$\int_{0}^{\infty} x^{a-1}(e^x-1)^{-1}dx $

Question

I need to show that I= $\int_{0}^{\infty} x^{a-1}(e^x-1)^{-1}dx = \Gamma(a) \times \Sigma n^{-a}$ where $a > 1$

I have no clue how to approach !

I am using $\Gamma(a) = \int_{0}^{\infty} t^{a-1}e^{-t}dt$

Answer 1

We have

\begin{align} I &= \int_0^\infty x^{a-1}(1 - e^{-x})^{-1}e^{-x}\, dx\\ &= \int_0^\infty x^{a-1}\sum_{n = 1}^\infty (e^{-x})^n\, dx\\ &= \int_0^\infty x^{a-1}\sum_{n = 1}^\infty e^{-nx}\, dx\\ &= \sum_{n = 1}^\infty \int_0^\infty x^{a-1} e^{-nx}\, dx\\ &= \sum_{n = 1}^\infty \frac{1}{n^{a}} \int_0^\infty x^{a-1}e^{-x}\, dx\\ &= \Gamma(a)\sum_{n = 1}^\infty n^{-a}. \end{align}

The interchange of sum and integral is justified by the monotone convergence theorem.

Answer 2

Hint: Write $$\frac{1}{e^x-1} = \frac{e^{-x}}{1-e^{-x}} = \sum_{k=1}^\infty e^{-kx}$$