Integral Representation of the Zeta Function: $\zeta(s)=\frac1{\Gamma(s)}\int_{0}^\infty \frac{x^{s-1}}{e^x-1}dx$

Recall that for $t>1$, $$\frac{1}{t-1}=\sum_{n=1}^\infty t^{-n}$$

Then substitute $t=e^{x}$ for $x>0$.

Then substituting $x=\frac{v}{n}$ in the $n$th term of the integral, you get:

$$\int_0^{\infty} x^{s-1}e^{-nx}\,dx=\frac{1}{n^s}\int_0^\infty v^{s-1}e^{-v}dv = \frac{1}{n^s}\Gamma(s)$$

$\newcommand{\dd}{\mathrm{d}}$

The Mellin transform of a function $f$ is given by

\begin{align} \left\{\mathscr{M}f\right\}(s)=\phi(s)=\int_0^{\infty}x^{s-1}f(x)\dd x \end{align}

For the function

\begin{align} f(x)=\frac{1}{e^x - 1} \end{align}

we have

\begin{align} \phi(s)&=\int_0^{\infty}\frac{x^{s-1}}{e^x - 1}\dd x \\ &=\sum_{n \geq1}\int_{0}^{\infty}x^{s-1}e^{-nx}\dd x \end{align}

Making the substitution $u=nx$ gives

\begin{align} \phi(s)&=\sum_{n \geq1}\frac{1}{n^s}\int_{0}^{\infty}u^{s-1}e^{-u}\dd u \\ &=\zeta(s)\Gamma(s) \end{align}

Equating this to the above integral gives the desired result:

\begin{align} \zeta(s)\Gamma(s)&=\int_{0}^{\infty}\frac{x^{s-1}}{e^x - 1}\dd x \\ \zeta(s)&=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}}{e^x - 1}\dd x \end{align}