Evaluate$ \int _0^{\infty} \frac{\operatorname{sech}^{k}(x)}{\cosh(2\pi/k)-\cos(2x)}\,dx $ for $k=1,2$

Question

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?

Answer 1

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 $$\small \int_0^{\infty } \frac{1}{\cosh ^2(x) (\cosh (\pi )-\cos (2 x))} \, dx=\text{csch}(\pi ) \left(2 \sum _{k=1}^{\infty } \frac{2 \pi k}{e^{2 \pi k}-1}+1\right)=\frac{1}{6} (3+\pi ) \text{csch}(\pi )$$