# Integral representation of the zeta function proof

$$\zeta(s) = \frac{1}{\Gamma(s)}\int^\infty_0\frac{t^{s-1}}{e^t-1}dt$$

$$\textit{proof}$$

Start by the integral representation

$$\int^\infty_0 \frac{e^{-t}t^{s-1}}{1-e^{-t}}\,dt$$

Using the power expansion

$$\frac{1}{1-e^{-t}} = \sum_{n=0}^\infty e^{-nt}$$

Hence we have

$$\int^\infty_0\,e^{-t}t^{s-1}\left(\sum_{n=0}^\infty e^{-nt}\right)\,dt$$

By swapping the series and integral

$$\sum_{n=0}^\infty\int^\infty_0\,t^{s-1}e^{-(n+1)t}\,dt = \Gamma(s) \sum_{n=0}^\infty \frac{1}{(n+1)^s}=\Gamma(s)\zeta(s)\,$$

This entry was posted in Gamma function, Zeta function and tagged , , . Bookmark the permalink.