\Gamma^{8m}}{\pi^{6m+6}},$
$\displaystyle\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{4m+3}\sinh((2n+1)\pi/2)}{\cosh^{3}((2n+1)\pi/2))}$
$\displaystyle\,=\displaystyle\frac{(4m+3)q_{4m+2}^{\prime}(1/2)}{2^{4m+7}(2m+1)}\cdot\displaystyle\frac{\Gamma^{8m+8}}{\pi^{6m+7}}.$

###### Proof.

We have the following classical result (see the bottom of page 47 of Hancock’s
book [6]):

${\rm sn}(iu,\sqrt{1-k^{2}})=i\displaystyle\frac{{\rm sn}(u,k)}{{\rm
cn}(u,k)},\quad{\rm dn}(iu,\sqrt{1-k^{2}})=\displaystyle\frac{{\rm
dn}(u,k)}{{\rm cn}(u,k)},$

where $i=\sqrt{-1}$ as usual. Therefore

$iu{\rm ds}(iu,\sqrt{1-k^{2}})-u{\rm ds}(u,k)=0.$ (8.89)

This implies that for all $m>0$

$q_{4m-2}(1-x)+q_{4m-2}(x)=0.$ (8.90)

Differentiating (8.89) with respect to $x$ we see that for all $m\geq 0$

$q_{4m}^{\prime}(1-x)+q_{4m}^{\prime}(x)=0.$ (8.91)

Differentiating (8.90) twice we get for all $m>0$

$q^{\prime\prime}_{4m-2}(1-x)+q^{\prime\prime}_{4m-2}(x)=0.$ (8.92)

Taking $x=1/2$ in (8.90)–(8.92) we have for all $m>0$

$q_{4m-2}(1/2)=0,\quad q^{\prime}_{4m}(1/2)=0,\quad
q^{\prime\prime}_{4m-2}(1/2)=0.$

Hence the theorem follows from Corollary 8.9 and (5.54). ∎

Note that the above theorem confirms some identities contained in [16,
Conjecture 4.1].

######  Example 8.11.

We have

$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{5}\sinh((2n-1)\pi/2)}{\cosh^{3}((2n-1)\pi/2))}=\displaystyle\frac{3\Gamma^{8}(1/4)}{64\pi^{8}}+\displaystyle\frac{\Gamma^{16}(1/4)}{3\cdot
2^{12}\pi^{12}},$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{7}\sinh((2n-1)\pi/2)}{\cosh^{3}((2n-1)\pi/2))}=\displaystyle\frac{3\Gamma^{16}(1/4)}{2^{10}\pi^{13}},$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{9}\sinh((2n-1)\pi/2)}{\cosh^{3}((2n-1)\pi/2))}=\displaystyle\frac{189\Gamma^{16}(1/4)}{5\cdot
2^{10}\pi^{14}}+\displaystyle\frac{9\Gamma^{24}(1/4)}{2^{16}\pi^{18}},$
$\displaystyle\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{(2n-1)^{11}\sinh((2n-1)\pi/2)}{\cosh^{3}((2n-1)\pi/2))}=\displaystyle\frac{153\Gamma^{24}(1/4)}{2^{14}\pi^{19}}.$

We can now prove a more precise evaluation of the second integral in Theorem
1.3.

######  Theorem 8.12.

For any positive integer $p$,

$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{4p+1}}{(\cos
x+\cosh x)^{2}}dx$
$\displaystyle=-\displaystyle\frac{(-1)^{p}(4p+1)q_{4p}(1/2)}{2^{6p+3}}\displaystyle\frac{\Gamma^{8p}(1/4)}{\pi^{2p}}+\displaystyle\frac{(-1)^{p}}{2^{6p+11}p}\big{(}4p\,q_{4p}(1/2)+q_{4p}^{\prime\prime}(1/2)\big{)}\displaystyle\frac{\Gamma^{8p+8}(1/4)}{\pi^{2p+4}}.$
(8.93)

