3

I observed that :

$$\int_0^\infty \frac{x}{e^x -1}\mathrm dx = \frac{π^2}{6} = \zeta(2)$$

$$\int_0^\infty \frac{x^2}{e^x -1}\mathrm dx = 2\zeta(3)$$

The first one can be evaluated by turning it into a geometric series and switching summation with integral. I derived second one by integration by parts and following same steps as in first one.

Question:

I want to know if it can be generalized as :

$$\int_0^\infty \frac{x^k}{e^x -1}\mathrm dx = k!\zeta(k+1)$$

a) for $k\in \mathbb N$

b) for (atleast for some) $k\in \mathbb R$

For $k\in\mathbb N$ , it suggests true due to pattern of integration by parts. But I am not sure. Can it be extended to some non-natural numbers ?

An_Elephant
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  • This is a classical result (e.g., http://dlmf.nist.gov/25.5.E1). Write $$ \frac{1}{{{\rm e}^x - 1}} = {\rm e}^{ - x} \frac{1}{{1 - {\rm e}^{ - x} }} = \sum\limits_{n = 1}^\infty {{\rm e}^{ - xn} } $$ and integrate term-by-term. – Gary Jun 14 '23 at 07:30
  • @Gary Thanks ! According to that, it holds for all numbers whose real part is greater than 1. – An_Elephant Jun 14 '23 at 07:32
  • @Gary I tried to see if it's a duplicate or not but the linked question didn't appeared . Thanks. – An_Elephant Jun 14 '23 at 07:34
  • @User0 Thanks Very much. I'll look it surely. – An_Elephant Jun 15 '23 at 10:54

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