ウォリス積分
lim m → ∞ m ∫ 0 π 2 sin m  θ d θ = lim m → ∞ m ∫ 0 π 2 cos m  θ d θ = π 2 {\displaystyle \lim _{m\to \infty }{\sqrt {m}}\int _{0}^{\tfrac {\pi }{2}}\sin ^{m}\theta \,d\theta =\lim _{m\to \infty }{\sqrt {m}}\int _{0}^{\tfrac {\pi }{2}}\cos ^{m}\theta \,d\theta ={\sqrt {\frac {\pi }{2}}}}
証明 — 先述の漸化式より
( m + 1 ) I m + 1 I m = m I m I m − 1 {\displaystyle (m+1)I_{m+1}I_{m}=mI_{m}I_{m-1}}
が成り立つ。故に数列 { m I m I m − 1 } {\displaystyle \{mI_{m}I_{m-1}\}} は定数列で
m I m I m − 1 = I 1 I 0 = π 2 . {\displaystyle mI_{m}I_{m-1}=I_{1}I_{0}={\frac {\pi }{2}}.