Take a look at the following integral:

$$\int_{-\infty}^{\infty}e^{-x^2}dx$$

You may recall that $f(x)=e^{-x^2}$ is a famous function which has no anti-derivative, which is true. However, just because it is not integratable does not mean that the function cannot be evaluated between bounds.

When inspecting the graph visually it does appear to converge (as opposed to an obvious divergence) so it is at least feasible that it may converge to a specific value.

The solution to solving this integral was discovered by Carl Friedrich Gauss, and is as follows:

$$\begin{align} \text{Let }I=\int_{-\infty}^{\infty} e^-x^2 \, dx \\ \end{align}$$

We now have the most peculiar first step:

$$\begin{align} \text{Therefore let }I^2=\int_{-\infty}^{\infty} e^-x^2 \, dx \int_{-\infty}^{\infty} e^-y^2 \, dy \\ \end{align}$$

Here we use the new variable $y$ for simplicity and convention.

$$\begin{align} \int_{-\infty}^{\infty} e^-x^2 \, dx \int_{-\infty}^{\infty} e^-y^2 \, dy = I^2\\ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2}e^{-y^2} \, dx \, dy = I^2 \\ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)} \, dx \, dy = I^2 \end{align}$$

Recall the polar coordinate conversion below, and that the bounds over $r$ are $[0,\infty]$ and over $\theta$ are $[0, 2\pi]$.

$$\begin{cases} r^2=x^2+y^2 \\ dy\,dx=r\,dr\,d\theta \\ \end{cases}$$

We now convert our integral to polar coordinates.

$$\begin{align} \int_{0}^{2\pi}\int_{0}^{\infty} e^{-r^2} r \, dr \, d\theta = I^2 \end{align}$$

In form we now have a very similar integral to that which we started with, but we have the critically important extra $r$ variable present. This now gives us the ability to integrate this function, which can be done easily by inspecting the derivative of $e^{-x^2}$. This is done simply with the chain rule.

$$\begin{align} \frac{d}{dx}[e^{-x^2}]=-2xe^{-x^2} \end{align}$$

This derivative is a simple factor of $-2$ difference from our current function value of $I^2$, so when we integrate over $r$ it will be simple to do so by including the reciprocal factor of $-\frac{1}{2}$. First we should integrate over theta because it will help keep the equation simple.

$$\begin{align} \int_{0}^{\infty}\int_{0}^{2\pi} r e^{-r^2} \, d\theta \, dr = I^2 \\ 2\pi \int_{0}^{\infty} re^{-r^{2}} \, dr = I^2 \\ \left( 2\pi*-\frac{1}{2} \right)*e^{-r^2}\Big|_{0}^{\infty}=I^2 \\ -\pi\ * (e^{-\infty^2}-e^{-0^2})=I^2 \\ -\pi\ * \left( \frac{1}{e^{\infty}} -1 \right)=I^2 \\ -\pi\ * \left( 0 -1 \right)=I^2 \\ \pi =I^2 \\ \sqrt{ \pi }=I\ \ \end{align}$$

Therefore:

$$\int_{-\infty}^{\infty} e^{-x^2} \, d =\sqrt{ \pi }$$

This is a quite surprising and pleasant result, and yet another area of mathematics where $e$ and $pi$ magically appear next to each other.

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