Wallis integrals

Short description: Family of mathematical integrals

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

## Contents

* 1 Definition, basic properties
* 2 Recurrence relation
* 3 Another relation to evaluate the Wallis' integrals
* 4 Equivalence
* 5 Deducing Stirling's formula
* 6 Deducing the Double Factorial Ratio
* 7 Evaluating the Gaussian Integral
* 8 Note
* 9 External links

## Definition, basic properties

The Wallis integrals are the terms of the sequence [math]\displaystyle{ (W_n)_{n \geq 0} }[/math] defined by

[math]\displaystyle{ W_n = \int_0^{\frac{\pi}{2}} \sin^n x \,dx, }[/math]

or equivalently,

[math]\displaystyle{ W_n = \int_0^{\frac{\pi}{2}} \cos^n x \,dx. }[/math]

The first few terms of this sequence are:

[math]\displaystyle{ W_0 }[/math] | [math]\displaystyle{ W_1 }[/math] | [math]\displaystyle{ W_2 }[/math] | [math]\displaystyle{ W_3 }[/math] | [math]\displaystyle{ W_4 }[/math] | [math]\displaystyle{ W_5 }[/math] | [math]\displaystyle{ W_6 }[/math] | [math]\displaystyle{ W_7 }[/math] | [math]\displaystyle{ W_8 }[/math] | ...  | [math]\displaystyle{ W_n }[/math]

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[math]\displaystyle{ \frac{\pi}{2} }[/math] | [math]\displaystyle{ 1 }[/math] | [math]\displaystyle{ \frac{\pi}{4} }[/math] | [math]\displaystyle{ \frac{2}{3} }[/math] | [math]\displaystyle{ \frac{3\pi}{16} }[/math] | [math]\displaystyle{ \frac{8}{15} }[/math] | [math]\displaystyle{ \frac{5\pi}{32} }[/math] | [math]\displaystyle{ \frac{16}{35} }[/math] | [math]\displaystyle{ \frac{35\pi}{256} }[/math] | ...  | [math]\displaystyle{ \frac{n-1}{n} W_{n-2} }[/math]

The sequence [math]\displaystyle{ (W_n) }[/math] is decreasing and has positive terms. In fact, for all [math]\displaystyle{ n \geq 0: }[/math]

* [math]\displaystyle{ W_n \gt 0, }[/math] because it is an integral of a non-negative continuous function which is not identically zero;
* [math]\displaystyle{ 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 \gt 0, }[/math] again because the last integral is of a non-negative continuous function.

Since the sequence [math]\displaystyle{ (W_n) }[/math] 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 [math]\displaystyle{ \sin^2 x = 1 - \cos^2 x }[/math], we have for all [math]\displaystyle{ n \geq 2 }[/math],

[math]\displaystyle{ \begin{align} \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{align} }[/math]

Integrating the second integral by parts, with:

* [math]\displaystyle{ v'(x)=\cos (x) \sin^{n-2}(x) }[/math], whose anti-derivative is [math]\displaystyle{ v(x) = \frac{1}{n-1} \sin^{n-1}(x) }[/math]
* [math]\displaystyle{ u(x)=\cos (x) }[/math], whose derivative is [math]\displaystyle{ u'(x) = - \sin(x), }[/math]

we have:

[math]\displaystyle{ \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. }[/math]

Substituting this result into equation (1) gives

[math]\displaystyle{ W_n = W_{n-2} - \frac{1}{n-1} W_n, }[/math]

and thus

[math]\displaystyle{ W_n = \frac{n-1}{n} W_{n-2}, \qquad\text{Equation (2)} }[/math]

for all [math]\displaystyle{ n \geq 2. }[/math]

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

* [math]\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}, }[/math]
* [math]\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)!}. }[/math]

## 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:

[math]\displaystyle{ \Beta(x,y)= \int_0^1 t^{x-1}(1-t)^{y-1}\,dt =\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} }[/math] for Re(x), Re(y) > 0
2. Euler integral of the second kind: the Gamma function:

[math]\displaystyle{ \Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt }[/math] for Re(z) > 0.

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

[math]\displaystyle{ \Beta(a,b)= 2\int_0^{\frac{\pi}{2}} \sin^{2a-1} u\cos^{2b-1} u\,du, }[/math]

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

[math]\displaystyle{ W_n = \frac{1}{2} \Beta\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)}. }[/math]

So, for odd [math]\displaystyle{ n }[/math], writing [math]\displaystyle{ n = 2p+1 }[/math], we have:

[math]\displaystyle{ 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)! }, }[/math]

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

[math]\displaystyle{ 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}. }[/math]

## Equivalence

* From the recurrence formula above [math]\displaystyle{ \mathbf{(2)} }[/math], we can deduce that

[math]\displaystyle{ \ W_{n + 1} \sim W_n }[/math] (equivalence of two sequences).

