Wallis' integrals

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 $(W_{n})_{n\geq 0}$ defined by

$W_{n}=\int _{0}^{\frac {\pi }{2}}\sin ^{n}x\,dx,$

or equivalently,

$W_{n}=\int _{0}^{\frac {\pi }{2}}\cos ^{n}x\,dx.$

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 $n\geq 0:$

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

${\begin{aligned}\int _{0}^{\frac {\pi }{2}}\sin ^{n}x\,dx&=\int _{0}^{\frac {\pi }{2}}(\sin ^{n-2}x)(1-\cos ^{2}x)\,dx\\&=\int _{0}^{\frac {\pi }{2}}\sin ^{n-2}x\,dx-\int _{0}^{\frac {\pi }{2}}\sin ^{n-2}x\cos ^{2}x\,dx.\qquad {\text{Equation (1)}}\end{aligned}}$

Integrating the second integral by parts, with:

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

we have:

$\int _{0}^{\frac {\pi }{2}}\sin ^{n-2}x\cos ^{2}x\,dx=\left[{\frac {\sin ^{n-1}x}{n-1}}\cos x\right]_{0}^{\frac {\pi }{2}}+{\frac {1}{n-1}}\int _{0}^{\frac {\pi }{2}}\sin ^{n-1}x\sin x\,dx=0+{\frac {1}{n-1}}W_{n}.$

Substituting this result into equation (1) gives

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

and thus

$W_{n}={\frac {n-1}{n}}W_{n-2},\qquad {\text{Equation (2)}}$

for all $n\geq 2.$

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:

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

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

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

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

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

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

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

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

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

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

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

Equivalence

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

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

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

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

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

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

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:

$W_{n}\sim {\sqrt {\frac {\pi }{2\,n}}}\quad$ (and consequently $\lim _{n\rightarrow \infty }{\sqrt {n}}\,W_{n}={\sqrt {\pi /2}}$ ).

Proof

For all $n\in \,\mathbb {N}$, let $u_{n}=(n+1)\,W_{n}\,W_{n+1}$.

It turns out that, $\forall n\in \mathbb {N} ,\,u_{n+1}=u_{n}$ because of equation $\mathbf {(2)}$. In other words $\ (u_{n})$ is a constant.

It follows that for all $n\in \,\mathbb {N}$, $u_{n}=u_{0}=W_{0}\,W_{1}={\frac {\pi }{2}}$.

Now, since $\ n+1\sim n$ and $\ W_{n+1}\sim W_{n}$, we have, by the product rules of equivalents, $\ u_{n}\sim n\,W_{n}^{2}$.

Thus, $\ n\,W_{n}^{2}\sim {\frac {\pi }{2}}$, from which the desired result follows (noting that $\ W_{n}>0$).

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

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

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:

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

In fact, letting $u/n=t$, the first inequality (in which $t\in [0,1]$) is equivalent to $1-t\leqslant e^{-t}$; whereas the second inequality reduces to $e^{-t}\leqslant (1+t)^{-1}$, which becomes $e^{t}\geqslant 1+t$. 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:

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

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, $\lim _{n\rightarrow +\infty }{\sqrt {n}}\;W_{n}={\sqrt {\pi /2}}$. So, it follows that $\int _{0}^{+\infty }e^{-x^{2}}dx={\sqrt {\pi }}/2$.

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.