Wallis' integrals explained

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} defined byW _{n} = \int ^{(\pi)/(2)} _{0} \sin ^{n} x dx, or equivalently,W _{n} = \int ^{(\pi)/(2)} _{0} \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}

-

(\pi)/(2)

1

(\pi)/(4)

(2)/(3)

(3 \pi)/(16)

(8)/(15)

(5 \pi)/(32)

(16)/(35)

(35 \pi)/(256)

...

(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 ^{(\pi)/(2)} _{0} \sin ^{n} xdx - \int ^{(\pi)/(2)} _{0} \sin ^{n+1} xdx = \int ^{(\pi)/(2)} _{0} (\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{align} \int ^{(\pi)/(2)} _{0} \sin ^{n} x dx & = \int ^{(\pi)/(2)} _{0} (\sin ^{n-2} x) ( 1- \cos ^{2} x) dx \\ & = \int ^{(\pi)/(2)} _{0} \sin ^{n-2} x dx - \int ^{(\pi)/(2)} _{0} \sin ^{n-2} x \cos ^{2} x dx. Equation(1) \end{align}

Integrating the second integral by parts, with:

v'(x) = \cos (x) \sin ^{n-2} (x) , whose anti-derivative is v(x) = (1)/(n -1) \sin ^{n-1} (x) u(x) = \cos (x) , whose derivative is u'(x) = - \sin(x), we have:\int ^{(\pi)/(2)} _{0} \sin ^{n-2} x \cos ^{2} x dx = \left[ (\sin ^{n-1} x)/(n -1) \cos x \right] ^{(\pi)/(2)} _{0} + (1)/(n -1) \int ^{(\pi)/(2)} _{0} \sin ^{n-1} x \sin x dx = 0 + (1)/(n -1) W _{n.}

Substituting this result into equation (1) gives

W _{n} = W _{n-2} - (1)/(n -1) W _{n,} and thusW _{n} = (n -1)/(n) W _{n-2} , 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} = (2 p -1)/(2 p) ⋅ (2 p -3)/(2 p -2) … (1)/(2) W _{0} = (( 2 p -1 )!!)/(( 2 p)!!) ⋅ (\pi)/(2) = (( 2 p)!)/(2 ^{2p} (p!) ^{2}) ⋅ (\pi)/(2) , W _{2p+1} = (2 p)/(2 p +1) ⋅ (2 p -2)/(2 p -1) … (2)/(3) W _{1} = (( 2 p)!!)/(( 2 p +1 )!!) = (2 ^{2p} (p!) ^{2})/(( 2 p +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:

\Beta(x,y) = \int ^{1} _{0} t ^{x-1} ( 1- t) ^{y-1} dt = (\Gamma(x)\Gamma(y))/(\Gamma(x + y)) for

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

\Gamma(z) = \int ^{inf ty} _{0} t ^{z-1} e ^{-t} dt for .If we make the following substitution inside the Beta function:\left\{\begin{matrix} t = \sin ^{2} u \ 1- t = \cos ^{2} u \ dt = 2 \sin u\cos u du\end{matrix}\right.,
we obtain:\Beta(a,b) = 2 \int ^{(\pi)/(2)} _{0} \sin ^{2a-1} u\cos ^{2b-1} udu, so this gives us the following relation to evaluate the Wallis integrals:W _{n} = (1)/(\Beta\left( 2) (n +1)/(, 2) (1)/(\right) = 2) (\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 = 2 p +1 , we have:W _{2p+1} = (\Gamma \left( p +1 \right) \Gamma \left( (1)/(2) \right))/(2 \Gamma \left( p +1 + (1)/(2) \right)) = (p! \Gamma \left( (1)/(2) \right))/(( 2 p +1 ) \Gamma \left( p + (1)/(2) \right)) = (2 ^{p} p!)/(( 2 p +1 )!!) = (2 ^{2p} (p!) ^{2})/(( 2 p +1 )!) , whereas for even n , writing n = 2 p and knowing that \Gamma\left(\tfrac{ 1 }{ 2 }\right) = \sqrt{\pi} , we get :W _{2p} = (\Gamma \left( p + (1)/(2) \right) \Gamma \left( (1)/(2) \right))/(2 \Gamma \left( p +1 \right)) = (( 2 p -1 )!! \pi)/(2 ^{p+1} p!) = (( 2 p)!)/(2 ^{2p} (p!) ^{2}) ⋅ (\pi)/(2) .

