Normalizing the Gaussian Integral: The Single-Variable Case

The goal here is to compute the integral $$ Z = \int_{-\infty}^{\infty}e^{-\frac{x^2}{2\sigma^2}}dx $$ where \(\sigma > 0\) is a real number.

Young man, in mathematics you don't understand things. You just get used to them.
— John von Neumann to Felix T. Smith

Sometimes, to solve a tough problem, it can be easier to solve a related problem and relate the solution to our original problem. We want to compute $$ Z = \int_{-\infty}^{\infty}e^{-\frac{x^2}{2\sigma^2}}dx. $$ Instead, we're going to first compute $$ Z^2 = \left(\int_{-\infty}^{\infty}e^{-\frac{x^2}{2\sigma^2}}dx\right)^2. $$

Starting from $$ Z^2 = \left(\int_{-a}^{a}e^{-\frac{x^2}{2\sigma^2}}dx\right)^2, $$ observe that $$ \begin{equation} \begin{aligned} Z^2 &= \left(\int_{-a}^{a}e^{-\frac{x^2}{2\sigma^2}}dx\right)^2 \\ &= \left(\int_{-a}^{a}e^{-\frac{x^2}{2\sigma^2}}dx\right)\left(\int_{-a}^{a}e^{-\frac{y^2}{2\sigma^2}}dy\right) \\ &= \int_{-a}^{a}\int_{-a}^{a}\exp\left({-\frac{1}{2\sigma^2}(x^2 + y^2)}\right)dxdy. \end{aligned} \end{equation} $$

Now the decisive step: we make the substitution \(x=r\cos\theta,\, y=r\sin\theta\). (When integrating, an expression like \(x^2 + y^2\) is screaming for a trigonometric substitution.)

At a minimum, \(r\) is zero at \(x=0,\,y=0\); it attains its maximum \(r = a\sqrt{2}\) at the points \((\pm a, \pm a)\). The angle \(\theta\), on the other hand, lies in \([0,2\pi)\).

The infinitesimal area element \(dxdy\) is now transformed into \(rdrd\theta\), so our integral then becomes $$ \begin{equation} \begin{aligned} Z^2 &= \int_{0}^{a\sqrt{2}}\int_{0}^{2\pi}\exp\left({-\frac{r^2}{2\sigma^2}}\right)rd\theta dr \end{aligned} \end{equation} $$

Now let's proceed to calculating the integral. The inner integral with respect to \(\theta\) is simple: $$ \int_{0}^{2\pi}\exp\left({-\frac{r^2}{2\sigma^2}}\right)rd\theta = 2\pi\exp\left({-\frac{r^2}{2\sigma^2}}\right)r $$ so we're left with $$ Z^2 = 2\pi \int_{0}^{a\sqrt{2}}\exp\left({-\frac{r^2}{2\sigma^2}}\right)rdr $$

This final integral is easily computed with the substitution \(u={r^2}/{2\sigma^2}\), so we end up with $$ Z^2 = 2\pi\sigma^2 (1 - \exp(-\frac{a^2}{\sigma^2})) $$ Now let's take the limit as \(a\rightarrow\infty\), which yields $$ \lim_{a\rightarrow\infty}Z^2 = 2\pi\sigma^2 (1 - \lim_{a\rightarrow\infty}\exp(-\frac{a^2}{\sigma^2})) = 2\pi\sigma^2 $$

Therefore, $$ Z = \int_{-\infty}^{\infty}e^{-\frac{x^2}{2\sigma^2}}dx = \sigma\sqrt{2\pi} $$

A mathematician is one to whom that is as obvious as that twice two makes four is to you. Liouville was a mathematician.
— Lord Kelvin (a physicist)