TITLE: A more general closed-form of an integral involving a square power of $\theta_4$ - function
QUESTION [6 upvotes]: $\textbf{Problem statement}$. Inspired by the computations at this nospoon
we introduce the following integral:
$$\int_{0}^{\infty }\frac{~\theta _{4}^{2}\left( \exp \left( -\pi \,y\,\beta
\right) \right) }{1+y^{2}}dy\;  \tag{1}\label{1}$$
which is as far as I know only calculated for $\beta =1$.
$$\int_{0}^{\infty }\frac{~\theta _{4}^{2}\left( \exp \left( -\pi \,y\right)
\right) }{1+y^{2}}dx=1   \tag{2}\label{2} $$
My goal is to calculate the integral of \eqref{1} for any $\beta $.
$\textbf{Ansatz}$.With the aid of the well-known representation of the square power of $ \theta_{4}$ Dieckmann
$$\theta _{4}^{2}\left( \exp \left( -\pi \,y\right) \right)
=1+2\sum_{k=1}^{\infty }(-1)^{k}~s{ech}(\pi k\,y)  \tag{3}\label{3}$$
and the transformation $y=\frac{x}{\beta }$, for \eqref{1} follows:
$$\frac{\pi }{2}+2\sum_{k=1}^{\infty }(-1)^{k}\int_{0}^{\infty }\frac{~\beta ~s%
{ech}(\pi k\,x)}{\beta ^{2}+x^{2}}\,dx  \tag{4}\label{4}$$
The integral in the sum:
$$\int_{0}^{\infty }\frac{\beta ~s{ech}(\pi k\,x)}{\beta ^{2}+x^{2}}%
\,dx=\int_{0}^{\infty }\frac{\beta ~}{\beta ^{2}+x^{2}}\frac{1\,}{\cosh
\left( \pi k\,x\right) }dx  \tag{5}\label{5}$$
is done in Sangchul Lee and returns the solution of \eqref{1}:
$$\mathcal{I}\left( \beta \right) =\frac{\pi }{2}+\sum_{k=1}^{\infty
}(-1)^{k}\left( \psi \left( \frac{k\,\beta \ }{2}+\frac{3}{4}\right) -\psi
\left( \frac{k\,\beta \ }{2}+\frac{1}{4}\right) \right) \;  \tag{6}\label{6}$$
Using nospoon
returns $\mathcal {I} \left(1\right) = 1 $. A proof is upon request.
For the readability, here some of the steps performed in Sangchul Lee.
Transformation of \eqref{5} with $y=\frac{x}{%
\beta }$ leads to:
$$\int_{0}^{\infty }\frac{\beta ~}{\beta ^{2}+x^{2}}\frac{1\,}{\cosh \left(
\pi k\,x\right) }dx=\int_{0}^{\infty }\frac{1~}{1+y^{2}}\frac{1\,}{\cosh
\left( a\,y\right) }dy  \tag{7}\label{7}$$
with $a=k~\beta \,\pi $. Transformation
with $z=$ $\frac{a\,y}{\pi }$ leads to:
$$\int_{0}^{\infty }\frac{1~}{1+y^{2}}\frac{1\,}{\cosh \left( a\,y\right) }dy=%
\frac{\pi a}{2}\int_{-\infty }^{\infty }\frac{dz}{\left( a^{2}+\pi
^{2}z^{2}\right) \cosh \left( \pi \,z\right) }  \tag{8}\label{8}$$
In the following, we need the Fourier transform of $f$
$$\widehat{f}\left( \xi \right) =\mathcal{F}\left[ f\left( z\right) \right]
=\int_{\mathcal{R}}f\left( z\right) \exp \left( -2\pi i\xi z\right) ~dz
\tag{9}\label{9}$$
Let
$$f\left( z\right) =s{ech}(\pi \,z),\;g\left( z\right) =\frac{1}{a^{2}+\pi
^{2}z^{2}}  \tag{10}\label{10}$$
and
$$\widehat{f}\left( \xi \right) =s{ech}(\pi \,\xi ),\;\widehat{g}\left( \xi
\right) =\frac{1}{a}\exp \left( -2a\left\vert \xi \right\vert \right) 
\tag{11}\label{11}$$
Also, if both $f$ and $g$ are in $L^{2}$, then
$$\int_{\mathcal{R}}\widehat{f}~g=\int_{\mathcal{R}}f~\widehat{g} \tag{12}\label{12}$$
results to
$$\frac{\pi }{2}+\pi a\sum_{k=1}^{\infty }(-1)^{k}\int_{-\infty }^{\infty }%
\frac{dz}{\left( a^{2}+\pi ^{2}z^{2}\right) \cosh \left( \pi \,z\right) }=%
\frac{\pi }{2}+2\pi \sum_{k=1}^{\infty }(-1)^{k}\int_{0}^{\infty }\frac{\exp
\left( -2\pi k\beta ~z\right) }{\cosh \left( \pi \,z\right) }dz  \tag{13}\label{13}$$
Further transformations then leads finally to the solution \eqref{6}.
