Evaluation of Integral

Evaluate the integral:

$$\int_0^{\infty} \frac{x^3}{e^x-1}\, {\rm d}x$$

Solution

Successively we have:

$$\begin{aligned}
\int_{0}^{\infty}\frac{x^3}{e^x-1}\, {\rm d}x &= \int_{0}^{\infty}e^{-x} x^3 \frac{1}{1-e^{-x}}\, {\rm d}x\\
 &=\int_{0}^{\infty}\sum_{n=0}^{\infty}x^3 e^{-x}e^{-nx}\, {\rm d}x \\
 &= \sum_{n=1}^{\infty}\int_{0}^{\infty}x^3 e^{-nx}\, {\rm d}x\\
 &\overset{u=nx}{=\! =\! =\! } \sum_{n=1}^{\infty}\frac{1}{n^4}\int_{0}^{\infty}u^3 e^{-u}\, {\rm d}u \\
 &=\zeta(4) \Gamma (4) \\
 &= \frac{\pi^4}{90}\cdot 3! \\
 &= \frac{\pi^4}{15}
\end{aligned}$$

In general the above solution also shows that:

$$\int_{0}^{\infty}\frac{x^s}{e^x-1}\,{\rm d}x = \zeta (s+1) \Gamma (s+1), \; s\geq 1$$

where $\zeta$ is the sixth letter of the Greek alphabet and of course the zeta Riemann function and $\Gamma$ is the Gamma Euler's function.