Calculation of integral $ \int_{0}^{\infty} \frac{x}{e^x - 1}\mathrm{d}x$

Question

Apparently the solution is trivial, since Landau-Lifshitz does not even lose a word on it. However, I have no idea, any suggestions?

$$ \int_{0}^{\infty} \frac{x}{e^x - 1} \mathrm{d}x$$

Answer 1

We have $$\int_{0}^{\infty}\frac{x}{e^{x}-1}dx=\int_{0}^{\infty}\frac{e^{-x}x}{1-e^{-x}}dx=\sum_{k\geq1}\int_{0}^{\infty}e^{-kx}xdx=\sum_{k\geq1}\frac{1}{k^{2}}=\color{red}{\frac{\pi^{2}}{6}}.$$ Note that in general holds

$$\zeta\left(s\right)=\frac{1}{\Gamma\left(s\right)}\int_{0}^{\infty}\frac{x^{s-1}}{e^{x}-1}dx,\, s>1.$$

Note that $$\sum_{k\geq1}\int_{0}^{\infty}e^{-kx}xdx\stackrel{kx=y}{=}\sum_{k\geq1}\frac{1}{k^{2}}\int_{0}^{\infty}e^{-y}ydy=\Gamma\left(2\right)\sum_{k\geq1}\frac{1}{k^{2}}=\sum_{k\geq1}\frac{1}{k^{2}}.$$ For a proof of $$\zeta\left(2\right)=\sum_{k\geq1}\frac{1}{k^{2}}=\frac{\pi^{2}}{6}$$ see here.