Now, for an equivalent integral representation of \eqref{1}, we first perform the sum in \eqref{13}:
$$\mathcal{I}\left( \beta \right) =\frac{\pi }{2}-2\pi \int_{0}^{\infty }\frac{%
s{ech}(\pi \,z)}{1+\exp \left( 2\pi \beta ~z\right) }dz  \tag{14}\label{14}$$
For $\beta =1$, the known value $\mathcal{I}\left( 1\right)
=1$ results. With the aid of Mathematica, 
further analytical expressions for some fixed $\beta$-values can be calculated. 
With \eqref{4} the following identity results:
$$\sum_{k=1}^{\infty }(-1)^{k}\int_{0}^{\infty }\frac{s{ech}(k~\pi \,z)}{\beta
^{2}+z^{2}}dz=-\frac{\pi }{\beta }\int_{0}^{\infty }\frac{s{ech}(\pi \,z)}{%
1+\exp \left( 2\pi \beta ~z\right) }dz  \tag{15}\label{15}$$
The integral form \eqref{14} can be transformed into other interesting expressions. 
Using the known identity Kim:
$$\,_{2}F_{1}\left( a,a;a+1;\frac{1}{2}\right) =2^{a-1}a~\left( \psi \left( 
\frac{a}{2}+\frac{1}{2}\right) -\psi \left( \frac{a}{2}\right) \right)
\tag{16}\label{16}$$
and \eqref{6}, these expressions can be reproduced and further identities can be derived.
For the readability, I omit lots of results, I've found so far. In case of interest, these results can be requested.
$\textbf{1st Question}$
$\textit{Does anybody know how to approach this sum \eqref{6}?} $
$\textit{ Where can I find out more about dealing with the sum?}$
$\textit{Is it possible to derive a simpler expression?}$
$\textbf{2st Question}$
$\textit{Can we find a closed form expression for $\mathcal{I}\left( \beta\right)$, at least for $\beta $ $\in \mathbb{N}$?, distinguishing even/odd $\beta $ ?}$
$\textbf{Bonus Q}$
$\textit{How can I proof the identity \eqref{15} with the help of the Poisson Summation Formula?}$
$\textit{Are there any further results can be obtained by doing so?}$

REPLY [2 votes]: To evaluate the series
\begin{equation}
S=\sum_{k=1}^{\infty
}(-1)^{k}\left( \psi \left( \frac{k\,\beta \ }{2}+\frac{3}{4}\right) -\psi
\left( \frac{k\,\beta \ }{2}+\frac{1}{4}\right) \right)
\end{equation}
we use the Mellin transform representation (Ederlyi 7.3.49 p.353)
\begin{equation}
\Psi(s+a)-\Psi(s+b)=\int_0^\infty x^{s-1}g(x)\,dx
\end{equation}
where
\begin{equation}
g(x)=\begin{cases}
\frac{x^b-x^a}{1-x} &\mbox{if } 0<x<1 \\
0 & \mbox{if } x> 1 \end{cases}
\end{equation}
and  $\Re s>-\Re a,-\Re b$. Here, with $s=k\beta/2$, we can express
\begin{align}
S&=\sum_{k=1}^{\infty
}(-1)^{k}\int_0^1 \frac{x^{1/4}-x^{3/4}}{1-x}x^{\frac{k\beta}{2}-1}\,dx\\
&=\sum_{k=1}^{\infty 
}(-1)^{k}\int_0^1 \frac{x^{\frac{k\beta}{2}-\frac{3}{4}}}{1+x^{1/2}}\,dx\\
&=4\sum_{k=1}^{\infty 
}(-1)^{k}\int_0^1 \frac{u^{2k\beta}}{1+u^2}\,du
\end{align}
By changing the order of summation with the DCT, we find
\begin{align}
S&=4\int_0^1\frac{du}{1+u^2}\sum_{k=1}^{\infty }(-1)^{k}u^{2k\beta}\\
&=-4\int_0^1\frac{u^{2\beta}}{(1+u^2)(1+u^{2\beta})}\,du\\
&=-4\int_1^\infty\frac{du}{(1+u^2)(1+u^{2\beta})}
\end{align}
and thus
\begin{equation}
I(\beta)=\frac{\pi}{2}-4\int_1^\infty\frac{du}{(1+u^2)(1+u^{2\beta})}
\end{equation}
Which is equivalent to the representation (14) in the OP.
