楕円積分
最初に示したものはヤコービの標準形であるが、ヤコービの標準形において積分変数 t = sin  θ {\displaystyle t=\sin {\theta }} と置けば(置換積分)幾らか簡単なルジャンドルの標準形が得られる。
F ( φ , k ) = ∫ 0 φ d θ 1 − k 2 sin 2  θ E ( φ , k ) = ∫ 0 φ 1 − k 2 sin 2  θ d θ Π ( a ; φ , k ) = ∫ 0 φ d θ ( 1 − a sin 2  θ ) 1 − k 2 sin 2  θ {\displaystyle {\begin{aligned}F(\varphi ,k)&=\int _{0}^{\varphi }{\frac {d\theta }{\sqrt {1-k^{2}\sin ^{2}\theta }}}\\E(\varphi ,k)&=\int _{0}^{\varphi }{\sqrt {1-k^{2}\sin ^{2}\theta }}~d\theta \\\Pi (a;\varphi ,k)&=\int _{0}^{\varphi }{\frac {d\theta }{(1-a\sin ^{2}\theta ){\sqrt {1-k^{2}\sin ^{2}\theta }}}}\end{aligned}}}
! k 2 n ∫ 0 π / 2 sin 2 n  θ d θ = π 2 − ∑ n = 1 ∞ ( 2 n − 1 ) ! ! ( 2 n − 1 ) ( 2 n ) ! ! k 2 n ( 2 n − 1 ) ! ! ( 2 n ) ! ! π 2 = π 2 ( 1 − ∑ n = 1 ∞ ( ( 2 n − 1 ) ! ! ( 2 n ) ! ! ) 2 k 2 n 2 n − 1 ) = π 2 ∑ n = 0 ∞ ( ( 2 n − 1 ) ! ! ( 2 n ) ! ! ) 2 k 2 n 1 − 2 n {\displaystyle {\begin{aligned}E(k)&=\int _{0}^{\pi /2}{\left(1-k^{2}\sin ^{2}\theta \right)^{\frac {1}{2}}}d\theta \\&=\int _{0}^{\pi /2}{\left(1-\sum _{n=1}^{\infty }{{\frac {(2n-1)!!}{(2n-1)2^{n}}}{\frac {(k^{2}\sin ^{2}\theta )^{n}}{n!}}}\right)}d\theta \\&={\frac {\pi }{2}}-\sum _{n=1}^{\infty }{{\frac {(2n-1)!
! ( 2 n ) ! ! k 2 n ∫ 0 π / 2 sin 2 n  θ d θ = π 2 + ∑ n = 1 ∞ ( 2 n − 1 ) ! ! ( 2 n ) ! ! k 2 n ( 2 n − 1 ) ! ! ( 2 n ) ! ! π 2 = π 2 ( 1 + ∑ n = 1 ∞ ( ( 2 n − 1 ) ! ! ( 2 n ) ! ! ) 2 k 2 n ) = π 2 ∑ n = 0 ∞ ( ( 2 n − 1 ) ! ! ( 2 n ) ! ! ) 2 k 2 n {\displaystyle {\begin{aligned}K(k)&=\int _{0}^{\pi /2}{\left(1-k^{2}\sin ^{2}\theta \right)^{-{\frac {1}{2}}}}d\theta \\&=\int _{0}^{\pi /2}{\left(1+\sum _{n=1}^{\infty }{{\frac {(2n-1)!!}{2^{n}}}{\frac {(k^{2}\sin ^{2}\theta )^{n}}{n!}}}\right)}d\theta \\&={\frac {\pi }{2}}+\sum _{n=1}^{\infty }{{\frac {(2n-1)!