Evaluating the improper integral $\int_{0}^{\infty} \frac{x^3}{e^{x}-1} dx$

Question

I read somewhere that

$$ \int_{0}^{\infty} \frac{x^3}{e^{x} - 1} = \frac{\pi^4}{15}$$

Does anyone see a way to prove this? My first idea was doing a contour integration and use the residue theorem, but that seems to be a lot of work so if anyone has a better idea I'd be glad to hear it.

Answer 1

For any $n\in\mathbb{N}^+$, $$ \int_{0}^{+\infty}\frac{x^n}{e^x-1}\,dx = \sum_{m\geq 1}\int_{0}^{+\infty}x^n e^{-mx}\,dx = n!\cdot\sum_{m\geq 1}\frac{1}{m^{n+1}} = n!\cdot\zeta(n+1).$$ In our case, $6\cdot\zeta(4)=\frac{\pi^4}{15}$.

Answer 2

HINT: Let $$e^x-1=t\implies x=\ln(t+1)$$$$e^x\ dx=dt\implies dx=\frac{dt}{t+1}$$ $$\int_{0}^{\infty}\frac{x^3}{e^x-1}dx=\int_{0}^{\infty}\frac{(\ln(t+1))^3}{t}\frac{dt}{t+1}$$