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:
Another relation to evaluate the Wallis' integrals
Wallis's integrals can be evaluated by using Euler integrals:
By examining , one obtains the following equivalence:
(and consequently ).
Proof
For all , let .
It turns out that, because of equation . In other words is a constant.
It follows that for all , .
Now, since and , we have, by the product rules of equivalents, .
Thus, , from which the desired result follows (noting that ).
Deducing Stirling's formula
Suppose that we have the following equivalence (known as Stirling's formula):
for some constant that we wish to determine. From above, we have
(equation (3))
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,
Deducing the Double Factorial Ratio
Similarly, from above, we have:
Expanding and using the formula above for double factorials, we get:
Simplifying, we obtain:
or
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 , 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:
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.
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