The Gaussian Integral

By way of revising some earlier topics that I've covered and of practising my LaTeX skills, I'm covering the evaluation of the Gaussian Integral: $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}\mathrm{d}x$$ $$\text{Let }I={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}\mathrm{d}x\text{ and }I={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-y^2}\mathrm{d}y$$ $$\text{Then }{I}^2={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-(x^2+y^2)}\mathrm{d}x\mathrm{d}y}$$ Changing to polar coordinates requires use of the Jacobian $r$, treated after the evaluation of the integral: $$I^2={\int }_0^{\mathrm{\infty }}r{e}^{-r^2}\mathrm{d}r{\int }_0^{2\pi }\mathrm{d}\theta$$ $$I^2=2\pi {\int }_0^{\mathrm{\infty }}r{e}^{-r^2}\mathrm{d}r$$ $$I^2=2\pi {[-\frac{1}{2}{e}^{-r^2}]}_0^{\mathrm{\infty }}=\pi$$ $$\text{Therefore }I=\sqrt{\pi }\text{ and }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}\mathrm{d}x=\sqrt{\pi }$$Now, about the Jacobian, let's suppose that $x$ and $y$ are functions of both $u$ and $v$, that is $x=g(u,v)$ and $y=h(u,v)$. The Jacobian is defined as: $$\frac{\partial (x,y)}{\partial (u,v)}=\left|\begin{array}{cc}{\displaystyle \frac{\partial x}{\partial u}}& {\displaystyle \frac{\partial x}{\partial v}}\\ {\displaystyle \frac{\partial y}{\partial u}}& {\displaystyle \frac{\partial y}{\partial v}}\end{array}\right|$$ Let's work with the specific change from rectangular to polar coordinates that is used in the solution to the Gaussian Integral.

We have $x=r\cos \theta$ and $y=r\sin \theta$

This means that ${\displaystyle \frac{\partial x}{\partial r}}=\cos \theta$ and ${\displaystyle \frac{\partial x}{\partial \theta }}=-r\sin \theta$
Also ${\displaystyle \frac{\partial y}{\partial r}}=\sin \theta$ and ${\displaystyle \frac{\partial y}{\partial \theta }}=r\cos \theta$. Thus: $$\frac{\partial (x,y)}{\partial (r,\theta )}=\left|\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right|$$ The determinant evaluates to $r{\cos }^2\theta +r{\sin }^2\theta =r$ (the value of the Jacobian).

The Gaussian Integral is closely connected with the normal distribution via the probability density function $\varphi (x)={\displaystyle \frac{1}{\sqrt{2\pi }}}{e}^{-\frac{1}{2}{x}^2}$, where $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\varphi (x)\mathrm{d}x={\displaystyle \frac{1}{\sqrt{2\pi }}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{1}{2}{x}^2}\mathrm{d}x=1$$ Wikipedia states that "in statistics one often uses the related error function, or $\text{erf}(x)$, defined as the probability of a random variable with normal distribution of mean 0 and variance 1/2 falling in the range $[-x,x]$"; that is: $$\text{erf}(x)={\displaystyle \frac{1}{\sqrt{\pi }}}{\int }_{-x}^x{e}^{-t^2}\mathrm{d}t$$Here is an informative video which explains why $\pi$ appears as the evaluation of the integral. It also explains how the double integral used to evaluate the integral is two dimensional and is finding the volume under a surface. By using the shell method to evaluate the integral, the need for the Jacobian is avoided: