# Integrate $$\int_{0}^{\infty} \frac{x^{4}}{(\mathrm{e}^{x}-1)^{2}} \mathrm{d} x$$

How to evaluate

\int\limits_{0}^{\infty} \frac{x^{4}}{(\mathrm{e}^{x}-1)^{2}} \mathrm{d} x

was a question on Mathematics Stack Exchange which for some inexplicable reason has been closed. Here is another solution.

Let

f(x) = \frac{1}{(\mathrm{e}^{x}-1)^{2}}

Then the Mellin transform of the function is

\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} \frac{x^{s-1}}{(\mathrm{e}^{x}-1)^{2}} \mathrm{d} x
= \Gamma(s)[\zeta(s-1) – \zeta(s)]

For $$s=5$$, we have

\int\limits_{0}^{\infty} \frac{x^{4}}{(\mathrm{e}^{x}-1)^{2}} \mathrm{d} x = \Gamma(5)[\zeta(4) – \zeta(5)]
= \frac{4\pi^{4}}{15} – 24\zeta(5)