A integral involving Riemann zeta function and Gamma function: $\int_{0}^{\infty}\frac{x^{s-1}}{e^{x}-1}\,dx=\zeta(s)\Gamma(s) $

Question

I need to prove this, today my Instructor solved an integral using this formula but didn't gave a proof $$\displaystyle \int_{0}^{\infty}\dfrac{x^{s-1}}{e^{x}-1}\,\mathrm dx=\zeta(s)\cdot\Gamma(s) $$ I tried to solve it using a series of $e^{x}$ but ended up nowhere.

Answer 1

We have $\int_{0}^{+\infty}z^{s-1}e^{-z}\,dz = \Gamma(s)$ for any $s>0$ by the very definition of the $\Gamma$ function.
Moreover $$ \frac{1}{e^x-1} = e^{-x}+e^{-2x}+e^{-3x}+\ldots $$ with uniform convergence over any compact subset of $\mathbb{R}^+$.
By the dominated convergence theorem it follows that $$\begin{eqnarray*} \int_{0}^{+\infty}\frac{x^{s-1}}{e^x-1}\,dx &=& \sum_{n\geq 1}\int_{0}^{+\infty}x^{s-1}e^{-nx}\,dx\\ &\stackrel{x\mapsto z/n}{=}& \sum_{n\geq 1}\frac{1}{n^s}\int_{0}^{+\infty}z^{s-1}e^{-z}\,dz\\&=&\Gamma(s)\sum_{n\geq 1}\frac{1}{n^s}\\&=&\Gamma(s)\,\zeta(s)\end{eqnarray*}$$ as wanted.