Legendreの倍数公式

ガンマ関数とベータ関数の関係式より

${\displaystyle B(2z)=\frac{\mathrm{\Gamma }(z)\mathrm{\Gamma }(z)}{\mathrm{\Gamma }(2z)}}$

また、ベータ関数の三角関数を用いた積分表示より

${\displaystyle B(z,z)=2{\int }_0^{\frac{\pi }{2}}si{n}^{2z-1}\theta co{s}^{2z-1}\theta d\theta }$

${\displaystyle =\frac{2}{2^{2z-1}}{\int }_0^{\frac{\pi }{2}}si{n}^{2z-1}2\theta d\theta }$（倍角公式）

${\displaystyle =\frac{1}{2^{2z-1}}{\int }_0^{\pi }si{n}^{2z-1}\varphi d\varphi }$($2\theta =\varphi$と置換)

${\displaystyle =\frac{2}{2^{2z-1}}{\int }_0^{\frac{\pi }{2}}si{n}^{2z-1}\varphi d\varphi }$(偶関数の性質)

${\displaystyle =\frac{1}{2^{2z-1}}B(z,\frac{1}{2})}$

${\displaystyle =\frac{1}{2^{2z-1}}\cdot \frac{\sqrt{\pi }\mathrm{\Gamma }(z)}{\mathrm{\Gamma }(z+\frac{1}{2})}}$

従って、最初の式と合わせて

${\displaystyle \frac{\mathrm{\Gamma }(z)\mathrm{\Gamma }(z)}{\mathrm{\Gamma }(2z)}=\frac{1}{2^{2z-1}}\cdot \frac{\mathrm{\Gamma }(z)\sqrt{\pi }}{\mathrm{\Gamma }(z+\frac{1}{2})}}$

${\displaystyle ⟺\mathrm{\Gamma }(2z)=\frac{2^{2z-1}}{\sqrt{\pi }}\mathrm{\Gamma }(z)\mathrm{\Gamma }(z+\frac{1}{2})}$