TITLE: How to find the definite integral $\int_0^\infty \frac{x}{\sinh ax}\;dx$
QUESTION [5 upvotes]: I'm trying to prove that
$$I:= \int_0^\infty \frac{x}{\sinh(ax)} dx = \frac{\pi^2}{4a^2}$$
Attempt: 
$$\sinh (ax) = \frac{1}{2}(e^{ax}-e^{-ax}) = \frac{1}{2}e^{-ax}(e^{2ax}-1)$$
Now I have $$\int_0^\infty \frac{x}{\sinh(ax)} dx = \int_0^\infty 2x \sum_{q=0}^\infty \frac{\frac{a^qx^q}{q!}}{e^{2ax}-1} dx$$. 
Substituting $y=2ax$ I get: $$I= \frac{1}{2a^2} \sum_{q=0}^\infty \frac{1}{2^qq!} \int_0^\infty \frac{y^{q+1}}{e^y-1}dy=\frac{1}{2a^2} \sum_{q=0}^\infty \frac{1}{2^qq!} \Gamma(q+2) \zeta(q+2)$$
Am I on the right way? What is with the infinite series; why it must have the value $\frac{\pi^2}{2}$? 
Every hints will be highly appreciated.

REPLY [3 votes]: We have:
$$ \int_{0}^{+\infty}\frac{x}{\sinh(a x)}\,dx = \frac{1}{a^2}\int_{0}^{+\infty}\frac{x}{\sinh x}\,dx =\frac{1}{a^2}\int_{1}^{+\infty}\frac{2\log t}{t^2+1}\,dt$$
hence:
$$ \int_{0}^{+\infty}\frac{x}{\sinh(a x)}\,dx = -\frac{2}{a^2}\int_{0}^{1}\frac{\log u}{1+u^2}\,du = \frac{2}{a^2}\cdot\frac{\pi^2}{8}=\color{red}{\frac{\pi^2}{4a^2}}.$$