ルジャンドルの関係式
d d k ( k ( 1 − k 2 ) d d k K ( k ) ) = E ( k ) − K ( k ) k − ( 1 − k 2 ) E ( k ) − ( 1 − k 2 ) K ( k ) k ( 1 − k 2 ) + 2 k K ( k ) = E ( k ) − K ( k ) − E ( k ) + ( 1 − k 2 ) K ( k ) k + 2 k K ( k ) = k K ( k ) {\displaystyle {\begin{aligned}{\frac {d}{dk}}\left(k(1-k^{2}){\frac {d}{dk}}K(k)\right)&={\frac {E(k)-K(k)}{k}}-(1-k^{2}){\frac {E(k)-(1-k^{2})K(k)}{k(1-k^{2})}}+2kK(k)\\&={\frac {E(k)-K(k)-E(k)+(1-k^{2})K(k)}{k}}+2kK(k)\\&=kK(k)\\\end{aligned}}}
が得られるが、ここで k ′ = 1 − k 2 {\displaystyle k'={\sqrt {1-k^{2}}}} とすれば
d d k ( k ( 1 − k 2 ) d d k K ( k ′ ) ) = d k ′ d k d d k ′ ( k ( 1 − k 2 ) d k ′ d k d d k ′ K ( k ′ ) ) = k 1 − k 2 d d k ′ ( k 2 1 − k 2 d d k ′ K ( k ′ ) ) = k k ′ d d k ( ( 1 − k ′ 2 ) k ′ d d k ′ K ( k ′ ) ) {\displaystyle {\begin{aligned}{\frac {d}{dk}}\left(k(1-k^{2}){\frac {d}{dk}}K(k')\right)&={\frac {dk'}{dk}}{\frac {d}{dk'}}\left(k(1-k^{2}){\frac {dk'}{dk}}{\frac {d}{dk'}}K(k')\right)\\&={\frac {k}{\sqrt {1-k^{2}}}}{\frac {d}{dk'}}\left(k^{2}{\sqrt {1-k^{2}}}{\frac {d}{dk'}}K(k')\right)\\&={\frac {k}{k'}}{\frac {d}{dk}}\left((1-k'^{2})k'{\frac {d}{dk'}}K(k')\right)\\\end{aligned}}}
k ( 1 − k 2 ) K ( k ) d 2 d k 2 ( k ( 1 − k 2 ) K ′ ( k ) ) − k ( 1 − k 2 ) K ′ ( k ) d 2 d k 2 ( k ( 1 − k 2 ) K ( k ) ) = 0 {\displaystyle {\sqrt {k(1-k^{2})}}K(k){\frac {d^{2}}{dk^{2}}}\left({\sqrt {k(1-k^{2})}}K'(k)\right)-{\sqrt {k(1-k^{2})}}K'(k){\frac {d^{2}}{dk^{2}}}\left({\sqrt {k(1-k^{2})}}K(k)\right)=0}
が成立する。積分して整理すると
k ( 1 − k 2 ) ( K ( k ) d d k K ′ ( k ) − K ′ ( k ) d d k K ( k ) ) = C {\displaystyle k(1-k^{2})\left(K(k){\frac {d}{dk}}K'(k)-K'(k){\frac {d}{dk}}K(k)\right)=C}
となり、これに
d d k K ( k ) = E ( k ) − ( 1 − k 2 ) K ( k ) k ( 1 − k 2 ) {\displaystyle {\frac {d}{dk}}K(k)={\frac {E(k)-(1-k^{2})K(k)}{k(1-k^{2})}}}
d d k K ′ ( k ) = d d k K ( 1 − k 2 ) = − k 1 − k 2 E ( 1 − k 2 ) − k 2 K ( 1 − k 2 ) 1 − k 2 k 2 = − E ′ ( k ) − k 2 K ′ ( k ) k ( 1 − k 2 ) {\displaystyle {\frac {d}{dk}}K'(k)={\frac {d}{dk}}K\left({\sqrt {1-k^{2}}}\right)={\frac {-k}{\sqrt {1-k^{2}}}}{\frac {E\left({\sqrt {1-k^{2}}}\right)-k^{2}K\left({\sqrt {1-k^{2}}}\right)}{{\sqrt {1-k^{2}}}\;k^{2}}}=-{\frac {E'(k)-k^{2}K'(k)}{k(1-k^{2})}}}