(0)

\begin{align*} \Gamma(z)\Gamma(1-z) & =-z\Gamma(z)\Gamma(-z)\\ & =-z\lim_{n\rightarrow\infty}n^{z}n!Q^{-1}(z,n+1)n^{-z}n!Q^{-1}(-z,n+1)\\ & =-z\lim_{n\rightarrow\infty}n!\prod_{k=0}^{n}\left(\frac{1}{(z+k)}\right)n!\prod_{k=0}^{n}\left(\frac{1}{(-z+k)}\right)\\ & =\frac{1}{z}\lim_{n\rightarrow\infty}\prod_{k=1}^{n}\frac{k}{(z+k)}\prod_{k=1}^{n}\frac{k}{(-z+k)}\\ & =\frac{1}{z}\prod_{k=1}^{\infty}\frac{k^{2}}{k^{2}-z^{2}}\\ & =\pi\left(\pi z\prod_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2}}\right)\right)^{-1}\\ & =\pi\sin^{-1}(\pi z) \end{align*}

(0)-2

\(0<\Re(z)<1\)とする。
\begin{align*} \Gamma(z)\Gamma(1-z) & =\int_{0}^{\infty}du\int_{0}^{\infty}dvu^{z-1}e^{-u}v^{-z}e^{-v}\\ & =\int_{0}^{\infty}vdt\int_{0}^{\infty}dv(tv)^{z-1}e^{-tv}v^{-z}e^{-v}\qquad,\qquad u=vt\\ & =\int_{0}^{\infty}dt\int_{0}^{\infty}dvt^{z-1}e^{-(t+1)v}\\ & =\int_{0}^{\infty}\frac{t^{z-1}}{t+1}dt\\ \\ & =\int_{0}^{\infty}du\int_{0}^{\infty}udtu^{z-1}e^{-t}(ut)^{-z}e^{-ut}\\ & =\int_{0}^{\infty}du\int_{0}^{\infty}dtt^{-z}e^{-(u+1)t} \end{align*} \[ \lim_{\epsilon\rightarrow0}\lim_{R\rightarrow\infty}\left(\int_{\epsilon}^{R}\frac{x^{z-1}}{x+1}dx+\int_{c(R,0,0\rightarrow2\pi)}\frac{\zeta^{z-1}}{\zeta+1}d\zeta+e^{2\pi iz}\int_{R}^{\epsilon}\frac{x^{z-1}}{x+1}dx+\int_{c(\epsilon,0,2\pi\rightarrow0)}\frac{\zeta^{z-1}}{\zeta+1}d\zeta\right)=2\pi iRes\left[\frac{\zeta^{z-1}}{\zeta+1},-1\right] \] 左辺第2項、第4項と右辺を計算すると、
\begin{align*} \lim_{R\rightarrow\infty}\left|\int_{c(R,0,0\rightarrow2\pi)}\frac{\zeta^{z-1}}{\zeta+1}d\zeta\right| & \leq\lim_{R\rightarrow\infty}\int_{c(R,0,0\rightarrow2\pi)}\left|\zeta^{z-2}\right|\left|\frac{\zeta}{\zeta+1}\right|\left|d\zeta\right|\\ & \leq\lim_{R\rightarrow\infty}R^{z-2}2\pi R\\ & =\lim_{R\rightarrow\infty}2\pi R^{z-1}\\ & =0 \end{align*} \begin{align*} \lim_{\epsilon\rightarrow0}\left|\int_{c(\epsilon,0,2\pi\rightarrow0)}\frac{\zeta^{z-1}}{\zeta+1}d\zeta\right| & \leq\lim_{\epsilon\rightarrow0}\int_{c(\epsilon,0,2\pi\rightarrow0)}\left|\zeta^{z-1}\right|\left|\frac{1}{\zeta+1}\right|\left|d\zeta\right|\\ & \leq\lim_{\epsilon\rightarrow0}\epsilon^{z-1}2\pi\epsilon\\ & =0 \end{align*} \begin{align*} 2\pi iRes\left[\frac{\zeta^{z-1}}{\zeta+1},-1\right] & =2\pi i(-1)^{z-1}\\ & =-2\pi i(-1)^{z}\\ & =-2\pi ie^{i\pi z} \end{align*} となるので、
\begin{align*} -2\pi ie^{i\pi z} & =\lim_{\epsilon\rightarrow0}\lim_{R\rightarrow\infty}\left(\int_{\epsilon}^{R}\frac{x^{z-1}}{x+1}dx+e^{2\pi iz}\int_{R}^{\epsilon}\frac{x^{z-1}}{x+1}dx\right)\\ & =\left(1-e^{2\pi iz}\right)\int_{0}^{\infty}\frac{x^{z-1}}{x+1}dx \end{align*} これより、
\begin{align*} \int_{0}^{\infty}\frac{x^{z-1}}{x+1}dx & =\frac{-2\pi ie^{i\pi z}}{1-e^{2\pi iz}}\\ & =\pi\frac{2i}{e^{i\pi z}-e^{-i\pi z}}\\ & =\pi\sin^{-1}(\pi z) \end{align*}