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:

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

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_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}$$

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: If we make the following substitution inside the Beta function: $$\quad \left\{\begin{matrix} t = \sin^2 u \\ 1-t = \cos^2 u \\ dt = 2\sin u\cos u du\end{matrix}\right.,$$ we obtain:
 * 1) Euler integral of the first kind: the Beta function:
 * $$\Beta(x,y)= \int_0^1 t^{x-1}(1-t)^{y-1}\,dt =\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ for $Re(x), Re(y) > 0$
 * 1) Euler integral of the second kind: the Gamma function:
 * $$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$$ for $Re(z) > 0$.
 * $$\Beta(a,b)= 2\int_0^{\frac{\pi}{2}} \sin^{2a-1} u\cos^{2b-1} u\,du,$$

so this gives us the following relation to evaluate the Wallis integrals:
 * $$ W_n = \frac{1}{2} \Beta\left(\frac{n+1}{2},\frac{1}{2}\right)=\frac{\Gamma\left(\tfrac{n+1}{2}\right)\Gamma\left(\tfrac{1}{2}\right)}{2\,\Gamma\left(\tfrac{n}{2}+1\right)}.$$

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 $$W_nW_{n+1}$$, one obtains the following equivalence:


 * $$W_n \sim \sqrt{\frac{\pi}{2\, n}}\quad$$ (and consequently $$\lim_{n \rightarrow \infty} \sqrt n\,W_n=\sqrt{\pi /2}$$ ).

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{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}$$

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: 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 $$t \mapsto e^t -1 -t$$).
 * $$\forall n\in \mathbb N^* \quad \forall u\in\mathbb R_+ \quad u\leqslant n\quad\Rightarrow\quad (1-u/n)^n\leqslant e^{-u}$$
 * $$\forall n\in \mathbb N^* \quad \forall u \in\mathbb R_+ \qquad e^{-u} \leqslant (1+u/n)^{-n} $$

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.