Integration of the normal density function

:"Main article: Normal distribution"

The probability density function for the normal distribution is given by

:$f(x;mu,sigma)=frac\{1\}\{sigmasqrt\{2pi\; ,\; exp\; left(\; -frac\{(x-\; mu)^2\}\{2sigma^2\}\; ight),$

where $mu$ is the mean and $sigma$ the standard deviation.

By the definition of a probability density function, $f$ must integrate to 1. That is,

:$I\; =\; int\_\{-infty\}^\{infty\}\; f(x),\; dx\; =\; 1.$

However, this integration is not straight-forward, since $f$ does not have an antiderivative in closed form. For the special case when $mu\; =\; 0$ and $sigma\; =\; 1$, one method is to pass to the related double integral

:$int\_\{-infty\}^\{infty\}\; int\_\{-infty\}^\{infty\}\; frac\{1\}\{2pi\}\; exp\; left(\; frac\{-x^2-y^2\}\{2\}\; ight)\; ,\; dx\; ,\; dy\; =\; I^2.$

This double integral in cartesian coordinates can be converted to the following integral in polar coordinates

:$int\_0^\{2pi\}\; int\_0^\{infty\}\; frac\{r\}\{2pi\}\; exp\; (-r^2/2)\; ,\; dr\; ,\; d\; heta\; =\; int\_0^\{infty\}\; r\; exp\; (-r^2/2)\; ,\; dr$

which can be evaluated using the substitution $u\; =\; -r^2/2$ to yield 1, the desired result.