ランデン変換
F ( α , k ) = ∫ φ = 0 α d φ 1 − k 2 sin 2  φ = ∫ φ = 0 α cos  φ d φ 1 − sin 2  φ 1 − k 2 sin 2  φ = ∫ θ = 0 β 2 1 + k cos  θ 1 − 4 k ( 1 + k ) 2 sin 2  θ ( 1 + 1 − 4 k ( 1 + k ) 2 sin 2  θ ) 2 + 2 1 − 4 k ( 1 + k ) 2 sin 2  θ − 4 1 + k sin 2  θ 1 + 1 − 4 k ( 1 + k ) 2 sin 2  θ 2 + 2 1 − 4 k ( 1 + k ) 2 sin 2  θ − 4 k 1 + k sin 2  θ 1 + 1 − 4 k ( 1 + k ) 2 sin 2  θ d θ = ∫ θ = 0 β 2 1 + k cos  θ 1 − 4 k ( 1 + k ) 2 sin 2  θ ( 1 + 1 − 4 k ( 1 + k ) 2 sin 2  θ ) 2 1 − sin 2  θ 2 + 2 1 − 4 k ( 1 + k ) 2 sin 2  θ − 4 k ( 1 + k ) 2 sin 2  θ ( 1 + 1 − 4 k ( 1 + k ) 2 sin 2  θ ) 2 d θ = 1 1 + k ∫ θ = 0 β 1 1 − 4 k ( 1 + k ) 2 sin 2  θ = 1 1 + k F ( β , 2 k 1 + k ) {\displaystyle {\begin{aligned}F\left(\alpha ,k\right)&=\int _{\phi =0}^{\alpha }{\frac {d\phi }{\sqrt {1-k^{2}\sin ^{2}\phi }}}\\&=\int _{\phi =0}^{\alpha }{\frac {\cos \phi {d\phi }}{{\sqrt {1-\sin ^{2}\phi }}{\sqrt {1-k^{2}\sin ^{2}\phi }}}}\\&=\int _{\theta =0}^{\beta }{\frac {\frac {{\frac {2}{1+k}}\cos \theta }{{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\left(1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)}}{{\frac {\sqrt {2+2{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}-{\frac {4}{1+k}}\sin ^{2}\theta }}{1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}\;{\frac {\sqrt {2+2{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}-{\frac {4k}{1+k}}\sin ^{2}\theta }}{1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}}}{d\theta }\\&=\int _{\theta =0}^{\beta }{\frac {\frac {{\frac {2}{1+k}}\cos \theta }{{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\left(1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)}}{\frac {2{\sqrt {1-\sin ^{2}\theta }}{\sqrt {2+2{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}{\left(1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)^{2}}}}{d\theta }\\&={\frac {1}{1+k}}\int _{\theta =0}^{\beta }{\frac {1}{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}\\&={\frac {1}{1+k}}F\left(\beta ,{\frac {2{\sqrt {k}}}{1+k}}\right)\end{aligned}}}
次の恒等式を楕円関数の下降ランデン変換という。