The Gaussian integral

Idea. One Gaussian integral is hard, but the square of it is an integral over the plane of $e^{-(x^2+y^2)}$, which is easy in polar coordinates because $rdr$ is the derivative of $\frac{r^2}{2}$.

Step 1: square it. $$I^2={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{-(x^2+y^2)}dxdy.$$

Step 2: polar coordinates. With $x^2+y^2=r^2$ and $dxdy=rdrd\theta$: $$I^2={\int }_0^{2\pi }{\int }_0^{\infty }{e}^{-r^2}rdrd\theta =2\pi \cdot \frac{1}{2}=\pi .$$

Step 3. $I>0$, so $I=\sqrt{\pi }\approx 1.772$.

Uses. Scaling gives $\int e^{-a{x}^2}dx=\sqrt{\pi /a}$, and with $a=\frac{1}{2}$ this is the $\sqrt{2\pi }$ in the normal density.