###### Proof.

By Cor. 4.4 we have

$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{4p+1}dx}{(\cos
x+\cosh x)^{2}}$
$\displaystyle=(-1)^{p}\displaystyle\frac{\pi^{4p+2}}{2^{2p+1}}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{4p+1}\sinh\left(\displaystyle\frac{2n+1}{2}\pi\right)}{\cosh^{3}\left(\displaystyle\frac{2n+1}{2}\pi\right)}-(-1)^{p}(4p+1)\displaystyle\frac{\pi^{4p+1}}{2^{2p+1}}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(2n+1)^{4p}}{\cosh^{2}\left(\displaystyle\frac{2n+1}{2}\pi\right)}$
$\displaystyle=(-1)^{p}\displaystyle\frac{\pi^{4p+2}}{2^{2p+1}}\displaystyle\frac{1}{4^{2p+5}p}\left\\{\begin{array}[]{l}\Gamma^{8}\big{(}4p\,q_{4p}(1/2)+q_{4p}^{\prime\prime}(1/2)\big{)}\\\
+2^{8}p(4p+1)q_{4p}(1/2)\pi^{4}\phantom{\displaystyle\frac{1}{1}}\end{array}\right\\}\displaystyle\frac{\Gamma^{8p}}{\pi^{6p+6}}-(-1)^{p}(4p+1)\displaystyle\frac{\pi^{4p+1}}{2^{2p+1}}\displaystyle\frac{q_{4p}(1/2)}{2^{4p+1}}\cdot\displaystyle\frac{\Gamma^{8p}}{\pi^{6p+1}}$
$\displaystyle=\displaystyle\frac{(-1)^{p}}{2^{6p+11}p}\left\\{\begin{array}[]{l}\Gamma^{8}\big{(}4p\,q_{4p}(1/2)+q_{4p}^{\prime\prime}(1/2)\big{)}\\\
-2^{8}p(4p+1)q_{4p}(1/2)\pi^{4}\phantom{\displaystyle\frac{1}{1}}\end{array}\right\\}\displaystyle\frac{\Gamma^{8p}}{\pi^{2p+4}}$

by Theorem 8.10. ∎

######  Example 8.13.

By Mathematica we can compute the following easily using the formulas in
Theorem 8.12:

$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{5}}{(\cos
x+\cosh
x)^{2}}dx=\displaystyle\frac{3\Gamma^{8}(1/4)}{2^{9}\pi^{2}}-\displaystyle\frac{\Gamma^{16}(1/4)}{3\cdot
2^{15}\pi^{6}},$
$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{9}}{(\cos
x+\cosh x)^{2}}dx=-\displaystyle\frac{189\Gamma^{16}(1/4)}{5\cdot
2^{15}\pi^{4}}+\displaystyle\frac{9\Gamma^{24}(1/4)}{2^{21}\pi^{8}},$
$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{13}}{(\cos
x+\cosh x)^{2}}dx=\displaystyle\frac{18711\Gamma^{24}(1/4)}{5\cdot
2^{21}\pi^{6}}-\displaystyle\frac{5301\Gamma^{32}(1/4)}{7\cdot
2^{27}\pi^{10}},$
$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{17}}{(\cos
x+\cosh x)^{2}}dx=-\displaystyle\frac{5544693\Gamma^{32}(1/4)}{5\cdot
2^{27}\pi^{8}}+\displaystyle\frac{233361\Gamma^{40}(1/4)}{2^{33}\pi^{12}},$
$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{21}}{(\cos
x+\cosh x)^{2}}dx=\displaystyle\frac{4233045663\Gamma^{40}(1/4)}{5\cdot
2^{33}\pi^{10}}-\displaystyle\frac{1940699817\Gamma^{48}(1/4)}{11\cdot
2^{39}\pi^{14}},$
$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{25}}{(\cos
x+\cosh x)^{2}}dx=-\displaystyle\frac{17651016517365\Gamma^{48}(1/4)}{13\cdot
2^{39}\pi^{12}}+\displaystyle\frac{283641186969\Gamma^{56}(1/4)}{2^{45}\pi^{16}},$
$\displaystyle\displaystyle\int_{0}^{\infty}\displaystyle\frac{x^{29}}{(\cos
x+\cosh x)^{2}}dx=\displaystyle\frac{20473552886711463\Gamma^{56}(1/4)}{5\cdot
2^{45}\pi^{14}}-\displaystyle\frac{854871935822163\Gamma^{64}(1/4)}{2^{51}\pi^{18}}.$

We conclude the paper by the following conjecture which is supported by
comparing the coefficients in Examples 6.4 and 8.13.

######  Conjecture 8.14.

For all positive integer $p$ let $i_{p},j_{p},g_{p}$ and $h_{p}$ be the
numbers appearing in Theorems 6.3 and 8.12. Then the following relations hold:

$\displaystyle
g_{p}=-\displaystyle\frac{2^{2p-1}-(-1)^{p}}{2^{2p-1}}i_{p}\quad\text{and}\quad
h_{p}=-\displaystyle\frac{2^{2p-1}+(-1)^{p}}{2^{2p-1}}j_{p}.$

Note that our approach to the two theorems are very different so that it is
highly likely that there is an intimate relation between the two we have not
discovered yet.

Acknowledgments. Ce Xu is supported by the National Natural Science Foundation
of China (Grant No. 12101008), the Natural Science Foundation of Anhui
Province (Grant No. 2108085QA01) and the University Natural Science Research
Project of Anhui Province (Grant No. KJ2020A0057). Jianqiang Zhao is supported
by the Jacobs Prize from The Bishop’s School.

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