Γ(n+1/2)を計算する

部分積分を用いる
$\begin{array}{rl}\mathrm{\Gamma }(s+1)& ={\displaystyle {\int }_0^{\mathrm{\infty }}{t}^s(-e^{-t}{)}^{\prime }dt}\\ & =[-t^s{e}^{-t}{]}_0^{\mathrm{\infty }}+s{\displaystyle {\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-t}dt}\\ & =s\mathrm{\Gamma }(s) (証明終)\end{array}$

定理$1$と$\mathrm{\Gamma }(1)=1$となることから自然数$n$について$\mathrm{\Gamma }(n+1)=n!$が成立する

$\begin{array}{rl}{\displaystyle \mathrm{\Gamma }(\frac{1}{2})}& ={\displaystyle {\int }_0^{\mathrm{\infty }}{t}^{-\frac{1}{2}}{e}^{-t}dt}\\ & ={\displaystyle {\int }_0^{\mathrm{\infty }}\frac{1}{y}{e}^{-y^2}2ydy(t=y^2と置換)}\\ & ={\displaystyle 2{\int }_0^{\mathrm{\infty }}{e}^{-y^2}dy}\\ & =\sqrt{\pi } (証明終)\end{array}$
但し最後の式変形でGauss積分を用いた

$\mathrm{\Gamma }(s+1)=s\mathrm{\Gamma }(s)$と${\displaystyle \mathrm{\Gamma }(\frac{1}{2})=\sqrt{\pi }}$を用いると
$$\begin{gathered} {\displaystyle \mathrm{\Gamma }(\frac{3}{2})=\mathrm{\Gamma }(1+\frac{1}{2})=\frac{1}{2}\mathrm{\Gamma }(\frac{1}{2})=\frac{1}{2}\sqrt{\pi }} \\ {\displaystyle \mathrm{\Gamma }(\frac{5}{2})=\mathrm{\Gamma }(1+\frac{3}{2})=\frac{3}{2}\mathrm{\Gamma }(\frac{3}{2})=\frac{3}{2}\frac{1}{2}\sqrt{\pi }} \\ {\displaystyle \mathrm{\Gamma }(\frac{7}{2})=\mathrm{\Gamma }(1+\frac{5}{2})=\frac{5}{2}\mathrm{\Gamma }(\frac{5}{2})=\frac{5}{2}\frac{3}{2}\frac{1}{2}\sqrt{\pi }} \end{gathered}$$\
以下帰納的に${\displaystyle \mathrm{\Gamma }(n+\frac{1}{2})=\frac{(2n-1)!!}{2^n}\sqrt{\pi }}$が得られる
ここで$(2n-1)!!={\displaystyle \frac{(2n)!}{2^{n}n!}}$を用いると
${\displaystyle \mathrm{\Gamma }(n+\frac{1}{2})={\displaystyle \frac{(2n)!}{4^{n}n!}\sqrt{\pi }}}$　　　(証明終)