Wallis' integrals

Template:Short description In mathematics, and more precisely in analysis, the Wallis integrals constitute a family of integrals introduced by John Wallis.

Definition, basic properties

The Wallis integrals are the terms of the sequence $(W_n{)}_{n\ge 0}$ defined by

$W_n=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n}xdx,$

or equivalently,

$W_n=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{n}xdx.$

The first few terms of this sequence are:

$W_0$	$W_1$	$W_2$	$W_3$	$W_4$	$W_5$	$W_6$	$W_7$	$W_8$	...	$W_n$
$\frac{\pi }{2}$	$1$	$\frac{\pi }{4}$	$\frac{2}{3}$	$\frac{3\pi }{16}$	$\frac{8}{15}$	$\frac{5\pi }{32}$	$\frac{16}{35}$	$\frac{35\pi }{256}$	...	$\frac{n-1}{n}{W}_{n-2}$

The sequence $(W_n)$ is decreasing and has positive terms. In fact, for all $n\ge 0:$

• $W_n>0,$ because it is an integral of a non-negative continuous function which is not identically zero;
• $W_n-W_{n+1}=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n}xdx-\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n+1}xdx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}({\sin }^{n}x)(1-\sin x)dx>0,$ again because the last integral is of a non-negative continuous function.

Since the sequence $(W_n)$ is decreasing and bounded below by 0, it converges to a non-negative limit. Indeed, the limit is zero (see below).

Recurrence relation

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

$\begin{array}{cc}\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n}xdx& =\underset{0}{\overset{\frac{\pi }{2}}{\int }}({\sin }^{n-2}x)(1-{\cos }^{2}x)dx\\ & =\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n-2}xdx-\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n-2}x{\cos }^{2}xdx.\text{Equation (1)}\end{array}$

Integrating the second integral by parts, with:

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

we have:

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

Substituting this result into equation (1) gives

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

and thus

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

for all $n\ge 2.$

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

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

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:

   $\mathrm{B}(x,y)=\underset{0}{\overset{1}{\int }}{t}^{x-1}(1-t{)}^{y-1}dt=\frac{\mathrm{\Gamma }(x)\mathrm{\Gamma }(y)}{\mathrm{\Gamma }(x+y)}$ for Template:Math

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

   $\mathrm{\Gamma }(z)=\underset{0}{\overset{\mathrm{\infty }}{\int }}{t}^{z-1}{e}^{-t}dt$ for Template:Math.

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

$\mathrm{B}(a,b)=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{2a-1}u{\cos }^{2b-1}udu,$

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

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

So, for odd $n$, writing $n=2p+1$, we have:

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

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

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

Equivalence

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

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

Indeed, for all $n\in \mathbb{N}$ :

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

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

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

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

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

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

Deducing Stirling's formula

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

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

for some constant $C$ that we wish to determine. From above, we have

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

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

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

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

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

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

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

Deducing the Double Factorial Ratio

Similarly, from above, we have:

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

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

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

Simplifying, we obtain:

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

or

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

Evaluating the Gaussian Integral

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

We first prove the following inequalities:

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

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

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

$\underset{0}{\overset{\sqrt{n}}{\int }}(1-x^2/n{)}^{n}dx⩽\underset{0}{\overset{\sqrt{n}}{\int }}{e}^{-x^2}dx⩽\underset{0}{\overset{+\mathrm{\infty }}{\int }}{e}^{-x^2}dx⩽\underset{0}{\overset{+\mathrm{\infty }}{\int }}(1+x^2/n{)}^{-n}dx$ for use with the sandwich theorem (as $n\to \mathrm{\infty }$).

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

As we have shown before, $\underset{n\to +\mathrm{\infty }}{\lim }\sqrt{n}{W}_n=\sqrt{\pi /2}$. So, it follows that $\underset{0}{\overset{+\mathrm{\infty }}{\int }}{e}^{-x^2}dx=\sqrt{\pi }/2$.

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

Note

The same properties lead to Wallis product, which expresses $\frac{\pi }{2}$ (see $\pi$) in the form of an infinite product.

External links

• Pascal Sebah and Xavier Gourdon. Introduction to the Gamma Function. In PostScript and HTML formats.