沃利斯乘積，又称沃利斯公式，由數學家約翰·沃利斯在1655年时发现。
今日多數的微積分教科書透過比較 ∫ 0 π sin n x d x {\displaystyle \int _{0}^{\pi }\sin ^{n}xdx} 在n是奇數或是偶數，甚至是接近無窮大的情況下，發現即使將n增加一就會發生不一樣的情形。在那時，微積分尚未存在，而且有關數學收斂的分析工具也還未俱全，所以完成這證明較現今有相當的難度。從現在來看，從欧拉公式中的正弦展開式得到此乘積是必然的結果。
在x = π/2時
先考慮不定積分 ∫ sin n x d x {\displaystyle \int \sin ^{n}xdx} 有
∫ sin n x d x {\displaystyle \int \sin ^{n}xdx}
= − ∫ sin n − 1 x d cos x {\displaystyle =-\int \sin ^{n-1}xd\cos x}
= − cos x sin n − 1 x + ∫ cos x d sin n − 1 x {\displaystyle =-\cos x\sin ^{n-1}x+\int \cos xd\sin ^{n-1}x}
= − cos x sin n − 1 x + ∫ ( n − 1 ) sin n − 2 x cos 2 x d x {\displaystyle =-\cos x\sin ^{n-1}x+\int (n-1)\sin ^{n-2}x\cos ^{2}xdx}
= − cos x sin n − 1 x + ( n − 1 ) ∫ sin n − 2 x ( 1 − sin 2 x ) d x {\displaystyle =-\cos x\sin ^{n-1}x+(n-1)\int \sin ^{n-2}x(1-\sin ^{2}x)dx}
= − cos x sin n − 1 x + ( n − 1 ) ∫ sin n − 2 x d x − ( n − 1 ) ∫ sin n x d x {\displaystyle =-\cos x\sin ^{n-1}x+(n-1)\int \sin ^{n-2}xdx-(n-1)\int \sin ^{n}xdx}
故
∫ sin n x d x = − 1 n cos x sin n − 1 x + n − 1 n ∫ sin n − 2 x d x {\displaystyle \int \sin ^{n}xdx=-{\frac {1}{n}}\cos x\sin ^{n-1}x+{\frac {n-1}{n}}\int \sin ^{n-2}xdx}
∫ 0 π 2 sin n x d x = n − 1 n ∫ 0 π 2 sin n − 2 x d x {\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}xdx={\frac {n-1}{n}}\int _{0}^{\frac {\pi }{2}}\sin ^{n-2}xdx}
對整數m
∫ 0 π 2 sin 2 m x d x {\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}
= 2 m − 1 2 m ∫ 0 π 2 sin 2 m − 2 x d x {\displaystyle ={\frac {2m-1}{2m}}\int _{0}^{\frac {\pi }{2}}\sin ^{2m-2}xdx}
= 2 m − 1 2 m 2 m − 3 2 m − 2 ∫ 0 π 2 sin 2 m − 4 x d x {\displaystyle ={\frac {2m-1}{2m}}{\frac {2m-3}{2m-2}}\int _{0}^{\frac {\pi }{2}}\sin ^{2m-4}xdx}
= . . . {\displaystyle =...}
= 2 m − 1 2 m 2 m − 3 2 m − 2 . . . 1 2 ∫ 0 π 2 sin 0 x d x {\displaystyle ={\frac {2m-1}{2m}}{\frac {2m-3}{2m-2}}...{\frac {1}{2}}\int _{0}^{\frac {\pi }{2}}\sin ^{0}xdx}
= 2 m − 1 2 m 2 m − 3 2 m − 2 . . . 1 2 π 2 {\displaystyle ={\frac {2m-1}{2m}}{\frac {2m-3}{2m-2}}...{\frac {1}{2}}{\frac {\pi }{2}}}
另一方面
∫ 0 π 2 sin 2 m + 1 x d x {\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}
= 2 m 2 m + 1 ∫ 0 π 2 sin 2 m − 1 x d x {\displaystyle ={\frac {2m}{2m+1}}\int _{0}^{\frac {\pi }{2}}\sin ^{2m-1}xdx}
