Consider instead the function $e^{-x^2/2-y^2/2}$ on the $x$, $y$-plane. What is the integral of this over the whole plane? Well you can do the $x$ and the $y$ integrals separately and get that the answer is the square of the Gaussian integral you wanted - so if we can do this new question then we will be able to do the one you wanted.

Okay - the whole function is very rotationally symmetric about the origin so we should change variables to plane polar coordinates $(r,t)$. The function becomes $e^{-r^2/2}$ and the area-element $\mathrm{dx}.\mathrm{dy}$ becomes $r.\mathrm{dr}.\mathrm{dt}$. So we have to integrate:

$r.e^{-r^2/2}\mathrm{dr}.\mathrm{dt}$ from $r=0$..infinity and $t=0$.. $2\pi$

Hopefully you will be able to do this integral and you'll see that the overall integral is $2\pi$.

So the original Gaussian integral had value equal to either $+\sqrt{2\pi }$ or $-\sqrt{2\pi }$. But it was the integral of an always positive function and so it must be the former.