\begin{eqnarray*} \lint_{-\infty}^\infty \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}} \, dx &=& \sqrt{ \left( \lint_{-\infty}^\infty \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}} \, dx \right)^2} \\ &=& \sqrt{ \lint_{-\infty}^\infty \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}} \, dx \lint_{-\infty}^\infty \frac{e^{-\frac{(y-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}} \, dy} \\ &=& \sqrt{ \lint_{-\infty}^\infty \lint_{-\infty}^\infty \frac{e^{-\frac{(x-\mu)^2+(y-\mu)^2}{2\sigma^2}}}{\sigma^2 2\pi} \, dx \, dy} \end{eqnarray*} Substitute $x^\prime=x-\mu$ and $y^\prime=y-\mu$ . Since the bounds are infinite, they do not change, and $dx^\prime=dx$ and $dy^\prime=dy$ . Thus, we have \begin{eqnarray*} \sqrt{\lint_{-\infty}^\infty \lint_{-\infty}^\infty \frac{e^{-\frac{(x-\mu)^2+(y-\mu)^2}{2\sigma^2}}}{\sigma^2 2\pi} \, dx \, dy} &=& \sqrt{\lint_{-\infty}^\infty \lint_{-\infty}^\infty \frac{e^{-\frac{(x^\prime)^2+(y^\prime)^2}{2\sigma^2}}}{\sigma^2 2\pi} \, dx^\prime \, dy^\prime}. \end{eqnarray*} Converting to polar coordinates, we obtain \begin{eqnarray*} \sqrt{\lint_{-\infty}^\infty \lint_{-\infty}^\infty \frac{e^{-\frac{(x^\prime)^2+(y^\prime)^2}{2\sigma^2}}}{\sigma^2 2\pi} \, dx^\prime \, dy^\prime} &=& \sqrt{\lint_0^\infty \lint_0^{2\pi} \frac{re^{-\frac{r^2}{2\sigma^2}}}{\sigma^2 2\pi} \, dr \, d\theta} \\ &=& \sqrt{\lint_0^{2\pi} \frac{d\theta}{2\pi}} \sqrt{\lint_0^\infty \frac{re^{-\frac{r^2}{2\sigma^2}}}{\sigma^2} \, dr} \\ &=& \sqrt{\frac{\theta}{2\pi}\bigg|_0^{2\pi}} \sqrt{\frac{1}{\sigma^2}\lint_{0}^\infty r e^{-\frac{r^2}{2\sigma^2}} \, dr} \\ &=& \sqrt{\frac{2\pi}{2\pi}} \sqrt{\frac{\sigma^2}{\sigma^2}\left( -e^{-\frac{r^2}{2\sigma^2}} \right)\bigg|_0^\infty} \\ &=& \sqrt{1} \sqrt{1}\\ &=& 1. \end{eqnarray*}