Integral in Normal Distribution

Normal distribution is defined by the so-called "hat function": $\displaystyle f (x)=\frac{1}{2\sqrt{\pi}}e^{-\frac{1}{2}x^{2}}$ The coefficient $\displaystyle \frac{1}{2\sqrt{\pi}}$ is so chosen as to insure that $\displaystyle \int_{-\infty}^{\infty} f(x)dx=1.$

To see how this come about, I'll compute the integral $\displaystyle I=\int_{-\infty}^{\infty} e^{-x^{2}}dx$. The idea is to convert the integral to a double integral by squaring and subsequently change to polar coordinates:

$\displaystyle \begin{array}{6,2} I\times I &= \int_{-\infty}^{\infty} e^{-x^{2}}dx \times\int_{-\infty}^{\infty} e^{-y^{2}}dy \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^{2}+y^{2}}dxdy \\ &= \int_{0}^{2\pi }\int_{0}^{\infty} re^{-r^{2}}drd\phi \\ &= \int_{0}^{2\pi }\left( -\frac{1}{2}e^{-r^{2}}\big|_{0}^{\infty} \right)d\phi \\ &= \int_{0}^{2\pi } \frac{1}{2}( 1 - 0)d\phi \\ &= 2\pi\times \frac{1}{2} = \pi, \end{array}$

implying that $\displaystyle I=\sqrt{\pi}$. The integral in the normal distribution is obtained by a change of variables and scale.