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!

• See heeehttps://math.stackexchange.com/questions/2136089/contour-integral-of-int-0-infty-fracx-sinh-x-operatornamedx?rq=1 Then by Cauchy you have $$0 = \int_{-\infty}^\infty \left(\frac{x}{\sinh(x)}\right)^3dx - \int_{\epsilon}^\infty\left(\frac{x+i\pi}{\sinh(x+i\pi)}\right)^3dx -\int_{C_\epsilon} \left(\frac{z}{\sinh(z)}\right)^3dz - \int_{-\infty}^{-\epsilon} \left(\frac{x+i\pi}{\sinh(x+i\pi)}\right)^3dx$$ where $C_\epsilon$ is counter-clockwise half semicircle under the pole at $i\pi.$ Dec 11 '17 at 19:46

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 $$\int_{0}^{+\infty}\left(\frac{x}{\sinh x}\right)^3\,dx = 24\sum_{n\geq 1}\frac{n(n+1)}{(2n+1)^4}=\color{red}{\frac{\pi^2(12-\pi^2)}{16}}.$$
\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 $$\int_0^\infty \frac{x^{n-1} }{e^x-1} dx=(n-1)! \zeta(n)$$.