Wallis' integrals

Recurrence relation

By means of integration by parts, a reduction formula can be obtained. Using the identity ${\displaystyle {\sin }^{2}x=1-{\cos }^{2}x}$, we have for all ${\displaystyle n\ge 2}$,

${\displaystyle \begin{array}{cc}{\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx& ={\int }_0^{\frac{\pi }{2}}({\sin }^{n-2}x)(1-{\cos }^{2}x)dx\\ & ={\int }_0^{\frac{\pi }{2}}{\sin }^{n-2}xdx-{\int }_0^{\frac{\pi }{2}}{\sin }^{n-2}x{\cos }^{2}xdx.\text{Equation (1)}\end{array}}$

Integrating the second integral by parts, with:

• ${\displaystyle v^{\prime }(x)=\cos (x){\sin }^{n-2}(x)}$, whose anti-derivative is ${\displaystyle v(x)=\frac{1}{n-1}{\sin }^{n-1}(x)}$
• ${\displaystyle u(x)=\cos (x)}$, whose derivative is ${\displaystyle u^{\prime }(x)=-\sin (x),}$

we have:

${\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n-2}x{\cos }^{2}xdx={\left[\frac{{\sin }^{n-1}x}{n-1}\cos x\right]}_0^{\frac{\pi }{2}}+\frac{1}{n-1}{\int }_0^{\frac{\pi }{2}}{\sin }^{n-1}x\sin xdx=0+\frac{1}{n-1}{W}_n.}$

Substituting this result into equation (1) gives

${\displaystyle W_n=W_{n-2}-\frac{1}{n-1}{W}_n,}$

and thus

${\displaystyle W_n=\frac{n-1}{n}{W}_{n-2},\text{Equation (2)}}$

for all ${\displaystyle n\ge 2.}$

This is a recurrence relation giving ${\displaystyle W_n}$ in terms of ${\displaystyle W_{n-2}}$. This, together with the values of ${\displaystyle W_0}$ and ${\displaystyle W_1,}$ give us two sets of formulae for the terms in the sequence ${\displaystyle (W_n)}$, depending on whether ${\displaystyle n}$ is odd or even:

• ${\displaystyle W_{2p}=\frac{2p-1}{2p}\cdot \frac{2p-3}{2p-2}\cdots \frac{1}{2}{W}_0=\frac{(2p-1)!!}{(2p)!!}\cdot \frac{\pi }{2}=\frac{(2p)!}{2^{2p}(p!{)}^2}\cdot \frac{\pi }{2},}$
• ${\displaystyle W_{2p+1}=\frac{2p}{2p+1}\cdot \frac{2p-2}{2p-1}\cdots \frac{2}{3}{W}_1=\frac{(2p)!!}{(2p+1)!!}=\frac{2^{2p}(p!{)}^2}{(2p+1)!}.}$

Another relation to evaluate the Wallis' integrals

Wallis's integrals can be evaluated by using Euler integrals:

1. Euler integral of the first kind: the Beta function:

   ${\displaystyle \mathrm{B}(x,y)={\int }_0^1{t}^{x-1}(1-t{)}^{y-1}dt=\frac{\mathrm{\Gamma }(x)\mathrm{\Gamma }(y)}{\mathrm{\Gamma }(x+y)}}$ for Re(x), Re(y) > 0

2. Euler integral of the second kind: the Gamma function:

   ${\displaystyle \mathrm{\Gamma }(z)={\int }_0^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}dt}$ for Re(z) > 0.

If we make the following substitution inside the Beta function: ${\displaystyle \{\begin{array}{c}t={\sin }^{2}u\\ 1-t={\cos }^{2}u\\ dt=2\sin u\cos udu\end{array},}$
we obtain:

${\displaystyle \mathrm{B}(a,b)=2{\int }_0^{\frac{\pi }{2}}{\sin }^{2a-1}u{\cos }^{2b-1}udu,}$

so this gives us the following relation to evaluate the Wallis integrals:

${\displaystyle W_n=\frac{1}{2}\mathrm{B}\left(\frac{n+1}{2},\frac{1}{2}\right)=\frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2\mathrm{\Gamma }\left(\frac{n}{2}+1\right)}.}$

So, for odd ${\displaystyle n}$, writing ${\displaystyle n=2p+1}$, we have:

${\displaystyle W_{2p+1}=\frac{\mathrm{\Gamma }\left(p+1\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2\mathrm{\Gamma }\left(p+1+\frac{1}{2}\right)}=\frac{p!\mathrm{\Gamma }\left(\frac{1}{2}\right)}{(2p+1)\mathrm{\Gamma }\left(p+\frac{1}{2}\right)}=\frac{2^{p}p!}{(2p+1)!!}=\frac{2^{2p}(p!{)}^2}{(2p+1)!},}$

whereas for even ${\displaystyle n}$, writing ${\displaystyle n=2p}$ and knowing that ${\displaystyle \mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi }}$, we get :

${\displaystyle W_{2p}=\frac{\mathrm{\Gamma }\left(p+\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2\mathrm{\Gamma }\left(p+1\right)}=\frac{(2p-1)!!\pi }{2^{p+1}p!}=\frac{(2p)!}{2^{2p}(p!{)}^2}\cdot \frac{\pi }{2}.}$

Equivalence

• From the recurrence formula above ${\displaystyle \mathbf{(}\mathbf{2}\mathbf{)}}$, we can deduce that

${\displaystyle W_{n+1}\sim W_n}$ (equivalence of two sequences).

