Gaussian integral

The area of the blue region is equal to the value of the Gaussian integral.

The Gaussian integral is the integral of the Gaussian function over the entire real number line.

Introduction

The Gaussian integral is the improper integral defined as

$\int\limits_{-\infty}^\infty e^{-x^2}dx$

The function $e^{-x^2}$ is known as the Gaussian function. Note how the graph takes the traditional bell-shape, the shape of the Laplace curve.

You can use several methods to show that the integrand, the Gaussian function, has no indefinite integral that can be expressed in elementary terms. In other words, the integral resists the tools of elementary calculus. Still, there are several methods to evaluate it (differentiation under the integral sign, polar integration, contour integration, square of an integral, etc.) We will demonstrate the polar integration method.

Evaluation

We denote the Gaussian integral as I for convenience. From the square of an integral, we have:

$I^2=\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty e^{-x^2}e^{-y^2}dy\,dx$

We will now express the integral in polar coordinates. We can easily do this, since $-x^2-y^2$ is merely $-r^2$ in polar coordinates. Hence, our integral becomes

$I^2=\int\limits_0^{2\pi}\int\limits_0^\infty e^{-r^2}r\,dr\,d\theta$

We evaluate the right hand side. The integrand now does have an elementary indefinite integral, which can be evaluated with the substitution $u=r^2$ as follows:

$\int\limits_0^\infty e^{-r^2}r\,dr=\dfrac12\int\limits_0^\infty e^{-u}du=\frac12\Big[-e^{-u}\Big]_0^\infty=\frac12$

This was the integral with respect to $r$ . We substitute it in our original integral:

$I^2=\int\limits_0^{2\pi}\dfrac12\,d\theta=\frac12\int\limits_0^{2\pi}d\theta=\pi$

Taking the square root of both sides finally yields the desired expression:

$I=\sqrt{\pi}$