One may check, for example, that, using the Laplace transform of $\operatorname{sech}^2$ (Ederlyi 4.9.8)
\begin{align}
I(1)&=\frac{\pi}{2}-4\int_1^\infty\frac{du}{(1+u^2)^2}\\
&=\frac{\pi}{2}-\int_0^\infty \frac{e^{-s}}{\cosh^2s}\,ds\\
&=\frac{\pi}{2}-\left[\frac{1}{2}\left( \Psi\left(\frac{3}{4} \right)-\Psi\left( \frac{1}{4} \right)\right)-1\right]\\
&=1
\end{align}
Unfortunately, this integral has probably no general closed form solution (see, for example,this answer for a similar integral).
However, explicit expressions can be found when $\beta$ is a positive rational number. With $\beta=p/(2q)$, with $p,q$ integers and $q\ge1$, changing $u=v^q$ one obtains
\begin{align}
J\left(\frac{p}{2q}\right)&=\int_1^\infty\frac{du}{(1+u^2)(1+u^{p/q})}\\
&=q\int_1^\infty\frac{v^{q-1}dv}{(1+v^{2q})(1+v^{p})}
\end{align}
It can be classically evaluated using the residue theorem by considering the function
\begin{equation}
f(z)=\frac{z^{q-1}}{(1+z^{2q})(1+z^{p})}\log (z-1)
\end{equation}
defined in the complex plane with a branch cut on the interval $[1,\infty)$ and by integrating it along the keyhole contour. One can express
\begin{equation}
I\left(\frac{p}{2q}\right)=\frac{\pi}{2}+4\sum_{z_j}\Re\left[\operatorname{Res}\left( f(z),z_j \right)\right]
\end{equation}
As an illustration, the case $\beta=2 n$ is calculated below. Here,
\begin{equation}
f(z)=\frac{\log (z-1)}{(1+z^{2})(1+z^{4n})}
\end{equation}
The poles $z_{\pm}=\pm i$ and $z_k=\exp(\frac{\pi}{4n}(2k+1)),k=0,1,\ldots,4p-1$ are simple. We have, using $z_\pm^2=-1)$,
\begin{align}
\operatorname{Res}\left( f(z),z_\pm \right)&=\frac{\log(z_\pm-1)}{2z_\pm(1+z^{4n}_\pm)}\\
&=\frac{z_\pm\log(z_\pm-1)}{2z_\pm^2(1+z^{4n}_\pm)}\\
&=\frac{\mp i\log(\pm i-1)}{4}
\end{align}
and thus
\begin{align}
\Re\operatorname{Res}\left( f(z),i \right)&=-\Re\left[\frac{i}{4}\left( \log\sqrt{2}+\frac{3i\pi}{4} \right)\right]\\
&=\frac{3\pi}{16}\\
\Re\operatorname{Res}\left( f(z),-i \right)&=\Re\left[\frac{i}{4}\left( \log\sqrt{2}+\frac{5i\pi}{4} \right)\right]\\
&=-\frac{5\pi}{16}
\end{align}
Similarly,
\begin{align}
\operatorname{Res}\left( f(z),z_k \right)&=\frac{\log(z_k-1)}{4nz_k^{4n-1}(1+z^{2}_k)}\\
&=\frac{z_k\log(z_k-1)}{4nz_k^{4n}(1+z^{2}_k)}\\
&=-\frac{\log(z_k-1)}{4n\left( z_k+\frac{1}{z_k} \right)}
\end{align}
and thus
\begin{align}
\Re\operatorname{Res}\left( f(z),z_k \right)&=-\frac{1}{16n}\frac{\log\left( 2\left( 1-\cos\left(\frac{\pi}{4n}(2k+1)  \right) \right) \right)}{\cos\left(\frac{\pi}{4n}(2k+1)  \right)}\\
&=-\frac{1}{8n}\frac{\log\left( 2\sin\left(\frac{\pi}{8n}(2k+1)  \right) \right) }{\cos\left(\frac{\pi}{4n}(2k+1)  \right)}
\end{align}
Finally,
\begin{equation}
I(2n)=-\frac{1}{2n}\sum_{k=0}^{4n-1}\frac{\log\left( 2\sin\left(\frac{\pi}{8n}(2k+1)  \right) \right) }{\cos\left(\frac{\pi}{4n}(2k+1)  \right)}
\end{equation}
The case $\beta=2n+1$ is slightly more complicated as the poles $\pm i$ are double.
\end{document}