TITLE: Is there a closed form expression for $\int_{- \infty}^\infty \int_{-\infty}^y \frac{1}{2 \pi} e^{-(1/2) ( x^2+y^2 )} \mathrm{d}x\,\mathrm{d}y$?
QUESTION [10 upvotes]: I have been trying to evaluate the integral:

$$\int_{- \infty}^\infty \int_{-\infty}^y \frac{1}{2 \pi} e^{-(1/2) ( x^2+y^2 )}\mathrm {d}x\,\mathrm{d}y$$

I know of course that the integral equals $1$ over $[-\infty,\infty] \times [-\infty,\infty]$ but I do not quite know how to handle the present case. Are there any tricks here? 
Thank you.

REPLY [4 votes]: Set $x=r\cos \theta,y=r\sin \theta$, then we have
$$\int_{- \infty}^\infty \int_{-\infty}^y  e^{-(1/2) ( x^2+y^2 )}\mathrm {d}x\,\mathrm{d}y=\int_{0}^\infty \left(\int_{-3\pi/4}^{\pi/4}  e^{-r^2/2}\mathrm {d}\theta\,\right)r\,\mathrm{d}r=\pi \int_{0}^\infty e^{-r^2/2}r\,\mathrm{d}r=\pi$$
So the original integral is equal to $(1/2)$.
This method as well as @MichaelHardy's method also works for the integrals like:
$$\int_{- \infty}^\infty \int_{-\infty}^{a y}  e^{-(1/2) ( x^2+y^2 )}\mathrm {d}x\,\mathrm{d}y, \text{  }a \in \mathbb{R}$$
The results are the same. This is because the function to be integrated ($e^{-r^2/2}$)  is rotational invariant (independent of $\theta$) and $x=a y$ is a straight line going through the origin and divides the plane into 2 halfs of equal size.