Erdős distinct distances problem

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

$W_n=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^n(x)dx,$

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

$W_n=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^n(x)dx$

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

$W_0=\frac{\pi }{2}\text{ and }{W}_1=1$

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

• $W_n>0$, because it is an integral of a non-negative continuous function which is not all zero in the integration interval
• $W_n-W_{n+1}=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^n(x)dx-\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n+1}(x)dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^n(x)[1-\sin (x)]dx⩾0$

(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

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

Noting that for all real $x$, ${\sin }^2(x)=1-{\cos }^2(x)$, we have, for all natural numbers $n⩾2$,

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^n(x)dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n-2}(x)[1-{\cos }^2(x)]dx$

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^n(x)dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n-2}(x)dx-\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n-2}(x){\cos }^2(x)dx$ (equation $\mathbf{(}\mathbf{1}\mathbf{)}$)

Integrating the second integral by parts, with:

• $u^{\prime }(x)=\cos (x){\sin }^{}(x)$, whose anti-derivative is $u(x)=\frac{1}{n-1}{\sin }^{n-1}(x)$
• $v(x)=\cos (x)$, whose derivative is $v^{\prime }(x)=-\sin (x)$

we have:

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n-2}(x){\cos }^2(x)dx={[\frac{1}{n-1}{\sin }^{n-1}(x)\cos (x)]}_0^{\frac{\pi }{2}}+\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{n-1}{\sin }^{n-1}(x)\sin (x)dx=0+\frac{1}{n-1}{W}_n$

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

$W_n=W_{n-2}-\frac{1}{n-1}{W}_n$

and thus

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

This gives the well-known identity:

$n{W}_n=(n-1)W_{n-2}$, valid for all $n⩾2$.

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 $n=2p$, $W_{2p}=\frac{2p-1}{2p}\times \frac{2p-3}{2p-2}\times \cdots \times \frac{1}{2}{W}_0=\frac{2p}{2p}\times \frac{2p-1}{2p}\times \frac{2p-2}{2p-2}\times \frac{2p-3}{2p-2}\times \cdots \times \frac{2}{2}\times \frac{1}{2}{W}_0=\frac{(2p)!}{2^{2p}(p!{)}^2}\frac{\pi }{2}$
• for $n=2p+1$, $W_{2p+1}=\frac{2p}{2p+1}\frac{2p-2}{2p-1}\cdots \frac{2}{3}{W}_1=\frac{2^{2p}(p!{)}^2}{(2p+1)!}$

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

Equivalence

• From the recurrence formula above $\mathbf{(}\mathbf{2}\mathbf{)}$, we can deduce that

$W_{n+1}\sim W_n$ (equivalence of two sequences).

Indeed, for all $n\in \mathbb{\mathbb{N}}$ :

$W_{n+2}⩽W_{n+1}⩽W_n$ (since the sequence is decreasing)

$\frac{W_{n+2}}{W_n}⩽\frac{W_{n+1}}{W_n}⩽1$ (since $W_n>0$)

$\frac{n+1}{n+2}⩽\frac{W_{n+1}}{W_n}⩽1$ (by equation $\mathbf{(}\mathbf{2}\mathbf{)}$).

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 $W_n{W}_{n+1}$, one obtains the following equivalence:

$W_n\sim \sqrt{\frac{\pi }{2n}}$ ( and consequently $\underset{n\to \mathrm{\infty }}{\lim }\sqrt{n}{W}_n=\sqrt{\pi /2}$ ).

Deducing Stirling's formula

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

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

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

• From above, we know that:

$W_{2p}\sim \sqrt{\frac{\pi }{4p}}=\frac{\sqrt{\pi }}{2}\frac{1}{\sqrt{p}}$ (equation $\mathbf{(}\mathbf{3}\mathbf{)}$)

• Expanding $W_{2p}$ and using the formula above for the factorials, we get:

$W_{2p}=\frac{(2p)!}{2^{2p}(p!{)}^2}\frac{\pi }{2}\sim \frac{C{\left(\frac{2p}{\mathrm{e}}\right)}^{2p}\sqrt{2p}}{2^{2p}{C}^2{\left(\frac{p}{\mathrm{e}}\right)}^{2p}{\left(\sqrt{p}\right)}^2}\frac{\pi }{2}$ and hence:

$W_{2p}\sim \frac{\pi }{C\sqrt{2}}\frac{1}{\sqrt{p}}$ (equation $\mathbf{(}\mathbf{4}\mathbf{)}$)

From $\mathbf{(}\mathbf{3}\mathbf{)}$ and $\mathbf{(}\mathbf{4}\mathbf{)}$, 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

The Gaussian integral can be evaluated through the use of Wallis' integrals.

