1. ## Probability Integral

In probability theory the important Gaussian Distribution often leads to the integral $X=\int_{-\infty}^{\infty} e^{-x^2} dx$.

In case you ever wondered where it comes from, here (informal but nice*) is a non-complex analysis demonstration.

We begin by considering the integral,
$Y=\iint_{\mathbb{R}^2} e^{-(x^2+y^2)} \ dA$

This is the integral over the entire plane $\mathbb{R}^2$.

Now there are two ways to get the entire plane.

Method 1: For $a,b>0$ we can create a rectangle $-a\leq x\leq a \mbox{ and }-b\leq x\leq b$. Now we make $a\to \infty \mbox{ and }b\to \infty$. That infinite rectangle will take the entire plane.
In other words, $Y=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)} \ dy \ dx$.
Now, the important realization is that this double integral has "seperated variables" meaning $f(x,y)=g(x)h(y)$ and we can write $\int_{-\infty}^{\infty} e^{-x^2} dx \cdot \int_{-\infty}^{\infty} e^{-y^2} dx$. Hence, $Y = X^2 \implies X = \sqrt{Y}$ (because it cannot be negative).

Method 2: We notice the expression $x^2+y^2$ in the double integral and try a polar coordinates substitution. If we let $r^2 = x^2+y^2$, we can let $0\leq \theta \leq 2\pi$ and $r\to \infty$. In other words, we draw a circle at the orgin with infinite radius, and this will take on values on the entire $\mathbb{R}^2$ plane.
Thus, we end up with after a change to polar coordinates,
$Y = \int_0^{\infty} \int_0^{2\pi} e^{-r^2} \ r d\theta \ dr = \lim_{r\to \infty} 2\pi \int_0^r re^{-r^2}dr = 2\pi \left( \frac{1}{2} \right) = \pi$

Thus, $Y=\pi$ which implies $X=\sqrt{\pi}$.

And hence,
$\int_{-\infty}^{\infty} e^{-x^2} \ dx = \sqrt{\pi}$

*)I am sure it can be made formal. But I do not know how do it because I am not familar with the theory of real multiple integration.

2. In my calculus textbook, a squeeze technique is used to prove the result via double integrals in polar coordinate.