算術幾何平均
| arg  ( b / a ) | ≠ π {\displaystyle |\arg(b/a)|\neq \pi } である複素数 a , b {\displaystyle a,\ b} について算術平均と調和平均を繰り返して得られる数列
a 0 = a , b 0 = b {\displaystyle a_{0}=a,\quad b_{0}=b} a n + 1 = a n + b n 2 , b n + 1 = 2 a n b n a n + b n ( n ≥ 0 ) {\displaystyle a_{n+1}={\frac {a_{n}+b_{n}}{2}},\quad b_{n+1}={\frac {2a_{n}b_{n}}{a_{n}+b_{n}}}\quad (n\geq 0)} の極限について AHM  ( a , b ) = lim n → ∞ a n = lim n → ∞ b n {\displaystyle \operatorname {AHM} (a,b)=\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }b_{n}}
x = tan  θ {\displaystyle x=\tan \theta } と置換すると、
I ( a , b ) = ∫ 0 ∞ d x ( a 2 + b 2 x 2 ) ( 1 + x 2 ) = ∫ 0 ∞ d x a 2 x 2 + b 2 x 2 + b 2 x 4 + a 2 = ∫ 0 ∞ d x ( a + b ) 2 x 2 + ( b x 2 − a ) 2 = 1 2 ∫ 0 ∞ d x x ( a + b 2 ) 2 + ( b x 2 − a 2 x ) 2 {\displaystyle {\begin{aligned}I(a,b)&=\int _{0}^{\infty }{\frac {dx}{\sqrt {(a^{2}+b^{2}x^{2})(1+x^{2})}}}\\&=\int _{0}^{\infty }{\frac {dx}{\sqrt {a^{2}x^{2}+b^{2}x^{2}+b^{2}x^{4}+a^{2}}}}\\&=\int _{0}^{\infty }{\frac {dx}{\sqrt {(a+b)^{2}x^{2}+(bx^{2}-a)^{2}}}}\\&={\frac {1}{2}}\int _{0}^{\infty }{\frac {dx}{x{\sqrt {\left(\displaystyle {\frac {a+b}{2}}\right)^{2}+\left(\displaystyle {\frac {bx^{2}-a}{2x}}\right)^{2}}}}}\\\end{aligned}}}