We first prove the following inequalities:

• $\mathrm{\forall }n\in {\mathbb{\mathbb{N}}}^{*}\mathrm{\forall }u\in {\mathbb{ℝ}}_{+}u⩽n\Rightarrow (1-u/n{)}^n⩽e^{-u}$
• $\mathrm{\forall }n\in {\mathbb{\mathbb{N}}}^{*}\mathrm{\forall }u\in {\mathbb{ℝ}}_{+}{e}^{-u}⩽(1+u/n{)}^{-n}$

In fact, letting $u/n=t$, the first inequality (in which $t\in [0,1]$) is equivalent to $1-t⩽e^{-t}$; whereas the second inequality reduces to $e^{-t}⩽(1+t{)}^{-1}$, which becomes $e^t⩾1+t$. 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:

$\underset{0}{\overset{\sqrt{n}}{\int }}(1-x^2/n{)}^{n}dx⩽\underset{0}{\overset{\sqrt{n}}{\int }}{e}^{-x^2}dx⩽\underset{0}{\overset{+\mathrm{\infty }}{\int }}{e}^{-x^2}dx⩽\underset{0}{\overset{+\mathrm{\infty }}{\int }}(1+x^2/n{)}^{-n}dx$ for use with the sandwich theorem (as $n\to \mathrm{\infty }$).

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, $\underset{n\to +\mathrm{\infty }}{\lim }\sqrt{n}{W}_n=\sqrt{\pi /2}$. So, it follows that $\underset{0}{\overset{+\mathrm{\infty }}{\int }}{e}^{-x^2}dx=\sqrt{\pi }/2$.

Remark: There are other methods of evaluating the Gaussian integral. Some of them are more direct.

Relation with the Beta and Gamma functions

One of the definitions of the Beta function reads:

$\mathrm{B}(x,y)=2\underset{0}{\overset{\pi /2}{\int }}(\sin \theta {)}^{2x-1}(\cos \theta {)}^{2y-1}d\theta ,\mathrm{R}\mathrm{e}(x)>0,\mathrm{R}\mathrm{e}(y)>0$

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:

$\mathrm{B}(\frac{n+1}{2},\frac{1}{2})=2\underset{0}{\overset{\pi /2}{\int }}(\sin \theta {)}^n(\cos \theta {)}^{0}d\theta =2\underset{0}{\overset{\pi /2}{\int }}(\sin \theta {)}^{n}d\theta =2W_n$

or equivalently,

$W_n=\frac{1}{2}\mathrm{B}(\frac{n+1}{2},\frac{1}{2})$.

Exploiting the identity relating the Beta function to Gamma function:

$\mathrm{B}(x,y)=\frac{\mathrm{\Gamma }(x)\mathrm{\Gamma }(y)}{\mathrm{\Gamma }(x+y)}$

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

$W_n=\frac{1}{2}\frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(\frac{n+1}{2}+\frac{1}{2})}=\frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2\mathrm{\Gamma }\left(\frac{n+2}{2}\right)}$

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

$W_{2p+1}=\frac{\mathrm{\Gamma }(p+1)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2\mathrm{\Gamma }(p+1+\frac{1}{2})}=\frac{p!\mathrm{\Gamma }\left(\frac{1}{2}\right)}{(2p+1)\mathrm{\Gamma }(p+\frac{1}{2})}=\frac{2^{p}p!}{(2p+1)!!}=\frac{4^p(p!{)}^2}{(2p+1)!}$

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

$W_{2p}=\frac{\mathrm{\Gamma }(p+\frac{1}{2})\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2\mathrm{\Gamma }(p+1)}=\frac{(2p-1)!!\pi }{2^{p+1}p!}=\frac{(2p)!}{4^p(p!{)}^2}\cdot \frac{\pi }{2}$

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. In PostScript and HTML formats.