Definite Integral to Infinity of x over Exponential of x minus One
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Theorem
- $\ds \int_0^\infty \frac x {e^x - 1} \rd x = \frac {\pi^2} 6$
Proof
| \(\ds \int_0^\infty \frac x {e^x - 1} \rd x\) | \(=\) | \(\ds \int_0^\infty \frac {x^{2 - 1} } {e^x - 1} \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \zeta 2 \map \Gamma 2\) | Integral Representation of Riemann Zeta Function in terms of Gamma Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\pi^2} 6 \times 1!\) | Basel Problem, Gamma Function Extends Factorial | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\pi^2} 6\) |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Exponential Functions: $15.79$