# Evaluate $$\int_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} dx$$

How to evaluate

\int\limits_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} dx

was a question posed at Mathematics Stack Exchange. Here is my solution to a more general problem.

We consider the more general integral
\begin{align}
\int\limits_{0}^{\infty} \frac{x^{n}}{\mathrm{e}^{ax}-1} dx
&= \int\limits_{0}^{\infty} \frac{\mathrm{e}^{-ax} x^{n}}{1-\mathrm{e}^{-ax}} dx \\
&= \int\limits_{0}^{\infty} \mathrm{e}^{-ax} x^{n} \sum\limits_{k = 0}^{\infty} \mathrm{e}^{-kax} dx \\
&= \sum\limits_{k = 0}^{\infty} \int\limits_{0}^{\infty} x^{n} \mathrm{e}^{-(a+ak)x} dx
\end{align}

To evaluate the integral, let $$(a+ak)x = z$$

\int\limits_{0}^{\infty} x^{n} \mathrm{e}^{-(a+ak)x} dx
= \frac{1}{a^{n+1}} \frac{1}{(1+k)^{n+1}} \int\limits_{0}^{\infty} z^{n} \mathrm{e}^{-z} dz
= \frac{1}{a^{n+1}} \frac{1}{(1+k)^{n+1}} \Gamma(n+1)

Now we have

\int\limits_{0}^{\infty} \frac{x^{n}}{\mathrm{e}^{ax}-1} dx
= \frac{1}{a^{n+1}} \Gamma(n+1) \sum\limits_{k = 1}^{\infty} \frac{1}{k^{n+1}}
= \frac{1}{a^{n+1}} \Gamma(n+1) \zeta(n+1)

The original problem is

\int\limits_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} dx
= \Gamma(4) \zeta(4) = 6\frac{\pi^{4}}{90} = \frac{\pi^{4}}{15}