ルジャンドルの関係式
) 2 ( ( 2 n ) 2 ( 2 n − 1 ) 2 ( 2 n − 2 + 1 ) − 1 2 n − 1 − 1 ) k 2 n = k 2 ( k d d k K ( k ) + K ( k ) ) + E ( k ) − K ( k ) {\displaystyle {\begin{aligned}k{\frac {d}{dk}}K(k)&=k{\frac {d}{dk}}{\frac {\pi }{2}}\left(1+\sum _{n=1}^{\infty }{\left({\frac {(2n-1)!!}{(2n)!!}}\right)^{2}k^{2n}}\right)\\&={\frac {\pi }{2}}\sum _{n=1}^{\infty }{\left({\frac {(2n-1)!!}{(2n)!!}}\right)^{2}\left({\frac {(2n)^{2}}{(2n-1)^{2}}}(2n-2+1)-{\frac {1}{2n-1}}-1\right)k^{2n}}\\&=k^{2}\left(k{\frac {d}{dk}}K(k)+K(k)\right)+E(k)-K(k)\end{aligned}}}
k ( 1 − k 2 ) d d k K ( k ) = E ( k ) − ( 1 − k 2 ) K ( k ) {\displaystyle k(1-k^{2}){\frac {d}{dk}}K(k)=E(k)-(1-k^{2})K(k)}
から、微分方程式