Today, after giving the definition of the normal distribution a student asked me a very interesting question: how does the number π emerge in the Gaussian curve? The question is legitimate, because the Gaussian curve does not show any relation to the circle curve. I improvised an brief answer, and I will try to give here the path of this fascinating adventure of π. Firstly, the distribution function of a normal distributed random variable with expectation μ and variance σ² is

$$\displaystyle\Phi(x)=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(t-\mu)^2}{2\sigma^2}}dt$$

and its limit is

$$\displaystyle\lim_{x\rightarrow\infty}\Phi(x)=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\sigma^{2}}}e^{-\frac{(t-\mu)^{2}}{2\sigma^{2}}}dt=1$$

This is a normalised form of the Poisson integral:

$$\displaystyle\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}$$

Now the question is why this equals $\sqrt{\pi}$. For this I would give 2 answers:

1. Path

Substitution x²=t gives

$$\displaystyle\int_0^\infty e^{-x^2} dx =\frac1 2\Gamma \left(\frac 1 2\right) = \frac{\sqrt{\pi}}{2}$$

And how is $\Gamma \left(\frac 1 2\right) = \sqrt{\pi}$?

This can be explained over the Gauss form of the Gamma function:

$$\displaystyle\Gamma (x) = \lim_{n\rightarrow\infty}\frac{n!n^{x-1}}{\prod_{i=0}^{n-1} x+i} = \lim_{n\rightarrow\infty}\frac{n!n^{x}}{\prod_{i=0}^{n} x+i}$$

These 2 different forms give at ½ when multiplied with each other the double of the Wallis product:

$$\displaystyle 2\lim_{n\rightarrow\infty} \prod_{k=1}^n \frac{k^2}{k^2 - \frac 1 4}=\pi$$

Now, why does $W(n):=\frac{k^2}{k^2 - \frac 1 4} \prod_{k=1}^n$ converge to $\frac \pi 2$ for $n\rightarrow\infty$ (Wallis product)? This is because of the recursive relation

$$\displaystyle \int_0^{\frac \pi 2} \sin^nxdx = \frac {n-1} n \int_0^{\frac \pi 2} \sin^{n-2}xdx$$

The recursive resolution of the terms of the following fraction yields:

$$\displaystyle1=\lim_{n\rightarrow\infty}\frac{\int_0^{\frac \pi 2}\sin^{2n+1}xdx}{\int_0^{\frac\pi 2}\sin^{2n}xdx }=\lim_{n\rightarrow\infty}W(n)\frac 2 \pi$$

and therefore the equation of the Wallis product.

We see here that π comes from the sine function to the approximation of the Wallis product, then as the value of Γ(½) and from there it evaluates the square of the Poisson integral. A long path!

2. Path

The second path that I will present is shorter and more direct. Because

$$\displaystyle\left(\int_{-\infty}^{\infty} e^{-x^2}dx\right)^2=\int_{R^2} e^{-x^2-y^2}dxdy$$

calculating the left side should give the expected result (π). This can be done by integrating over a circle with centre in 0 and radius a (B(a)) and then by transformation to polar coordinates:

$$\displaystyle\lim_{a\rightarrow\infty}\int_{B\left(a\right)}e^{-x^{2}-y^{2}}dxdy=\lim_{r\rightarrow\infty}\int_{0}^{2\pi}\int_{0}^{r}re^{-r^{2}}drd\phi=\pi$$

References:

Forster Otto: Analysis 1; Differential- und Integralrechnung einer Veränderlichen, 6,. verbesserte Auflage, o.O., vieweg, 2001.

Κουνιας Στρατής, χρόνης Μωυσιάδης: Θεωρία Πιθανοτήτων Ι, Κλασική πιθανότητα, μονοδιάστατες κατανομές, Θεσσαλονίκη, Εκδόσης Ζήτη, 1999.

PS. Wikipedia has an interesting article about the calculation of the Gaussian integral.