A random variable
\[X\]
is said to be modelled by a Normal distribution \[N( \mu , \sigma^2 )\]
if \[X \sim \frac{1}{ \sqrt{2 \pi}} e^{- \frac{x^2}{2 \sigma^2}}\]
Integration is via a roundabout route.
\[I_x =\frac{1}{ \sigma \sqrt{2 \pi}} \int^{\infty}_{- \infty} e^{- \frac{x^2}{2 \sigma^2}} dx\]
\[I_y =\frac{1}{ \sigma \sqrt{2 \pi}} \int^{\infty}_{- \infty} e^{- \frac{y^2}{2 \sigma^2}} dy\]
\[\begin{equation} \begin{aligned} I_x I_y &= \frac{1}{\sigma \sqrt{2 \pi}} \int^{\infty}_{- \infty} e^{- \frac{x^2}{2 \sigma^2}} dx \frac{1}{\sigma \sqrt{2 \pi}} \int^{\infty}_{- \infty} e^{- \frac{y^2}{2 \sigma^2}} dy \\ &= \frac{1}{ 2 \sigma^2 \pi} \int^{\infty}_{- \infty} \int^{\infty}_{- \infty} e^{- \frac{(x^2+y^2)}{2 \sigma^2}} dxdy \end{aligned} \end{equation} \]
Now transform to polar coordinates, obtaining
\[\begin{equation} \begin{aligned} I_x I_y &= \frac{1}{ 2 \sigma^2 \pi} \int^{\infty}_{0} \int^{2 \pi}_{0} e^{- \frac{ r^2}{2 \sigma^2}} r d \theta dr \\ &= \frac{1}{\sigma^2} \int^{\infty}_{0} e^{- \frac{ r^2}{2 \sigma^2}} r dr \\ &= [ -e^{- \frac{r^2}{2 \sigma^2}}]^{\infty}_0 \\ &= 0-(-1)=1 \end{aligned} \end{equation} \]
hence
\[I_x =I_y =1 \]