TITLE: Evaluate$ \int _0^{\infty} \frac{\operatorname{sech}^{k}(x)}{\cosh(2\pi/k)-\cos(2x)}\,dx $ for $k=1,2$
QUESTION [6 upvotes]: For $k=1,2$, I would like to evaluate
$$
\int _0^{\infty} \frac{\operatorname{sech}^{k}(x)}{\cosh(2\pi/k)-\cos(2x)}\,dx
$$
It's my first time to ask question. I've tried multiple variations of this, but none of them seem to work. Any ideas?

REPLY [5 votes]: By combining $\frac{\sinh (a \pi )}{\cosh (a \pi )-\cos (2 x)}=2 \sum _{k=1}^{\infty } e^{-a\pi k} \cos (2 k x)+1$, $\int_0^{\infty } \frac{\cos (2 k x)}{\cosh (x)} \, dx=\frac{1}{2} \pi  \text{sech}(\pi  k)$ and $\sum _{k=1}^{\infty } \text{sech}(\pi  k)=\frac{8 \sqrt{\pi }}{\Gamma \left(-\frac{1}{4}\right)^2}-\frac{1}{2}$ (all of which are rather elementary) one have
$$\scriptsize \int_0^{\infty } \frac{1}{\cosh (x) (\cosh (2 \pi )-\cos (2 x))} \, dx=\text{csch}(2 \pi ) \left(2 \pi  \sum _{k=1}^{\infty } e^{-\pi k}-\pi  \sum _{k=1}^{\infty } \text{sech}(\pi  k)+\frac{\pi }{2}\right)=\pi  \text{csch}(2 \pi ) \left(\coth \left(\frac{\pi }{2}\right)-\frac{8 \sqrt{\pi }}{\Gamma \left(-\frac{1}{4}\right)^2}\right)$$
Moreover by using $\int_0^{\infty } \frac{\cos (2 k x)}{\cosh ^2(x)} \, dx=\pi  k \text{csch}(\pi  k)$ and $\sum _{k=1}^{\infty } \frac{k}{e^{2 \pi  k}-1}=\frac{1}{24}-\frac{1}{8 \pi }$ one have the other
$$\small \int_0^{\infty } \frac{1}{\cosh ^2(x) (\cosh (\pi )-\cos (2 x))} \, dx=\text{csch}(\pi ) \left(2 \sum _{k=1}^{\infty } \frac{2 \pi  k}{e^{2 \pi k}-1}+1\right)=\frac{1}{6} (3+\pi ) \text{csch}(\pi )$$