What is the integral of $e^{-x^2/2}$ over $\mathbb{R}$ What is the integral of

$$\int_{-\infty}^{\infty}e^{-x^2/2}dx\,?$$

My working is here:
= $-e^(-1/2x^2)/x$ from negative infinity to infinity.
What is the value of this? Not sure how to carry on from here.  Thank you.

$$\left(\int\limits_{-\infty}^\infty e^{-\frac12x^2}dx\right)^2=\int\limits_{-\infty}^\infty e^{-\frac12x^2}dx\int\limits_{-\infty}^\infty e^{-\frac12y^2}dy=\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty e^{-\frac12(x^2+y^2)}dxdy=$$
Change now to polar coordinates:
$$=\int\limits_0^{2\pi}\int\limits_0^\infty re^{-\frac12r^2}drd\theta=\left.-2\pi e^{-\frac12r^2}\right|_0^\infty=2\pi$$
So your integral equals $\;\sqrt{2\pi}\;$
