A more general closed-form of an integral involving a square power of $\theta_4$ - function

$\textbf{Problem statement}$. 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 $?}$

$\textbf{Bonus Q}$

$\textit{How can I proof the identity \eqref{15} with the help of the Poisson Summation Formula?}$

$\textit{Are there any further results can be obtained by doing so?}$