Wallis’ Pi |

There is a remarkable expression for the number $\pi$ as an infinite product. Starting with definite integral $\int\limits_{0}^{\frac{\pi}{2}} \sin^{m}(x)\;dx, m=0,1,2,3,4...$ , we derive it as follows:

$\int\limits_{0}^{\frac{\pi}{2}} \sin^m(x)\;dx$

$=\int\limits_{0}^{\frac{\pi}{2}}\sin^{m-1}(x)\cdot\sin(x)\;dx$

$= \int\limits_{0}^{\frac{\pi}{2}} \sin^{m-1}(x)\cdot(-\cos(x))'\;dx$

By $\int\limits_{a}^{b} u\cdot v'\;dx = u\cdot v \bigg|_{b}^{a}- \int\limits_{b}^{a} u' \cdot v\;dx,$

$= \sin^{m-1}(x)\cdot(-\cos(x))\bigg|_{0}^{\frac{\pi}{2}} - \int\limits_{0}^{\frac{\pi}{2}}(\sin^{m-1}(x))'\cdot(-\cos(x))\;dx$

$= 0 - \int\limits_{0}^{\frac{\pi}{2}}(m-1)\sin^{m-2}(x)\cos(x)\cdot(-\cos(x))\;dx$

$= +\int\limits_{0}^{\frac{\pi}{2}}(m-1)\sin^{m-2}(x)\cdot\cos^2(x)\;dx$

$\overset{\cos^2(x) = 1-\sin^2(x)}{=}\int\limits_{0}^{\frac{\pi}{2}}(m-1)\sin^{m-2}(x)\cdot (1-\sin^{2}(x))\;dx$

$= \int\limits_{0}^{\frac{\pi}{2}}(m-1)\sin^{m-2}(x) - (m-1)\sin^{m}(x)\;dx$

$= (1-m)\int\limits_{0}^{\frac{\pi}{2}}\sin^{m}(x)\;dx + (m-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{m}(x)\;dx.$

That is,

$m\int\limits_{0}^{\frac{\pi}{2}}\sin^{m}(x)\;dx = (m-1)\int\limits_{0}^{\frac{\pi}{2}}\sin^{m-2}(x)\;dx.$

And so,

$\int\limits_{0}^{\frac{\pi}{2}}\sin^{m}(x) \; dx= \frac{m-1}{m}\int\limits_{0}^{\frac{\pi}{2}}\sin^{m-2}(x)\;dx.\quad\quad\quad(1)$

By repeated application of (1) we have the following values for $I_{m} = \int\limits_{0}^{\frac{\pi}{2}} \sin^{m}(x)\;dx$ :

For even $m$ ,

$I_{2n} = \left(\prod\limits_{k=0}^{n-1}\frac{2n-2k-1}{2n-2k}\right)I_0\overset{I_0 = \int\limits_{0}^{\frac{\pi}{2}}\;dx = \frac{\pi}{2}}{\implies}I_{2n}= \left(\prod\limits_{k=0}^{n-1}\frac{2n-2k-1}{2n-2k}\right)\cdot \frac{\pi}{2}.$

Similarly, for odd $m$ ,

$I_{2n+1} = \left(\prod\limits_{k=0}^{n-1}\frac{2n+1-2k-1}{2n+1-2k}\right)\cdot I_{1}\overset{I_1 = \int\limits_{0}^{\frac{\pi}{2}}\sin^{1}(x)\;dx = 1}{\implies} I_{2n+1}= \prod\limits_{k=0}^{n-1}\frac{2n-2k}{2n-2k+1}.$

i.e.,

$I_{2n} = \frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-5}{2n-4}\cdot\frac{2n-7}{2n-6}\cdot...\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}\quad\quad\quad(2)$

$I_{2n+1} =\frac{2n}{2n+1}\cdot\frac{2n-2}{2n-1}\cdot\frac{2n-4}{2n-3}\cdot\frac{2n-6}{2n-5}\cdot ...\cdot\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3}\quad\quad\quad\quad\quad(3)$

