Evaluate $\int_0^\infty \left(\frac{x}{\sinh x}\right)^3dx$ I need to evaluate $$\int_0^\infty \left(\frac{x}{\sinh x}\right)^3dx$$
I know that I need to use the residue theorem to solve it, but I don't understand how to choose contour properly.
Thank you for any help!
 A: You do not need to use the residue theorem.
By substituting $x=\log t$ we get
$$ \int_{1}^{+\infty}\frac{8t^2\log^3(t)}{(t^2-1)^3}\,dt = \int_{0}^{1}\frac{-8t^2\log^3(t)}{(1-t^2)^3}\,dt $$
where
$$ \frac{t^2}{(1-t^2)^3}=\sum_{n\geq 1}\frac{n(n+1)}{2}\,t^{2n} $$
and 
$$ \int_{0}^{1}t^{2n}(-\log^3 t)\,dt =\frac{6}{(2n+1)^4}$$
lead to
$$ \int_{0}^{+\infty}\left(\frac{x}{\sinh x}\right)^3\,dx = 24\sum_{n\geq 1}\frac{n(n+1)}{(2n+1)^4}=\color{red}{\frac{\pi^2(12-\pi^2)}{16}}. $$
A: Integrate by parts
\begin{align}
I& = \int_0^\infty \left(\frac{x}{\sinh x}\right)^3dx\\
&= -\int_0^\infty x^3 \text{csch} \>x \>d(\coth x)\\
&= -3\int_0^\infty x^2 d(\text{csch} \>x) 
 -\int_0^\infty x^3 \text{csch} \>x \coth^2 xdx\\
&= 6\int_0^\infty x \>\text{csch} \>x \>dx
 -\int_0^\infty x^3 \text{csch} \>x \>dx - I\\
 &= 3\int_0^\infty x \>\text{csch} \>x \>dx
 -\frac12 \int_0^\infty x^3 \text{csch} \>x \>dx\\
&= \frac92 \int_0^\infty \frac{x}{e^x-1} dx
 - \frac{15}{16}\int_0^\infty \frac{x^3}{e^x-1} dx\\
&= \frac92 \zeta(2) - \frac{45}8\zeta(4)
= \frac92 \cdot \frac{\pi^2}6- \frac{45}8\cdot \frac{\pi^4}{90}
=\frac{3\pi^2}4- \frac{\pi^4}{16}\\
\end{align}
Note $\int_0^\infty \frac{x^{n-1} }{e^x-1} dx=(n-1)! \zeta(n)$.
