(0)

\begin{align*} \Gamma(2z) & =\frac{\Gamma(z)\Gamma(z)}{B(z,z)}\\ & =\Gamma^{2}(z)\left(\int_{0}^{1}t^{z-1}(1-t)^{z-1}dt\right)^{-1}\\ & =\Gamma^{2}(z)\left(\frac{1}{4^{z-1}}\int_{0}^{1}\left\{ 1-\left(1-2t\right)^{2}\right\} ^{z-1}dt\right)^{-1}\\ & =\Gamma^{2}(z)\left(\frac{2}{4^{z-1}}\int_{0}^{\frac{1}{2}}\left\{ 1-\left(1-2t\right)^{2}\right\} ^{z-1}dt\right)^{-1}\\ & =\Gamma^{2}(z)\left(-\frac{2}{4^{z}}\int_{1}^{0}\left(1-x\right)^{z-1}x^{-\frac{1}{2}}dx\right)^{-1}\qquad,\qquad1-2t=\sqrt{x}\\ & =\Gamma^{2}(z)\left(\frac{1}{2^{2z-1}}\int_{0}^{1}x^{\frac{1}{2}-1}\left(1-x\right)^{z-1}dx\right)^{-1}\\ & =\Gamma^{2}(z)\left(\frac{1}{2^{2z-1}}B\left(\frac{1}{2},z\right)\right)^{-1}\\ & =\Gamma^{2}(z)\left(\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(z\right)}{2^{2z-1}\Gamma\left(z+\frac{1}{2}\right)}\right)^{-1}\\ & =\frac{2^{2z-1}}{\sqrt{\pi}}\Gamma(z)\Gamma\left(z+\frac{1}{2}\right) \end{align*}

(0)-2

\(2z\in\mathbb{N}\)のときの証明
\(m\in\mathbb{N}\)とする。
\begin{align*} \Gamma(2m) & =\left(2m-1\right)!!\left(2m-2\right)!!\\ & =\left(2^{m}\frac{\left(m-\frac{1}{2}\right)!}{\sqrt{\pi}}\right)2^{m-1}\left(m-1\right)!\\ & =\frac{2^{2m-1}}{\sqrt{\pi}}\Gamma\left(m\right)\Gamma\left(m+\frac{1}{2}\right) \end{align*} \begin{align*} \Gamma\left(2\left(m+\frac{1}{2}\right)\right) & =\Gamma(2m+1)\\ & =\left(2m-1\right)!!\left(2m\right)!!\\ & =\left(2^{m}\frac{\left(m-\frac{1}{2}\right)!}{\sqrt{\pi}}\right)2^{m}m!\\ & =2^{m}\frac{\Gamma\left(m+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)}2^{m}\Gamma\left(m+1\right)\\ & =\frac{2^{2\left(m+\frac{1}{2}\right)-1}}{\sqrt{\pi}}\Gamma\left(m+\frac{1}{2}\right)\Gamma\left(m+\frac{1}{2}+\frac{1}{2}\right) \end{align*} これより\(2z\in\mathbb{N}\)のとき成立する。