Equivalence

• From the recurrence formula above

(2) , we can deduce thatW _{n} \sim W _{n} (equivalence of two sequences).

Indeed, for all

n \in N :W _{n} \leqslant W _{n} \leqslant W _{n} (since the sequence is decreasing)(W _{n})/(W _{n}) \leqslant (W _{n})/(W _{n}) \leqslant 1 (since W _{n} > 0 )(n +1)/(n +2) \leqslant (W _{n})/(W _{n}) \leqslant 1 (by equation (2) ).

By the sandwich theorem, we conclude that

(W _{n})/(W _{n}) \to 1 , and hence W _{n} \sim W _{n} .

• By examining

W _{nW} _{n+1} , one obtains the following equivalence:W _{n} \sim \sqrt{ (\pi)/(2 n) }\quad (and consequently \lim _{n} \sqrt nW _{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( (n)/(e) \right) ^{n,} for some constant C that we wish to determine. From above, we haveW _{2p} \sim \sqrt{ (\pi)/(4 p) } = \frac (equation (3))

Expanding

W _{2p} and using the formula above for the factorials, we get\begin{align} W _{2p} & = (( 2 p)!)/(⋅ 2 ^{2p} (p!) ^{2}) (\pi)/(2) \\ &\sim (C \left( (2 p)/(e) \right) ^{2p} \sqrt{ 2 p)/() }\cdot\frac \\ &= \frac. \text \end

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

(\pi)/(C\sqrt{ 2 p) } \sim \frac.Solving for C gives C = \sqrt{ 2 \pi}. In other words,n! \sim \sqrt{ 2 \pi n} \left( (n)/(e) \right) ^{n.}

Deducing the Double Factorial Ratio

Similarly, from above, we have:

W _{2p} \sim \sqrt{ (\pi)/(4 p) } = \frac\sqrt.Expanding W _{2p} and using the formula above for double factorials, we get:W _{2p} = (( 2 p -1 )!!)/(( 2 p)!!) ⋅ (\pi)/(2) \sim (1)/(\sqrt{ 2) (\pi)/(p) }.Simplifying, we obtain:(( 2 p -1 )!!)/(( 2 p)!!) \sim (1)/(\sqrt{\pi p) },or(( 2 p)!!)/(( 2 p -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 N ^{*} \forall u\in R _{+} u\leqslant n ⇒ ( 1- u/n) ^{n\leqslant} e ^{-u} \forall n\in N ^{*} \forall u \in R _{+} e ^{-u} \leqslant ( 1+ u/n) ^{-n} In fact, letting u/n = t ,the first inequality (in which t \in [ 0 , 1 ] ) isequivalent to 1- t\leqslant e ^{-t} ;whereas the second inequality reduces toe ^{-t} \leqslant ( 1+ t) ^{-1} ,which becomes e ^{t\geqslant} 1+ t .These 2 latter inequalities follow from the convexity of theexponential function(or from an analysis of the function t \mapsto e ^{t} -1 - t ).

Letting

u = x ^{2} andmaking use of the basic properties of improper integrals(the convergence of the integrals is obvious),we obtain the inequalities:\int ^{\sqrt n} _{0} ( 1- x ^{2/n)} ^{n} dx \leqslant \int ^{\sqrt n} _{0} e ^{- x ^{2}} dx \leqslant \int ^{+inf ty} _{0} e ^{- x ^{2}} dx \leqslant \int ^{+inf ty} _{0} ( 1+ x ^{2/n)} ^{-n} dx for use with the sandwich theorem (as n \to inf ty ).

The first and last integrals can be evaluated easily usingWallis' integrals.For the first one, let

x = \sqrt n \sint (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 →} \sqrt n W _{n=\sqrt{\pi} / 2 } . So, it follows that\int ^{+inf ty} _{0} 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

(\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.