Showing that $\int_0^{\infty} x^3 / (e^x-1)\,\mathrm{d}x = \pi^4 / 15$

Asked Sep 07 '18 at 13:41

Active Sep 07 '18 at 13:54

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I tried to solve the following by the $u(x)v(x)\,\mathrm{d}x$ method but am stuck at the very first step that I can't take any one of the two as $u$ or $v$.

$$\int_0^{\infty}\frac{x^3}{e^x-1}\,dx=\frac{\pi^4}{15}$$

What is the step by step process to solve this integration?

edited Sep 07 '18 at 13:54

asked Sep 07 '18 at 13:41

Gagan Saggu

1

Took a stab at editing for you. Please confirm this is your question. – Adrian Keister Sep 07 '18 at 13:45
Thanks for editing,Sir. – Gagan Saggu Sep 07 '18 at 13:51
1

See this previous answer: https://math.stackexchange.com/questions/99843/contour-integral-for-x3-ex-1 – AloneAndConfused Sep 07 '18 at 13:54
$\Gamma(4)\zeta(4)$ – Rohan Shinde Sep 07 '18 at 14:08
It is equal to $\Gamma (4) \zeta (4) =3! \frac {\pi^4} {90}=\frac {\pi^4} {15}$. – little o Sep 07 '18 at 14:15

Showing that $\int_0^{\infty} x^3 / (e^x-1)\,\mathrm{d}x = \pi^4 / 15$

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