Chapter 4, Problem 194SE

To determine

Show that $\frac{1}{2\pi }\left({\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-\frac{1}{2}u{y}^2}dy}\right)\left({\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-\frac{1}{2}u{x}^2}}dx\right)=\frac{1}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-\left(\frac{1}{2}\right)u\left(x^2+y^2\right)}}}dxdy=\frac{1}{u}$.