\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:
