How to compute $\sum_{n=0}^{\infty} \frac{(-1)^n}{\cosh((n+\frac{1}{2}) \pi)}$?
My attempt was trying to consider the Mellin transform of $\frac{1}{\cosh(x)}$ and use the inverse Mellin transform as a representation of $\frac{1}{\cosh(x)}$
$$\frac{1}{\cosh(x)}=\mathcal{M}^{-1}\Big(2\Gamma(s)\beta(s)\Big)(x) $$ Which implies $$\sum_{n=0}^{\infty} \frac{(-1)^n}{\cosh((n+\frac{1}{2}) \pi)}=\frac{1}{2\pi i}\int_{\sigma - i\infty}^{\sigma +i\infty}2^{s+1} \frac{\Gamma(s)\beta^2(s)}{\pi^s}\, ds$$
This integral didn't lead me anywhere.
I did realize the series may be similar to a particular value of theta functions, but I couldn't find any that arrives to this sum.
Any help is appreciated
From this website (See "Hyperbolic series involving the lemniscate functions") we have:
$$\sum _{n=0}^{\infty } \frac{(-1)^n}{\cosh \left(\left(n+\frac{1}{2}\right) \pi \right)}=\\\sum _{m=0}^{\infty } (-1)^m \text{sech}\left(\left(\frac{1}{2}+m\right) \pi \right)=\\\frac{i \log (1+\text{sech}(\pi ))}{\pi }-\frac{i \psi _{e^{-4 \pi }}^{(0)}\left(\frac{1}{4}-\frac{i}{4}\right)}{\pi }+\frac{2 i \psi _{e^{-2 \pi }}^{(0)}\left(\frac{1}{4}-\frac{i}{4}\right)}{\pi }+\frac{i \psi _{e^{-2 \pi }}^{(0)}\left(\frac{1}{2}-\frac{i}{2}\right)}{\pi }-\frac{2 i \psi _{e^{-\pi }}^{(0)}\left(\frac{1}{2}-\frac{i}{2}\right)}{\pi }$$
Mathematica code:
(I*Log[1 + Sech[Pi]])/Pi - (I*QPolyGamma[0, 1/4 - I/4, E^(-4*Pi)])/Pi + ((2*I)*QPolyGamma[0, 1/4 - I/4, E^(-2*Pi)])/Pi + (I*QPolyGamma[0, 1/2 - I/2, E^(-2*Pi)])/Pi - ((2*I)*QPolyGamma[0, 1/2 - I/2, E^(-Pi)])/Pi