Since for $0<x<\frac{\pi}{2}, 0< \sin(x) < 1,$ we have $\sin^{2n-1}(x) > \sin^{2n}(x) > \sin^{2n+1}(x).$ It means

$\int\limits_{0}^{\frac{\pi}{2}}\sin^{2n-1}(x) > \int\limits_{0}^{\frac{\pi}{2}}\sin^{2n}(x)\;dx > \int\limits_{0}^{\frac{\pi}{2}}\sin^{2n+1}(x)\;dx>0$

or,

$I_{2n-1} > I_{2n} > I_{2n+1}>0.$

Hence,

$\frac{I_{2n-1}}{I_{2n+1}}>\frac{I_{2n}}{I_{2n+1}} > 1.\quad\quad\quad(4)$

Moreover, we have

$I_{2n+1} \overset{(1)}{=} \frac{(2n+1)-1}{2n+1}I_{(2n+1)-2} = \frac{2n}{2n+1} I_{2n-1},$

so that

$\frac{I_{2n-1}}{I_{2n+1}} = \frac{2n+1}{2n}.\quad\quad\quad(5)$

And,

$\frac{I_{2n}}{I_{2n+1}} \overset{(2), (3)}{=} \frac{(2n+1)\cdot(2n-1)^2\cdot (2n-3)^2...7^2\cdot 5^2\cdot 3^2}{(2n)^2\cdot (2n-2)^2\cdot (2n-4)^2\cdot ...\cdot 6^2\cdot 4^2\cdot 2^2}\cdot \frac{\pi}{2}$

$= \frac{(2n)^2\cdot (2n-1)^2\cdot (2n-2)^2\cdot (2n-3)^2\cdot (2n-4)^2\cdot...\cdot 7^2\cdot 6^2\cdot 5^2\cdot 4^2\cdot 3^2\cdot 2^2}{(2n)^4(2n-2)^4\cdot (2n-4)^4...6^4\cdot 4^4\cdot 2^4}\cdot (2n+1)\cdot\frac{\pi}{2}$

$= \frac{\left((2n)\cdot (2n-1)\cdot (2n-2)\cdot (2n-3)\cdot (2n-4)\cdot...\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\right)^2}{(2^4\cdot n^4)\cdot (2^4\cdot(n-1)^4) \cdot (2^4\cdot (n-2)^4)\cdot...\cdot (2^4\cdot 3^4)\cdot (2^4\cdot 2^4)\cdot 2^4}\cdot (2n+1)\cdot\frac{\pi}{2}$

$= \frac{((2n)!)^2}{\underbrace{2^4\cdot 2^4\cdot ...\cdot 2^4}_{n 2^4s}\cdot (n\cdot(n-1)\cdot(n-2)\cdot ...\cdot 3\cdot 2\cdot 1)^4}(2n+1)\cdot\frac{\pi}{2}$

$= \frac{((2n)!)^2\cdot (2n+1)}{2^{4n}\cdot(n!)^4}\cdot \frac{\pi}{2}$

gives

$\frac{I_{2n}}{I_{2n+1}} = \frac{((2n)!)^2(2n+1)}{2^{4n}(n!)^4}\cdot\frac{\pi}{2}.\quad\quad\quad(6)$

Substituting (5) and (6) into (4) yields

$\frac{2n+1}{2n} >\frac{((2n)!)^2(2n+1)}{2^{4n}(n!)^4}\cdot\frac{\pi}{2}>1.$

Since $\lim\limits_{n \rightarrow \infty} \frac{2n+1}{2n} = 1, \lim\limits_{n \rightarrow \infty}1 = 1,$ by Squeeze Theorem for Sequences,

$\lim\limits_{n \rightarrow \infty}\frac{((2n)!)^4(2n+1)}{2^{4n}(n!)^4} \cdot \frac{\pi}{2}= 1\implies \lim\limits_{n \rightarrow \infty}\frac{((2n)!)^2(2n+1)}{2^{4n}(n!)^4}=\frac{2}{\pi}.$