מכפלת ואליס

על בסיס מכפלת אוילר לפונקציית סינוס

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

על בסיס אינטגרציה

נגדיר ${\displaystyle I_{n}=\int \limits _{0}^{\pi }\sin(x)^{n}dx}$ .

נבצע אינטגרציה בחלקים:

${\displaystyle {\begin{aligned}u&=\sin(x)^{n-1}\quad \Rightarrow \quad du=(n-1)\sin(x)^{n-2}\cos(x)dx\\dv&=\sin(x)dx\quad \Rightarrow \quad v=-\cos(x)\\\Rightarrow ~I_{n}&=\int \limits _{0}^{\pi }\sin(x)^{n}dx=\int \limits _{0}^{\pi }udv=uv{\Big |}_{x=0}^{x=\pi }-\int \limits _{0}^{\pi }vdu\\{}&=-\sin(x)^{n-1}\cos(x){\Big |}_{x=0}^{x=\pi }-\int \limits _{0}^{\pi }-\cos(x)(n-1)\sin(x)^{n-2}\cos(x)dx\\{}&=0+(n-1)\int \limits _{0}^{\pi }\cos(x)^{2}\sin(x)^{n-2}dx,~n>1\\{}&=(n-1)\int \limits _{0}^{\pi }{\bigl (}1-\sin(x)^{2}{\bigr )}\sin(x)^{n-2}dx\\{}&=(n-1)\int \limits _{0}^{\pi }\sin(x)^{n-2}dx-(n-1)\int \limits _{0}^{\pi }\sin(x)^{n}dx\\{}&=(n-1)I_{n-2}-(n-1)I_{n}\\{}&={\frac {n-1}{n}}I_{n-2}\\\Rightarrow ~{\frac {I_{n}}{I_{n-2}}}&={\frac {n-1}{n}}\\\Rightarrow ~{\frac {I_{2n-1}}{I_{2n+1}}}&={\frac {2n+1}{2n}}\end{aligned}}}$

נשתמש במשוואה פונקציונלית הזו בדרך הבאה:

${\displaystyle {\begin{aligned}I_{0}&=\int \limits _{0}^{\pi }dx=x{\Big |}_{0}^{\pi }=\pi \\I_{1}&=\int \limits _{0}^{\pi }\sin(x)dx=-\cos(x){\Big |}_{0}^{\pi }=-\cos(\pi )+\cos(0)=2\\I_{2n}&=\int \limits _{0}^{\pi }\sin(x)^{2n}dx={\frac {2n-1}{2n}}I_{2n-2}={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}I_{2n-4}\end{aligned}}}$

נחזור על התהליך:

${\displaystyle {\begin{aligned}={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}\cdot {\frac {2n-5}{2n-4}}\cdots {\frac {5}{6}}\cdot {\frac {3}{4}}\cdot {\frac {1}{2}}I_{0}=\pi \prod _{k=1}^{n}{\frac {2k-1}{2k}}\\I_{2n+1}=\int \limits _{0}^{\pi }\sin(x)^{2n+1}dx={\frac {2n}{2n+1}}I_{2n-1}={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}I_{2n-3}\end{aligned}}}$

נחזור על התהליך:

${\displaystyle {\begin{aligned}={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}\cdot {\frac {2n-4}{2n-3}}\cdots {\frac {6}{7}}\cdot {\frac {4}{5}}\cdot {\frac {2}{3}}I_{1}=2\prod _{k=1}^{n}{\frac {2k}{2k+1}}\\\sin(x)^{2n+1}\leq \sin(x)^{2n}\leq \sin(x)^{2n-1},~0\leq x\leq \pi \\\Rightarrow ~I_{2n+1}\leq I_{2n}\leq I_{2n-1}\\\Rightarrow ~1\leq {\frac {I_{2n}}{I_{2n+1}}}\leq {\frac {I_{2n-1}}{I_{2n+1}}}={\frac {2n+1}{2n}}\end{aligned}}}$

ניתן לראות על פי כלל הסנדוויץ' כי:

${\displaystyle {\begin{aligned}\Rightarrow ~\lim _{n\to \infty }{\frac {I_{2n}}{I_{2n+1}}}=1\\\lim _{n\to \infty }{\frac {I_{2n}}{I_{2n+1}}}={\frac {\pi }{2}}\lim _{n\to \infty }\prod _{k=1}^{n}\left({\frac {2k-1}{2k}}\cdot {\frac {2k+1}{2k}}\right)=1\\\Rightarrow ~{\frac {\pi }{2}}=\prod _{k=1}^{\infty }\left({\frac {2k}{2k-1}}\cdot {\frac {2k}{2k+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdots \end{aligned}}}$