= 2 m 2 m + 1 2 m − 2 2 m − 1 ∫ 0 π 2 sin 2 m − 3 x d x {\displaystyle ={\frac {2m}{2m+1}}{\frac {2m-2}{2m-1}}\int _{0}^{\frac {\pi }{2}}\sin ^{2m-3}xdx}
= 2 m 2 m + 1 2 m − 2 2 m − 1 . . . 2 3 ∫ 0 π 2 sin x d x {\displaystyle ={\frac {2m}{2m+1}}{\frac {2m-2}{2m-1}}...{\frac {2}{3}}\int _{0}^{\frac {\pi }{2}}\sin xdx}
= 2 m 2 m + 1 2 m − 2 2 m − 1 . . . 2 3 {\displaystyle ={\frac {2m}{2m+1}}{\frac {2m-2}{2m-1}}...{\frac {2}{3}}}
兩式相除得
∫ 0 π 2 sin 2 m x d x ∫ 0 π 2 sin 2 m + 1 x d x = 2 m − 1 2 m 2 m − 3 2 m − 2 . . . 1 2 π 2 2 m 2 m + 1 2 m − 2 2 m − 1 . . . 2 3 {\displaystyle {\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}={\frac {{\frac {2m-1}{2m}}{\frac {2m-3}{2m-2}}...{\frac {1}{2}}{\frac {\pi }{2}}}{{\frac {2m}{2m+1}}{\frac {2m-2}{2m-1}}...{\frac {2}{3}}}}}
π 2 = 2 1 2 3 4 3 4 5 . . . 2 m 2 m − 1 2 m 2 m + 1 ∫ 0 π 2 sin 2 m x d x ∫ 0 π 2 sin 2 m + 1 x d x = ∫ 0 π 2 sin 2 m x d x ∫ 0 π 2 sin 2 m + 1 x d x ∏ n = 1 m 2 n 2 n − 1 ⋅ 2 n 2 n + 1 {\displaystyle {\frac {\pi }{2}}={\frac {2}{1}}{\frac {2}{3}}{\frac {4}{3}}{\frac {4}{5}}...{\frac {2m}{2m-1}}{\frac {2m}{2m+1}}{\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}={\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}\prod _{n=1}^{m}{\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}}
又因為
1 = ∫ 0 π 2 sin 2 m + 1 x d x ∫ 0 π 2 sin 2 m + 1 x d x < ∫ 0 π 2 sin 2 m x d x ∫ 0 π 2 sin 2 m + 1 x d x < ∫ 0 π 2 sin 2 m − 1 x d x ∫ 0 π 2 sin 2 m + 1 x d x = 2 m + 1 2 m {\displaystyle 1={\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}<{\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}<{\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m-1}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}={\frac {2m+1}{2m}}}
由夾擠定理知
lim m → ∞ 1 = lim m → ∞ 2 m + 1 2 m = 1 {\displaystyle \lim _{m\to \infty }1=\lim _{m\to \infty }{\frac {2m+1}{2m}}=1}
π 2 = ∏ n = 1 ∞ 2 n 2 n − 1 ⋅ 2 n 2 n + 1 = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋅ 8 7 ⋅ 8 9 ⋯ {\displaystyle {\frac {\pi }{2}}=\prod _{n=1}^{\infty }{\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots }
我們可將上述的正弦乘積式化為泰勒级数：
在n是奇數或是偶數，甚至是接近無窮大的情況下，發現即使將n增加一就會發生不一樣的情形。在那時，微積分尚未存在，而且有關數學收斂的分析工具也還未俱全，所以完成這證明較現今有相當的難度。從現在來看，從欧拉公式中的正弦展開式得到此乘積是必然的結果。
有
