How to calculate this integral with contour integration?

Question

Studying contour integration, calculus of residues and complex integration, I came across this integral:

$$I(k)=\int_{0}^{\infty} \frac{\cos(kx)-1}{\sinh^2 (x)} \; dx \;\; \mbox{where} \;k \in \mathbb{R} $$

Could you help me? Thanks

Answer 1

We may assume $k>0$ without loss of generality.
We have $\frac{\sinh(x)}{x}=\prod_{n\geq 1}\left(1+\frac{x^2}{n^2 \pi^2}\right)$ and by applying $-\frac{d^2}{dx^2}\log(\cdot)$ to both sides

$$ \frac{1}{\sinh(x)^2}=\frac{1}{x^2}+\sum_{n\geq 1}\frac{2(x^2-n^2\pi^2)}{(x^2+n^2\pi^2)^2}\tag{A} $$ By the residue theorem $$ \int_{0}^{+\infty}(\cos(kx)-1)\frac{2(x^2-n^2\pi^2)}{(x^2+n^2\pi^2)^2}\,dx=\text{Re}\,2\pi i\,\text{Res}\left(\frac{(x^2-n^2\pi^2)(e^{kix}-1)}{(x^2+n^2\pi^2)^2},x=\pi n i\right) $$ equals $-\pi k e^{-\pi k n}$, hence the original integral equals $$\int_{0}^{+\infty}\frac{\cos(kx)-1}{x^2}\,dx -\pi k\sum_{n\geq 1}e^{-\pi k n} = -\pi k\left(\frac{1}{2}+\frac{1}{e^{\pi k}-1}\right)=-\frac{\pi k}{2}\,\coth\left(\frac{\pi k}{2}\right).\tag{B}$$