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?
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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$$\frac{1}{e^{x}-1}=\frac{e^{-x}}{1-e^{-x}}=\sum_{n=0}^{\infty} e^{-(n+1)x}$$
Can you finish the rest?