In mathematics, and more precisely in analysis, the Wallis integrals constitute a family of integrals introduced by John Wallis.
The Wallis integrals are the terms of the sequence defined by
or equivalently,
The first few terms of this sequence are:
The sequence is decreasing and has positive terms. In fact, for all
Since the sequence is decreasing and bounded below by 0, it converges to a non-negative limit. Indeed, the limit is zero (see below).
By means of integration by parts, a reduction formula can be obtained. Using the identity , we have for all ,
Integrating the second integral by parts, with:
we have:
Substituting this result into equation (1) gives
and thus
for all
This is a recurrence relation giving in terms of . This, together with the values of and give us two sets of formulae for the terms in the sequence , depending on whether is odd or even:
Wallis's integrals can be evaluated by using Euler integrals:
If we make the following substitution inside the Beta function: <br> we obtain:
so this gives us the following relation to evaluate the Wallis integrals:
So, for odd , writing , we have:
whereas for even , writing and knowing that , we get :
Suppose that we have the following equivalence (known as Stirling's formula):
for some constant that we wish to determine. From above, we have
Expanding and using the formula above for the factorials, we get
From (3) and (4), we obtain by transitivity:
Solving for gives In other words,
Similarly, from above, we have:
Expanding and using the formula above for double factorials, we get:
Simplifying, we obtain:
or
The Gaussian integral can be evaluated through the use of Wallis' integrals.
We first prove the following inequalities:
In fact, letting , the first inequality (in which ) is equivalent to ; whereas the second inequality reduces to , which becomes . These 2 latter inequalities follow from the convexity of the exponential function (or from an analysis of the function ).
Letting and making use of the basic properties of improper integrals (the convergence of the integrals is obvious), we obtain the inequalities:
for use with the sandwich theorem (as ).
The first and last integrals can be evaluated easily using Wallis' integrals. For the first one, let (t varying from 0 to ). Then, the integral becomes . For the last integral, let (t varying from to ). Then, it becomes .
As we have shown before, . So, it follows that .
Remark: There are other methods of evaluating the Gaussian integral. Some of them are more direct.
The same properties lead to Wallis product, which expresses (see ) in the form of an infinite product.