Integral $ \int_{-\infty}^{\infty} e^{ \frac{-u^2}{2}} du$

Question

Could you explain what rules are used to compute this integral? $ \int_{-\infty}^{\infty} e^{ \frac{-u^2}{2}} du = \sqrt{2\pi}$

Answer 1

$$\begin{align} \int_{\Bbb R^2}e^{-\frac{x^2+y^2}{2}}dydx&=\int_0^\infty\!\!\!\int_0^{2\pi}\rho e^{-\frac{\rho^2}2}d\theta d\rho=\\ &=2\pi\int_0^\infty \rho e^{-\frac{\rho^2}2}d\rho=2\pi \end{align}$$

If we let $I=\int_{\Bbb R}e^{-\frac {x^2}2}dx$, Fubini's theorem and the former computation tell us that $I^2=2\pi$, so $I=\sqrt{2\pi}$.