沃利斯乘积

今日多数的微积分教科书透过比较${\displaystyle \int _{0}^{\pi }\sin ^{n}xdx}$ 在n是奇数或是偶数，甚至是接近无穷大的情况下，发现即使将n增加一就会发生不一样的情形。在那时，微积分尚未存在，而且有关数学收敛的分析工具也还未俱全，所以完成这证明较现今有相当的难度。从现在来看，从欧拉公式中的正弦展开式得到此乘积是必然的结果。

${\displaystyle {\frac {\sin(x)}{x}}=\left(1-{\frac {x^{2}}{\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{4\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{9\pi ^{2}}}\right)\cdots =\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{n^{2}\pi ^{2}}}\right),}$ 

在x = π/2时

${\displaystyle {\frac {2}{\pi }}=\prod _{n=1}^{\infty }\left(1-{\frac {1}{4n^{2}}}\right)=\left(1-{\frac {1}{2^{2}}}\right)\left(1-{\frac {1}{2^{2}\cdot 4}}\right)\left(1-{\frac {1}{2^{2}\cdot 9}}\right)\cdots }$ 

${\displaystyle {\begin{aligned}{\frac {\pi }{2}}&{}=\prod _{n=1}^{\infty }\left({\frac {4n^{2}}{4n^{2}-1}}\right)\\&{}=\prod _{n=1}^{\infty }{\frac {(2n)(2n)}{(2n-1)(2n+1)}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots \end{aligned}}}$ 

先考虑不定积分${\displaystyle \int \sin ^{n}xdx}$ 有

${\displaystyle \int \sin ^{n}xdx}$ 

${\displaystyle =-\int \sin ^{n-1}xd\cos x}$ 

${\displaystyle =-\cos x\sin ^{n-1}x+\int \cos xd\sin ^{n-1}x}$ 

${\displaystyle =-\cos x\sin ^{n-1}x+\int (n-1)\sin ^{n-2}x\cos ^{2}xdx}$ 

${\displaystyle =-\cos x\sin ^{n-1}x+(n-1)\int \sin ^{n-2}x(1-\sin ^{2}x)dx}$ 

${\displaystyle =-\cos x\sin ^{n-1}x+(n-1)\int \sin ^{n-2}xdx-(n-1)\int \sin ^{n}xdx}$ 

故

${\displaystyle \int \sin ^{n}xdx=-{\frac {1}{n}}\cos x\sin ^{n-1}x+{\frac {n-1}{n}}\int \sin ^{n-2}xdx}$ 

${\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}xdx={\frac {n-1}{n}}\int _{0}^{\frac {\pi }{2}}\sin ^{n-2}xdx}$ 

对整数m

${\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}$ 

${\displaystyle ={\frac {2m-1}{2m}}\int _{0}^{\frac {\pi }{2}}\sin ^{2m-2}xdx}$ 

${\displaystyle ={\frac {2m-1}{2m}}{\frac {2m-3}{2m-2}}\int _{0}^{\frac {\pi }{2}}\sin ^{2m-4}xdx}$ 

${\displaystyle =...}$ 

${\displaystyle ={\frac {2m-1}{2m}}{\frac {2m-3}{2m-2}}...{\frac {1}{2}}\int _{0}^{\frac {\pi }{2}}\sin ^{0}xdx}$ 

${\displaystyle ={\frac {2m-1}{2m}}{\frac {2m-3}{2m-2}}...{\frac {1}{2}}{\frac {\pi }{2}}}$ 

另一方面

${\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}$ 

${\displaystyle ={\frac {2m}{2m+1}}\int _{0}^{\frac {\pi }{2}}\sin ^{2m-1}xdx}$ 

${\displaystyle ={\frac {2m}{2m+1}}{\frac {2m-2}{2m-1}}\int _{0}^{\frac {\pi }{2}}\sin ^{2m-3}xdx}$ 

${\displaystyle =...}$ 

${\displaystyle ={\frac {2m}{2m+1}}{\frac {2m-2}{2m-1}}...{\frac {2}{3}}\int _{0}^{\frac {\pi }{2}}\sin xdx}$ 

${\displaystyle ={\frac {2m}{2m+1}}{\frac {2m-2}{2m-1}}...{\frac {2}{3}}}$ 

两式相除得

${\displaystyle {\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}={\frac {{\frac {2m-1}{2m}}{\frac {2m-3}{2m-2}}...{\frac {1}{2}}{\frac {\pi }{2}}}{{\frac {2m}{2m+1}}{\frac {2m-2}{2m-1}}...{\frac {2}{3}}}}}$ 

故

${\displaystyle {\frac {\pi }{2}}={\frac {2}{1}}{\frac {2}{3}}{\frac {4}{3}}{\frac {4}{5}}...{\frac {2m}{2m-1}}{\frac {2m}{2m+1}}{\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}={\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}\prod _{n=1}^{m}{\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}}$ 

又因为

${\displaystyle 1={\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}<{\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}<{\frac {\int _{0}^{\frac {\pi }{2}}\sin ^{2m-1}xdx}{\int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}xdx}}={\frac {2m+1}{2m}}}$ 

由夹挤定理知

${\displaystyle \lim _{m\to \infty }1=\lim _{m\to \infty }{\frac {2m+1}{2m}}=1}$ 

故

${\displaystyle {\frac {\pi }{2}}=\prod _{n=1}^{\infty }{\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots }$ 

我们可将上述的正弦乘积式化为泰勒级数：

${\displaystyle x\left(1-{\frac {x^{2}}{\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{4\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{9\pi ^{2}}}\right)\cdots =x-{\frac {1}{3!}}x^{3}+{\frac {1}{5!}}x^{5}-\cdots }$