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 $(W_n)_$ defined by
:$W_n = \int_0^ \sin^n x \,dx,$
or equivalently,
:$W_n = \int_0^ \cos^n x \,dx.$
The first few terms of this sequence are:

The sequence $(W_n)$ is decreasing and has positive terms. In fact, for all $n \geq 0:$
*$W_n > 0,$ because it is an integral of a non-negative continuous function which is not identically zero;
*$W_n - W_ = \int_0^ \sin^n x\,dx - \int_0^ \sin^ x\,dx = \int_0^ (\sin^n x)(1 - \sin x )\,dx > 0,$ 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 $n \geq 2$,

:$\begin \int_0^ \sin^n x \,dx &= \int_0^ (\sin^ x) (1-\cos^2 x) \,dx \\ &= \int_0^ \sin^ x \,dx - \int_0^ \sin^ x \cos^2 x \,dx. \qquad\text \end$

Integrating the second integral by parts, with:
:* $v'(x)=\cos (x) \sin^(x)$, whose anti-derivative is $v(x) = \frac \sin^(x)$
:* $u(x)=\cos (x)$, whose derivative is $u'(x) = - \sin(x),$
we have:
:\int_0^ \sin^ x \cos^2 x \,dx = \left \frac \cos x \right0^ + \frac\int_0^ \sin^ x \sin x \,dx = 0 + \frac W_n.

Substituting this result into equation (1) gives
:$W_n = W_ - \frac W_n,$
and thus
:$W_n = \frac W_, \qquad\text$
for all $n \geq 2.$

This is a recurrence relation giving $W_n$ in terms of $W_$. 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:

* $W_=\frac \cdot \frac \cdots \frac W_0 = \frac \cdot \frac = \frac \cdot \frac,$
* $W_=\frac \cdot \frac \cdots \frac W_1 = \frac = \frac.$

Another relation to evaluate the Wallis' integrals

Wallis's integrals can be evaluated by using Euler integrals:
#''Euler integral of the first kind'': the Beta function:
#:$\Beta(x,y)= \int_0^1 t^(1-t)^\,dt =\frac$ for
#''Euler integral of the second kind'': the Gamma function:
#:$\Gamma(z) = \int_0^\infty t^ e^\,dt$ for .
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:
:$\Beta(a,b)= 2\int_0^{\frac{\pi}{2 \sin^{2a-1} u\cos^{2b-1} u\,du,$
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:
*$\forall n\in \mathbb N^* \quad \forall u\in\mathbb R_+ \quad u\leqslant n\quad\Rightarrow\quad (1-u/n)^n\leqslant e^{-u}$
*$\forall n\in \mathbb N^* \quad \forall u \in\mathbb R_+ \qquad e^{-u} \leqslant (1+u/n)^{-n}$
In fact, letting $u/n=t$,
the first inequality (in which t \in ,1/math>) 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''. I
PostScript
an

formats.

{{DEFAULTSORT:Wallis' Integrals
Integrals