- Prove that for \(\operatorname { R e } ( s ) > 1\) ，
\[ \zeta (s) = \frac {1}{\Gamma (s)} \int_ {0} ^ {\infty} \frac {x ^ {s - 1}}{e ^ {x} - 1} d x. \]
[Hint: Write \(\textstyle 1 / ( e ^ { x } - 1 ) = \sum _ { n = 1 } ^ { \infty } e ^ { - n x } . ]\)
Solution
Let \(s=\sigma+i\tau\) with \(\sigma>1\). Since \[ \frac1{e^x-1}=\sum_{n=1}^\infty e^{-nx},\qquad x>0, \] and \[ \sum_{n=1}^\infty\int_0^\infty e^{-nx}x^{\sigma-1}\,dx =\Gamma(\sigma)\sum_{n=1}^\infty n^{-\sigma}<\infty, \] Fubini’s theorem permits termwise integration. Hence \[ \int_0^\infty \frac{x^{s-1}}{e^x-1}\,dx =\sum_{n=1}^\infty\int_0^\infty e^{-nx}x^{s-1}\,dx. \] With \(u=nx\), \[ \int_0^\infty e^{-nx}x^{s-1}\,dx =n^{-s}\int_0^\infty e^{-u}u^{s-1}\,du =n^{-s}\Gamma(s). \] Therefore \[ \int_0^\infty \frac{x^{s-1}}{e^x-1}\,dx =\Gamma(s)\sum_{n=1}^\infty n^{-s} =\Gamma(s)\zeta(s), \] which gives \[ \boxed{\zeta(s)=\frac1{\Gamma(s)}\int_0^\infty\frac{x^{s-1}}{e^x-1}\,dx}, \qquad \Re s>1. \]