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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.

user159517
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  • You can expand $\frac{1}{e^x-1} = e^{-x} \cdot \frac{1}{1-e^{-x}}$ into a geometric series. (The monotone convergence theorem says that's fine.) Then recall the $\Gamma$ integral and $\zeta$. – Daniel Fischer Oct 18 '15 at 19:30
  • $\large 6\zeta(4)=\frac{\pi^4}{15}$ is the given result – Peter Oct 18 '15 at 19:35
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    Here : https://en.wikipedia.org/wiki/Riemann_zeta_function , you can see the integral representation of the zeta-function. – Peter Oct 18 '15 at 19:38

2 Answers2

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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, $6\cdot\zeta(4)=\frac{\pi^4}{15}$.

Jack D'Aurizio
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HINT: Let $$e^x-1=t\implies x=\ln(t+1)$$$$e^x\ dx=dt\implies dx=\frac{dt}{t+1}$$ $$\int_{0}^{\infty}\frac{x^3}{e^x-1}dx=\int_{0}^{\infty}\frac{(\ln(t+1))^3}{t}\frac{dt}{t+1}$$

Harish Chandra Rajpoot
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