Indeed, for all ${\displaystyle n\in \mathbb{N}}$ :

${\displaystyle W_{n+2}⩽W_{n+1}⩽W_n}$ (since the sequence is decreasing)

${\displaystyle \frac{W_{n+2}}{W_n}⩽\frac{W_{n+1}}{W_n}⩽1}$ (since ${\displaystyle W_n>0}$)

${\displaystyle \frac{n+1}{n+2}⩽\frac{W_{n+1}}{W_n}⩽1}$ (by equation ${\displaystyle \mathbf{(}\mathbf{2}\mathbf{)}}$).

By the sandwich theorem, we conclude that ${\displaystyle \frac{W_{n+1}}{W_n}\to 1}$, and hence ${\displaystyle W_{n+1}\sim W_n}$.

• By examining ${\displaystyle W_n{W}_{n+1}}$, one obtains the following equivalence:

${\displaystyle W_n\sim \sqrt{\frac{\pi }{2n}}}$ (and consequently ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\sqrt{n}{W}_n=\sqrt{\pi /2}}$ ).

Deducing Stirling's formula

Suppose that we have the following equivalence (known as Stirling's formula):

${\displaystyle n!\sim C\sqrt{n}{\left(\frac{n}{e}\right)}^n,}$

for some constant ${\displaystyle C}$ that we wish to determine. From above, we have

${\displaystyle W_{2p}\sim \sqrt{\frac{\pi }{4p}}=\frac{\sqrt{\pi }}{2\sqrt{p}}}$ (equation (3))

Expanding ${\displaystyle W_{2p}}$ and using the formula above for the factorials, we get

${\displaystyle \begin{array}{cc}{W}_{2p}& =\frac{(2p)!}{2^{2p}(p!{)}^2}\cdot \frac{\pi }{2}\\ & \sim \frac{C{\left(\frac{2p}{e}\right)}^{2p}\sqrt{2p}}{2^{2p}{C}^2{\left(\frac{p}{e}\right)}^{2p}{\left(\sqrt{p}\right)}^2}\cdot \frac{\pi }{2}\\ & =\frac{\pi }{C\sqrt{2p}}.\text{ (equation (4))}\end{array}}$

From (3) and (4), we obtain by transitivity:

${\displaystyle \frac{\pi }{C\sqrt{2p}}\sim \frac{\sqrt{\pi }}{2\sqrt{p}}.}$

Solving for ${\displaystyle C}$ gives ${\displaystyle C=\sqrt{2\pi }.}$ In other words,

${\displaystyle n!\sim \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n.}$

Deducing the Double Factorial Ratio

Similarly, from above, we have:

${\displaystyle W_{2p}\sim \sqrt{\frac{\pi }{4p}}=\frac{1}{2}\sqrt{\frac{\pi }{p}}.}$

Expanding ${\displaystyle W_{2p}}$ and using the formula above for double factorials, we get:

${\displaystyle W_{2p}=\frac{(2p-1)!!}{(2p)!!}\cdot \frac{\pi }{2}\sim \frac{1}{2}\sqrt{\frac{\pi }{p}}.}$

Simplifying, we obtain:

${\displaystyle \frac{(2p-1)!!}{(2p)!!}\sim \frac{1}{\sqrt{\pi p}},}$

or

${\displaystyle \frac{(2p)!!}{(2p-1)!!}\sim \sqrt{\pi p}.}$

Evaluating the Gaussian Integral

The Gaussian integral can be evaluated through the use of Wallis' integrals.

We first prove the following inequalities:

• ${\displaystyle \mathrm{\forall }n\in {\mathbb{N}}^{\ast }\mathrm{\forall }u\in {\mathbb{R}}_{+}u⩽n\Rightarrow (1-u/n{)}^n⩽e^{-u}}$
• ${\displaystyle \mathrm{\forall }n\in {\mathbb{N}}^{\ast }\mathrm{\forall }u\in {\mathbb{R}}_{+}{e}^{-u}⩽(1+u/n{)}^{-n}}$

In fact, letting ${\displaystyle u/n=t}$, the first inequality (in which ${\displaystyle t\in [0,1]}$) is equivalent to ${\displaystyle 1-t⩽e^{-t}}$; whereas the second inequality reduces to ${\displaystyle e^{-t}⩽(1+t{)}^{-1}}$, which becomes ${\displaystyle e^t⩾1+t}$. These 2 latter inequalities follow from the convexity of the exponential function (or from an analysis of the function ${\displaystyle t\mapsto e^t-1-t}$).

Letting ${\displaystyle u=x^2}$ and making use of the basic properties of improper integrals (the convergence of the integrals is obvious), we obtain the inequalities:

${\displaystyle {\int }_0^{\sqrt{n}}(1-x^2/n{)}^{n}dx⩽{\int }_0^{\sqrt{n}}{e}^{-x^2}dx⩽{\int }_0^{+\mathrm{\infty }}{e}^{-x^2}dx⩽{\int }_0^{+\mathrm{\infty }}(1+x^2/n{)}^{-n}dx}$ for use with the sandwich theorem (as ${\displaystyle n\to \mathrm{\infty }}$).

The first and last integrals can be evaluated easily using Wallis' integrals. For the first one, let ${\displaystyle x=\sqrt{n}\sin t}$ (t varying from 0 to ${\displaystyle \pi /2}$). Then, the integral becomes ${\displaystyle \sqrt{n}{W}_{2n+1}}$. For the last integral, let ${\displaystyle x=\sqrt{n}\tan t}$ (t varying from ${\displaystyle 0}$ to ${\displaystyle \pi /2}$). Then, it becomes ${\displaystyle \sqrt{n}{W}_{2n-2}}$.

As we have shown before, ${\displaystyle \underset{n\to +\mathrm{\infty }}{lim}\sqrt{n}{W}_n=\sqrt{\pi /2}}$. So, it follows that ${\displaystyle {\int }_0^{+\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }/2}$.

Remark: There are other methods of evaluating the Gaussian integral. Some of them are more direct.