Wallis' integrals

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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 (W_n)_{\,n\, \in\, \mathbb{N}\,} defined by:

 W_n = \int_0^{\frac{\pi}{2}} \sin^n(x)\,dx,

or equivalently (through a substitution: x = \frac{\pi}{2} - t):

 W_n = \int_0^{\frac{\pi}{2}} \cos^n(x)\,dx

In particular, the first few terms of this sequence are:

W_0 W_1 W_2 W_3 W_4 W_5 W_6 W_7 W_8 ...
\frac{\pi}{2} 1 \frac{\pi}{4} \frac{2}{3} \frac{3\pi}{16} \frac{8}{15} \frac{5\pi}{32} \frac{16}{35} \frac{35\pi}{256} ...

The sequence \ (W_n) is decreasing and has strictly positive terms. In fact, for all n \in\, \mathbb{N} :

  • \ W_n > 0, because it is an integral of a non-negative continuous function which is not all zero in the integration interval
  • 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 \geqslant 0
(by the linearity of integration and because the last integral is an integral of a non-negative function within the integration interval)

Note: Since the sequence \ (W_n) is decreasing and bounded below by 0, it converges to a non-negative limit. Indeed, the limit is non-zero (see below).

Recurrence relation, evaluating the Wallis' integrals[edit]

By means of integration by parts, an interesting recurrence relation can be obtained:

Noting that for all real x, \quad \sin^2(x) = 1-\cos^2(x), we have, for all natural numbers n \geqslant 2,
\int_0^{\frac{\pi}{2}} \sin^{n}(x)\,dx = \int_0^{\frac{\pi}{2}} \sin^{n-2}(x) \left[1-\cos^2(x)\right]\,dx
\int_0^{\frac{\pi}{2}} \sin^{n}(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 (equation \mathbf{(1)})

Integrating the second integral by parts, with:

we have:

\int_0^{\frac{\pi}{2}} \sin^{n-2}(x) \cos^2(x)\,dx = \left[ \frac{1}{n-1} \sin^{n-1}(x) \cos(x)\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \ \frac{1}{n-1} \sin^{n-1}(x) \sin(x)\,dx = 0 + {1\over {n-1}}\,W_{n}

Substituting this result into \mathbf{(1)} gives:

W_n=W_{n-2} - {1\over {n-1}}\,W_{n}

and thus

 \qquad \left(1+ \frac{1}{n-1}\right)W_n=W_{n-2} (equation \mathbf{(2)})

This gives the well-known identity:

n\,W_n = (n-1)\,W_{n-2}\qquad \,, valid for all n \geqslant 2\qquad \,.

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.

  • for \quad n=2\,p, \quad W_{2\,p}=\frac{2\,p-1}{2\,p}\times\frac{2\,p-3}{2\,p-2}\times\cdots\times\frac{1}{2}\,W_0=\frac{2\,p}{2\,p}\times\frac{2\,p-1}{2\,p}\times\frac{2\,p-2}{2\,p-2}\times\frac{2\,p-3}{2\,p-2}\times\cdots\times\frac{2}{2}\times\frac{1}{2}\,W_0 = \frac{(2\,p)!}{2^{2\,p}\, (p!)^2} \frac{\pi}{2}
  • for \quad n=2\,p+1, \quad W_{2\,p+1}=\frac{2\,p}{2\,p+1}\,\frac{2\,p-2}{2\,p-1}\cdots\frac{2}{3}\,W_1=\frac{2^{2\,p}\, (p!)^2}{(2\,p +1)!}~

Note that all the even terms are irrational, whereas the odd terms are all rational.

Another relation to evaluate the Wallis'integrals[edit]

Wallis's integrals can be evaluated by using Euler integral :

  1. Euler integral of the first kind: the Beta function:
    \Beta(x,y)= \int_0^1t^{x-1}(1-t)^{y-1}\,dt =\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}
  2. Euler integral of the second kind: the Gamma function:
    \Gamma(z) = \int_0^\infty  t^{z-1}\,e^{-t}\,dt

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

We know that \Gamma(\tfrac{1}{2})=\sqrt \pi, so this gives us the following relation to evaluate the Wallis'integrals:

 W_n ={\frac{1}{2}}\Beta(\frac{n+1}{2},\frac{1}{2})=\frac{\sqrt \pi}{2}\frac{\Gamma(\tfrac{n+1}{2})}{\Gamma(\tfrac{n}{2}+1)}.


