Infinite series involving sech squared PART II

In this post we prove the following infinite sum

$\begin{align*} \sum_{n=-\infty}^\infty \frac{1}{\cosh^2(\pi n )}=\frac{1}{\pi}+\frac{\Gamma^4\left( \frac14\right)}{8 \pi^3} \end{align*}$

Click here for the proof.

We relied in a few previous established results:

$\begin{align*} \vartheta_4^4(q)=1+8\sum_{n=1}^\infty \frac{(-1)^n n q^n}{1+q^n} \end{align*}$

$\begin{align*} \ln \vartheta_2\left(e^{-s \pi}\right)-\ln 2 +\frac{\pi s}{4} &=\frac{1}{2}\sum_{n=1}^\infty \frac{1-\tanh(n \pi s)}{n} \end{align*}$

$\begin{align*} \sum_{n=-\infty}^\infty n^2 e^{-\pi n^2 } &=\frac{1}{4 \pi} \sum_{n=-\infty}^\infty e^{-\pi n^2 } \end{align*}$

$\begin{align*} \vartheta_3(e^{-\pi})= \sum_{n=-\infty}^\infty e^{-\pi n^2}=\frac{\Gamma\left(\frac{1}{4} \right)}{\pi^{3/4}\sqrt{2}} \end{align*}$

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