Definite Integral to Infinity of x over Exponential of x minus One

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Theorem

$\displaystyle \int_0^\infty \frac x {e^x - 1} \rd x = \frac {\pi^2} 6$


Proof

\(\displaystyle \int_0^\infty \frac x {e^x - 1} \rd x\) \(=\) \(\displaystyle \int_0^\infty \frac {x^{2 - 1} } {e^x - 1} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \map \zeta 2 \map \Gamma 2\) Integral Representation of Riemann Zeta Function in terms of Gamma Function
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^2} 6 \times 1!\) Basel Problem, Gamma Function Extends Factorial
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^2} 6\)

$\blacksquare$

Sources