Indeed, for all [math]\displaystyle{ n \in\, \mathbb{N} }[/math] :
[math]\displaystyle{ \ W_{n + 2} \leqslant W_{n + 1} \leqslant W_n }[/math] (since the sequence is decreasing)
[math]\displaystyle{ \frac{W_{n + 2}}{W_n} \leqslant \frac{W_{n + 1}}{W_n} \leqslant 1 }[/math] (since [math]\displaystyle{ \ W_n \gt 0 }[/math])
[math]\displaystyle{ \frac{n + 1}{n + 2} \leqslant \frac{W_{n + 1}}{W_n} \leqslant 1 }[/math] (by equation [math]\displaystyle{ \mathbf{(2)} }[/math]).
By the sandwich theorem, we conclude that [math]\displaystyle{ \frac{W_{n + 1}}{W_n} \to 1 }[/math], and hence [math]\displaystyle{ \ W_{n + 1} \sim W_n }[/math].

* By examining [math]\displaystyle{ W_nW_{n+1} }[/math], one obtains the following equivalence:

[math]\displaystyle{ W_n \sim \sqrt{\frac{\pi}{2\, n}}\quad }[/math] (and consequently [math]\displaystyle{ \lim_{n \rightarrow \infty} \sqrt n\,W_n=\sqrt{\pi /2} }[/math] ).

Proof

For all [math]\displaystyle{ n \in\, \mathbb{N} }[/math], let [math]\displaystyle{ u_n = (n + 1)\, W_n\, W_{n + 1} }[/math].

It turns out that, [math]\displaystyle{ \forall n\in \N,\, u_{n + 1} = u_n }[/math] because of equation [math]\displaystyle{ \mathbf{(2)} }[/math]. In other words [math]\displaystyle{ \ (u_n) }[/math] is a constant.

It follows that for all [math]\displaystyle{ n \in\, \mathbb{N} }[/math], [math]\displaystyle{ u_n = u_0 = W_0\, W_1 = \frac{\pi}{2} }[/math].

Now, since [math]\displaystyle{ \ n + 1 \sim n }[/math] and [math]\displaystyle{ \ W_{n + 1} \sim W_n }[/math], we have, by the product rules of equivalents, [math]\displaystyle{ \ u_n \sim n\, W_n^2 }[/math].

Thus, [math]\displaystyle{ \ n\, W_n^2 \sim \frac{\pi}{2} }[/math], from which the desired result follows (noting that [math]\displaystyle{ \ W_n \gt 0 }[/math]).

## Deducing Stirling's formula

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

[math]\displaystyle{ n! \sim C \sqrt{n}\left(\frac{n}{e}\right)^n, }[/math]

for some constant [math]\displaystyle{ C }[/math] that we wish to determine. From above, we have

[math]\displaystyle{ W_{2p} \sim \sqrt{\frac{\pi}{4p}} = \frac{\sqrt{\pi}}{2\sqrt{p}} }[/math] (equation (3))

Expanding [math]\displaystyle{ W_{2p} }[/math] and using the formula above for the factorials, we get

[math]\displaystyle{ \begin{align} 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{align} }[/math]

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

[math]\displaystyle{ \frac{\pi}{C\sqrt{2p}} \sim \frac{\sqrt{\pi}}{2\sqrt{p}}. }[/math]

Solving for [math]\displaystyle{ C }[/math] gives [math]\displaystyle{ C = \sqrt{2\pi}. }[/math] In other words,

[math]\displaystyle{ n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n. }[/math]

## Deducing the Double Factorial Ratio

Similarly, from above, we have:

[math]\displaystyle{ W_{2p} \sim \sqrt{\frac{\pi}{4p}} = \frac{1}{2}\sqrt{\frac{\pi}{p}}. }[/math]

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

[math]\displaystyle{ W_{2p} = \frac{(2p-1)!!}{(2p)!!} \cdot \frac{\pi}{2} \sim \frac{1}{2}\sqrt{\frac{\pi}{p}}. }[/math]

Simplifying, we obtain:

[math]\displaystyle{ \frac{(2p-1)!!}{(2p)!!} \sim \frac{1}{\sqrt{\pi \, p}}, }[/math]

[math]\displaystyle{ \frac{(2p)!!}{(2p-1)!!} \sim \sqrt{\pi\, p}. }[/math]

## Evaluating the Gaussian Integral

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

We first prove the following inequalities:

* [math]\displaystyle{ \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} }[/math]
* [math]\displaystyle{ \forall n\in \mathbb N^* \quad \forall u \in\mathbb R_+ \qquad e^{-u} \leqslant (1+u/n)^{-n} }[/math]

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

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

[math]\displaystyle{ \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 }[/math] for use with the sandwich theorem (as [math]\displaystyle{ n \to \infty }[/math]).

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

As we have shown before, [math]\displaystyle{ \lim_{n\rightarrow +\infty} \sqrt n\;W_n=\sqrt{\pi /2} }[/math]. So, it follows that [math]\displaystyle{ \int_0^{+\infty} e^{-x^2} dx = \sqrt{\pi} /2 }[/math].

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 [math]\displaystyle{ \frac{\pi}{2}\, }[/math] (see [math]\displaystyle{ \pi }[/math]) in the form of an infinite product.

## External links

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

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