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Question Number 134291 by Lordose last updated on 02/Mar/21

Prove \int_{0}^{\infty} \frac{x^{a}}{1 + e^{x}} dx = (1 - 2^{- a}) ζ (a + 1) Γ (a + 1)

Answered by Dwaipayan Shikari last updated on 02/Mar/21

I (a) = \int_{0}^{\infty} \frac{x^{a}}{1 + e^{x}} d x = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \int_{0}^{\infty} e^{- n x} x^{a} d x

= \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \frac{1}{n^{a + 1}} Γ (a + 1)

= Γ (a + 1) ζ (a + 1) (1 - \frac{1}{2^{a + 1 - 1}}) = a! ζ (a + 1) (1 - 2^{- a})