Chapter 14.3, Problem 63E

Probability

The value of the integral

$I={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{\raisebox{1ex}{$-x^2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}dx$

Is required in the normal probability density function.

a) Use polar coordinates to evaluate the improper integral.

$\begin{array}{l}{I}^2=\left({\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{\raisebox{1ex}{$-x^2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}dx\right)\left({\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{\raisebox{1ex}{$-y^2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}dy\right)\\ I^2={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{\raisebox{1ex}{$-(x^2+y^2)$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}dA}\end{array}$

b) Use the result of part (a) to determine I.