I haven't found out a good contour to evaluate the following integral.
$\displaystyle\int_{0}^{\infty}\dfrac{x\cos ax}{\sinh x}dx$. Well, a natural way should consider the complex integral $\displaystyle\int_{\Gamma}\dfrac{ze^{iaz}}{\sinh z}dz$.
However, I don't know how to find a suitable contour $\Gamma$ to surround the zeros of $\sinh z$ which are $z=\pi ni $ for $n \in \mathbb{Z}$.
I'd really appreciate any help.
Best regards.
You may prove first that: $$ \int_{0}^{+\infty}\sin(ax)e^{-nx}\,dx = \frac{a}{a^2+n^2} = \frac{1}{2}\left(\frac{1}{a+ni}+\frac{1}{a-ni}\right) \tag{1}$$ holds by integration by parts, then by differentiating with respect to $a$: $$ \int_{0}^{+\infty}x\cos(ax)e^{-nx}\,dx = \frac{n^2-a^2}{(n^2+a^2)^2} = -\frac{1}{2}\left(\frac{1}{(a+in)^2}+\frac{1}{(a-in)^2}\right)\tag{2}$$ follows, and: $$ \sum_{n\geq 0}\left(\frac{1}{(a+i(2n+1))^2}+\frac{1}{(a-i(2n+1))^2}\right) = -\frac{\pi^2}{4}\text{sech}^2\left(\frac{\pi a}{2}\right) \tag{3}$$ is a consequence of the Weierstrass product for the $\cosh$ function, since $\text{sech}(z)^2 = \frac{d}{dz^2}\log\cosh(z)$. By $(2)$ and $(3)$ you may easily compute your integral: just expand $\frac{1}{\sinh}$ as a geometric series.
