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: $$\int_{-\infty}^\infty e^{-x^2} \mathrm{d}x$$ $$\text{Let } I= \int_{-\infty}^\infty e^{-x^2} \mathrm{d}x \, \text{ and }  \,I= \int_{-\infty}^\infty e^{-y^2} \mathrm{d}y$$ $$\text{Then } I^2=\displaystyle \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} \mathrm{d}x \,\mathrm{d}y$$ Changing to polar coordinates requires use of the Jacobian $r$, treated after the evaluation of the integral: $$I^2= \int_0^\infty re^{-r^2} \mathrm{d}r \int_0^{2 \pi} \mathrm{d}\theta$$ $$I^2= 2\pi\int_0^\infty  r \, e^{-r^2} \mathrm{d}r$$ $$I^2= 2 \pi\, \left[ - \frac{1}{2} \, e^{-r^2} \right ]_0 ^ \infty=\pi$$ $$\text{Therefore }I=\sqrt{\pi} \, \text{ and } \, \int_{-\infty}^\infty e^{-x^2} \mathrm{d}x=\sqrt{\pi}$$ 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: $$\begin{equation} \frac{\partial (x,y)}{\partial (u,v)} = \left| \begin{array}{cc} \dfrac{\partial x}{\partial u}& \dfrac{\partial x}{\partial v} \\ \dfrac{\partial y}{\partial u}& \dfrac{\partial y}{\partial v} \\ \end{array} \right| \end{equation}$$ 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 $\dfrac{\partial x}{\partial r}=\cos \theta$ and $\dfrac{\partial x}{\partial \theta}=-r \sin \theta$
Also $\dfrac{\partial y}{\partial r}=\sin \theta$ and $\dfrac{\partial y}{\partial \theta}=r \cos \theta$. Thus: $$\begin{equation} \frac{\partial (x,y)}{\partial (r,\theta)} = \left| \begin{array}{cc} \cos \theta & -r \sin \theta \\ \sin \theta  &r \cos \theta\\ \end{array} \right| \end{equation}$$ 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 $\phi (x)=\dfrac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2}x^2}$, where $$\int_{-\infty} ^\infty \phi(x) \mathrm{d}x=\dfrac{1}{\sqrt{2 \pi}}\int_{-\infty} ^\infty e^{-\frac{1}{2}x^2} \mathrm{d}x =1$$ 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: $$\text{erf}(x)=\dfrac{1}{\sqrt{\pi}}\int_{-x}^x e^{-t^2} \mathrm{d}t$$ 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: