超幾何関数
) 2 k 2 n = π 2 ⋅ 2 F 1 [ 1 2 , 1 2 1 ; k 2 ] E ( k ) = π 2 ∑ n = 0 ∞ ( ( 2 n − 1 ) ! ! ( 2 n ) ! ! ) 2 k 2 n 1 − 2 n = π 2 ⋅ 2 F 1 [ 1 2 , − 1 2 1 ; k 2 ] {\displaystyle {\begin{aligned}K(k)&={\frac {\pi }{2}}\sum _{n=0}^{\infty }{\left({\frac {(2n-1)!!}{(2n)!!}}\right)^{2}k^{2n}}={\frac {\pi }{2}}\cdot {_{2}F_{1}}\left[{\begin{matrix}{\frac {1}{2}},{\frac {1}{2}}\\1\end{matrix}};k^{2}\right]\\E(k)&={\frac {\pi }{2}}\sum _{n=0}^{\infty }{\left({\frac {(2n-1)!!}{(2n)!!}}\right)^{2}{\frac {k^{2n}}{1-2n}}}={\frac {\pi }{2}}\cdot {_{2}F_{1}}\left[{\begin{matrix}{\frac {1}{2}},-{\frac {1}{2}}\\1\end{matrix}};k^{2}\right]\\\end{aligned}}}
}}z^{n}\\&={\frac {\Gamma (c)}{\Gamma (a)\Gamma (c-a)}}\sum _{n=0}^{\infty }\mathrm {B} (a+n,c-a){\frac {(b)_{n}}{n!}}z^{n}\\&={\frac {\Gamma (c)}{\Gamma (a)\Gamma (c-a)}}\sum _{n=0}^{\infty }\left(\int _{0}^{1}t^{a+n-1}(1-t)^{c-a-1}dt\right){\frac {(b)_{n}}{n!}}z^{n}\\&={\frac {\Gamma (c)}{\Gamma (a)\Gamma (c-a)}}\int _{0}^{1}t^{a-1}(1-t)^{c-a-1}\left(\sum _{n=0}^{\infty }{\frac {(b)_{n}}{n!}}(tz)^{n}\right)dt\\&={\frac {\Gamma (c)}{\Gamma (a)\Gamma (c-a)}}\int _{0}^{1}t^{a-1}(1-t)^{c-a-1}(1-tz)^{-b}dt\\\end{aligned}}}
として導かれる。