Γ関数とζ関数の関係式を一つ紹介

$Re(s)>1となる複素数sについて{\displaystyle {\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{e^x-1}dx=\mathrm{\Gamma }(s)\zeta (s)}$
この式から$\zeta (s)={\displaystyle \frac{1}{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{e^x-1}dx}$となり、これは$\zeta$の定義式でもあります。

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$\begin{array}{rl}{\displaystyle {\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{e^x-1}dx}& ={\displaystyle {\int }_0^{\mathrm{\infty }}\frac{x^{s-1}{e}^{-x}}{1-e^{-x}}dx}\\ & ={\displaystyle {\int }_0^{\mathrm{\infty }}{x}^{s-1}{e}^{-x}\sum _{k=0}^{\mathrm{\infty }}{e}^{-kx}dx}\\ & ={\displaystyle {\int }_0^{\mathrm{\infty }}\sum _{k=1}^{\mathrm{\infty }}{e}^{-kx}{x}^{s-1}dx}\\ & ={\displaystyle \sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-kx}{x}^{s-1}dx}\\ & ={\displaystyle \sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-y}(\frac{y}{k}{)}^{s-1}\frac{dy}{k}}\\ & ={\displaystyle \sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^s}{\int }_0^{\mathrm{\infty }}{e}^{-y}{y}^{s-1}dy}\\ & ={\displaystyle \sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^s}\mathrm{\Gamma }(s)}\\ & =\mathrm{\Gamma }(s)\zeta (s) （証明終）\end{array}$