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.

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