  • 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 \quad\lim_{n \rightarrow \infty} \sqrt n\,W_n=\sqrt{\pi /2}\quad ).

Deducing Stirling's formula[edit]

Suppose that we have the following equivalence (known as Stirling's formula)

\ n\,! \sim C\, \sqrt{n}\left(\frac{n}{\mathrm{e}}\right)^n, where \ C \in \R^*.

We now want to determine the value of this constant \ C using the formula for W_{2\, p}.

  • From above, we know that:
W_{2\, p} \sim \sqrt{\frac{\pi}{4\, p}} = \frac{\sqrt{\pi}}{2}\, \frac{1}{\sqrt{p}} (equation \mathbf{(3)})
  • Expanding W_{2\,p} and using the formula above for the factorials, we get:
W_{2\,p}=\frac{(2\,p)!}{2^{2\,p}\, (p\,!)^2}\, \frac{\pi}{2} \sim \frac{C\, \left(\frac{2\, p}{\mathrm{e}}\right)^{2p}\, \sqrt{2\, p}}{2^{2p}\, C^2\,  \left(\frac{p}{\mathrm{e}}\right)^{2p}\, \left(\sqrt{p}\right)^2}\, \frac{\pi}{2} and hence:
W_{2\,p} \sim \frac{\pi}{C\, \sqrt{2}}\, \frac{1}{\sqrt{p}} (equation \mathbf{(4)})
From \mathbf{(3)} and \mathbf{(4)}, we obtain, by transitivity,
\frac{\pi}{C\, \sqrt{2}}\, \frac{1}{\sqrt{p}} \sim \frac{\sqrt{\pi}}{2}\, \frac{1}{\sqrt{p}}, which gives :
\frac{\pi}{C\, \sqrt{2}} = \frac{\sqrt{\pi}}{2}, and hence C = \sqrt{2\, \pi}.
We have thus proved Stirling's formula:
\ n\,! \sim \sqrt{2\, \pi\, n}\, \left(\frac{n}{\mathrm{e}}\right)^n.

Evaluating the Gaussian Integral[edit]

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 \quad 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).

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.

Relation with the Beta and Gamma functions[edit]

One of the definitions of the Beta function reads:

\Beta(x,y) =
\qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0 \!

Putting x = \frac{n+1}{2}, y = \frac{1}{2} into this equation gives us an expression of the Wallis' integrals in terms of the Beta function:

\Beta \left( \frac{n+1}{2},\frac{1}{2} \right) =
= 2\int_0^{\pi/2}(\sin\theta)^{n}\,d\theta
= 2 W_n

or equivalently,

W_n = \frac{1}{2} \Beta \left( \frac{n+1}{2},\frac{1}{2} \right)

Exploiting the identity relating the Beta function to Gamma function:


We can rewrite the above in terms of the Gamma function:

    = \frac{1}{2} \frac{\Gamma \left( \frac{n+1}{2} \right)
                        \Gamma \left( \frac{1}{2} \right)
                    \Gamma \left( \frac{n+1}{2} + \frac{1}{2} \right)
    =             \frac{\Gamma \left( \frac{n+1}{2} \right)
                        \Gamma \left( \frac{1}{2} \right)
                    2 \, \Gamma \left( \frac{n+2}{2} \right)

So, for odd n, writing n = 2p+1, we have:

    =             \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! 
    =             \frac{4^p \; (p!)^2

whereas for even n, writing n = 2p, we get:

    =             \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)!
                    4^p \; (p!)^2


The same properties lead to Wallis product, which expresses \frac{\pi}{2}\, (see \pi) in the form of an infinite product.

External links[edit]

  • Pascal Sebah and Xavier Gourdon. Introduction to the Gamma Function. In PostScript and HTML formats.