# How to find the definite integral $\int_0^\infty \frac{x}{\sinh ax}\;dx$

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.

• Have you tried parameter differentiation on $a$ ? – Gabriel Romon Mar 1 '15 at 18:37
• Have you tried integration by parts? – user207710 Mar 1 '15 at 18:44
• Yes, I have. But in many cases suitable series expansions can be helpful. – kryomaxim Mar 1 '15 at 18:44
• In general, $~\displaystyle\int_0^\infty\frac{x^{k-1}}{\sinh x}~dx~=~\Big(2-2^{1-k}\Big)~\Gamma\big(k\big)~\zeta\big(k\big)~$ – Lucian Mar 1 '15 at 20:26

We may write \begin{align} \int_0^{+\infty} \frac{x}{\sinh (ax)} \:dx&=2\int_0^{+\infty} \frac{x}{e^{ax} - e^{-ax}} \:dx\\\\ &=2\int_0^{+\infty} \frac{x}{1 - e^{-2ax}} e^{-ax}\:dx\\\\ &=2\int_0^{+\infty} x \sum_{n=0}^{\infty}e^{-a(2n+1)x}dx\\\\ &=2\sum_{n=0}^{\infty}\int_0^{+\infty} x \:e^{-a(2n+1)x}dx\\\\ &=\frac{2}{a^2}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\\\\ &=\frac{2}{a^2}\times\frac{\pi^2}{8}\\\\ &=\frac{\pi^2}{4a^2}. \end{align}
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: $$\int_{0}^{+\infty}\frac{x}{\sinh(a x)}\,dx = -\frac{2}{a^2}\int_{0}^{1}\frac{\log u}{1+u^2}\,du = \frac{2}{a^2}\cdot\frac{\pi^2}{8}=\color{red}{\frac{\pi^2}{4a^2}}.$$