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Show that

$$ \int^{\infty}_{0} \frac{x^{3} \, dx}{e^{x}-1} = \frac{\pi^{4}}{15} $$

by expanding the integrand in powers of $e^{-x} $ and integrating term by term.

Could anyone help with this one?

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  • $\begingroup$ In general, $~\displaystyle\int_0^\infty\frac{x^n}{e^{x}-1}~dx ~=~ n!~\zeta(n+1),~$ and $~\displaystyle\int_0^\infty\frac{x^n}{e^{x}+1}~dx ~=~ n!~\eta(n+1).~$ See the Riemann $\zeta$ and Dirichlet $\eta$ functions for more information. $\endgroup$
    – Lucian
    Apr 17, 2015 at 6:01

1 Answer 1

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$$\frac{1}{e^{x}-1}=\frac{e^{-x}}{1-e^{-x}}=\sum_{n=0}^{\infty} e^{-(n+1)x}$$

Can you finish the rest?

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