Family of mathematical integrals
In mathematics, and more precisely in analysis, the Wallis integrals constitute a family of integrals introduced by John Wallis.
Definition, basic properties
[edit]
The Wallis integrals are the terms of the sequence
defined by
![{\displaystyle W_{n}=\int _{0}^{\frac {\pi }{2}}\sin ^{n}x\,dx,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47647776ba61c70a5ab19c93c9b8df8f0db95192)
or equivalently,
![{\displaystyle W_{n}=\int _{0}^{\frac {\pi }{2}}\cos ^{n}x\,dx.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a8c0fb116c5e020a52251d1adf084a5900f0811)
The first few terms of this sequence are:
|
|
|
|
|
|
|
|
|
...
|
|
|
|
|
|
|
|
|
|
|
...
|
|
The sequence
is decreasing and has positive terms. In fact, for all
because it is an integral of a non-negative continuous function which is not identically zero;
again because the last integral is of a non-negative continuous function.
Since the sequence
is decreasing and bounded below by 0, it converges to a non-negative limit. Indeed, the limit is zero (see below).
Recurrence relation
[edit]
By means of integration by parts, a reduction formula can be obtained. Using the identity
, we have for all
,
![{\displaystyle {\begin{aligned}\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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18fabd81c0df96f21541e936b4bee0a3337b2ab5)
Integrating the second integral by parts, with:
, whose anti-derivative is ![{\displaystyle v(x)={\frac {1}{n-1}}\sin ^{n-1}(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edacce4cee9416ab6c1ec070926900a23f2793d5)
, whose derivative is ![{\displaystyle u'(x)=-\sin(x),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ce7758371c64eed5762b0c830814bbae1f7fc9f)
we have:
![{\displaystyle \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}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9bc041e854a7886c226ea1ef97744ddfec878b4)
Substituting this result into equation (1) gives
![{\displaystyle W_{n}=W_{n-2}-{\frac {1}{n-1}}W_{n},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab22dae9328b209579e36afc017bd0efb329f74b)
and thus
![{\displaystyle W_{n}={\frac {n-1}{n}}W_{n-2},\qquad {\text{Equation (2)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89ff369dd95d8d037056152fd92e48172f5e9990)
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:
![{\displaystyle 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}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e64a2405d899e3945ce92fd81f638f9ae686ea07)
![{\displaystyle 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)!}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1856b66b85f50b7de264352efdceaa5a397d5328)
Another relation to evaluate the Wallis' integrals
[edit]
Wallis's integrals can be evaluated by using Euler integrals:
- Euler integral of the first kind: the Beta function:
for Re(x), Re(y) > 0
- Euler integral of the second kind: the Gamma function:
for Re(z) > 0.
If we make the following substitution inside the Beta function:
![{\displaystyle \quad \left\{{\begin{matrix}t=\sin ^{2}u\\1-t=\cos ^{2}u\\dt=2\sin u\cos udu\end{matrix}}\right.,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bf307563edb190d24cad7e619d7460def0f7d2d)
we obtain:
![{\displaystyle \mathrm {B} (a,b)=2\int _{0}^{\frac {\pi }{2}}\sin ^{2a-1}u\cos ^{2b-1}u\,du,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00be5a324fdbb56e365d69ee4e744e464a3d55fa)
so this gives us the following relation to evaluate the Wallis integrals:
![{\displaystyle W_{n}={\frac {1}{2}}\mathrm {B} \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)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0bf50a9a90404c2491f3de05431e60ae279d468f)
So, for odd
, writing
, we have:
![{\displaystyle 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)!}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa388aa4e577f72019bf7f19660e4662b5759277)
whereas for even
, writing
and knowing that
, we get :
![{\displaystyle 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}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef949047464423028a6277e13aa68497e443a611)
- From the recurrence formula above
, we can deduce that
(equivalence of two sequences).
- Indeed, for all
:
(since the sequence is decreasing)
(since
)
(by equation
).
- By the sandwich theorem, we conclude that
, and hence
.
- 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
).
Suppose that we have the following equivalence (known as Stirling's formula):
![{\displaystyle n!\sim C{\sqrt {n}}\left({\frac {n}{e}}\right)^{n},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd7b806955b7a9524f975e805a8a896563d392c4)
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
![{\displaystyle {\begin{aligned}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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88e46f118d7cf59937c3e8bc7fa12bbd3307b86c)
From (3) and (4), we obtain by transitivity:
![{\displaystyle {\frac {\pi }{C{\sqrt {2p}}}}\sim {\frac {\sqrt {\pi }}{2{\sqrt {p}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22c4251b94fb54e4f7446a7279551d7c2d67626c)
Solving for
gives
In other words,
![{\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b3c28f23e205ed542a2b9bbeff5c56db3881877)
Deducing the Double Factorial Ratio
[edit]
Similarly, from above, we have:
![{\displaystyle W_{2p}\sim {\sqrt {\frac {\pi }{4p}}}={\frac {1}{2}}{\sqrt {\frac {\pi }{p}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/481a64ed9fe0bba5c4cb2fe8c48498ee1645d470)
Expanding
and using the formula above for double factorials, we get:
![{\displaystyle W_{2p}={\frac {(2p-1)!!}{(2p)!!}}\cdot {\frac {\pi }{2}}\sim {\frac {1}{2}}{\sqrt {\frac {\pi }{p}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6debce8493ad6e5f976e1b648360d8cbfaad5c22)
Simplifying, we obtain:
![{\displaystyle {\frac {(2p-1)!!}{(2p)!!}}\sim {\frac {1}{\sqrt {\pi \,p}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d174bd0c9b466f99f99117570e9eec5e80b3436e)
or
![{\displaystyle {\frac {(2p)!!}{(2p-1)!!}}\sim {\sqrt {\pi \,p}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9fea7c2865aea93ff6aeae100200e150702c109b)
Evaluating the Gaussian Integral
[edit]
The Gaussian integral can be evaluated through the use of Wallis' integrals.
We first prove the following inequalities:
![{\displaystyle \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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8d809a6e196a53cfde30077d994a36df1da613a)
![{\displaystyle \forall n\in \mathbb {N} ^{*}\quad \forall u\in \mathbb {R} _{+}\qquad e^{-u}\leqslant (1+u/n)^{-n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5aeda4c85f248f9381080ef840aac8206f55b628)
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.
- Pascal Sebah and Xavier Gourdon. Introduction to the Gamma Function. In PostScript and HTML formats.