# What is int_-oo^oo e^(-x^2)dx?

Nov 29, 2015

${\int}_{- \infty}^{\infty} {e}^{- {x}^{2}} \mathrm{dx} = \sqrt{\pi}$

#### Explanation:

The error function $\text{erf} \left(x\right)$ is defined as follows:

$\text{erf} \left(x\right) = \frac{2}{\sqrt{\pi}} {\int}_{0}^{x} {e}^{- {t}^{2}} \mathrm{dt}$

This non-elementary function is suitably scaled so that:

${\lim}_{x \to - \infty} \text{erf} \left(x\right) = - 1$

${\lim}_{x \to + \infty} \text{erf} \left(x\right) = 1$

So

${\int}_{- \infty}^{\infty} {e}^{- {x}^{2}} \mathrm{dx} = \frac{\sqrt{\pi}}{2} \left({\lim}_{x \to + \infty} \text{erf"(x) - lim_(x->-oo) "erf} \left(x\right)\right)$

$= \frac{\sqrt{\pi}}{2} \left(1 - \left(- 1\right)\right) = \sqrt{\pi}$