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}{\sigma\sqrt{2\pi}} \, \exp \left( -\frac{(x- \mu)^2}{2\sigma^2} \right),$$

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


 * $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{2\pi} \exp \left( \frac{-x^2-y^2}{2} \right) \, dx \, dy = I^2.$$

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


 * $$\int_0^{2\pi} \int_0^{\infty} \frac{r}{2\pi} \exp (-r^2/2) \, dr \, d\theta = \int_0^{\infty} r \exp (-r^2/2) \, dr$$

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