# Gaussian Integral DRAFT

Let,

\begin{aligned} I & = \int^{\infty}_{-\infty} e^{-\frac 1 2 x^2} dx\\ J & = \int^{\infty}_{0} e^{-x^2} dx\\ K & = \int^{-\infty}_{\infty} e^{-\pi x^2}dx \end{aligned}

Theorem: $I = \sqrt{2\pi}$, and, equivalently, $J = \sqrt{\pi}/2$ and $K = 1$.

## Alternative Proof by Change of Variables

According to the first proof, we have

$J^2 = \int^{\infty}_{0} (\int^{\infty}_{0} e^{-(x^2+y^2)} dx)dy\\$

Set $x = yt$

\begin{aligned} J^2 & = \int^{\infty}_{0} (\int^{\infty}_{0} e^{-y^2(1+t^2)} y\ dt)dy\\ & = \int^{\infty}_{0} (\int^{\infty}_{0} y\ e^{-y^2(1+t^2)}\ dy)dt\ \end{aligned}

Interchange of integral justifies, by Fubini's theorem for improper Riemann integrals, integration by parts brah brah brah

$\int^{\infty}_{0} y\ e^{-ay^2}\ dy = \frac 1 {2a}$ $J^2 = \int^{\infty}_{0} \frac {dt}{2(t^2+1)} = \frac 1 2 \cdot \frac \pi 2 = \frac \pi 4$



## Proof by Gamma Function

$n! = \int^{\infty}_{0} t^n e^{-t}{\mathrm d}t$ Gamma function

$\Gamma (x) = \int^{\infty}_{0} t^x e^{-t} \frac {\mathrm d t} t,$ $\frac {\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = \int^{1}_{0} t^{x-1}(1-t)^{y-1}\mathrm d t$