Is there a closed form expression for $\int_{- \infty}^\infty \int_{-\infty}^y \frac{1}{2 \pi} e^{-(1/2) ( x^2+y^2 )} \mathrm{d}x\,\mathrm{d}y$? 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.
 A: Jack D'Aurizio's answer is good, but since you said in comments under it that you wanted a different point of view, let's try this:
\begin{align}
u & = (\cos45^\circ)x-(\sin45^\circ)y = \tfrac{\sqrt{2}}2 x - \tfrac{\sqrt{2}}2 y \\
v & = (\sin45^\circ)x+(\cos45^\circ)y = \tfrac{\sqrt{2}}2 x + \tfrac{\sqrt{2}}2 y
\end{align}
This is just a $45^\circ$ rotation of the coordinate system, suggested by the fact that your boundary line $y=x$ is just a $45^\circ$ rotation of one of the coordinate axes.
Then simplify $u^2+v^2$ and find that it comes down to $x^2+y^2$.
Solving the system of two equations above for $x$ and $y$, one gets
\begin{align}
x & = \phantom{-}\tfrac{\sqrt{2}}2 u + \tfrac{\sqrt{2}}2 v \\
y & = -\tfrac{\sqrt{2}}2 u + \tfrac{\sqrt{2}}2 v 
\end{align}
By trivial algebra, the condition that $x\le y$ now becomes $u\le0$.
If you know about Jacobians, you get
$$
du\,dv = \left|\frac{\partial(u,v)}{\partial(x,y)}\right|\,dx\,dy = \left|\frac{\partial u}{\partial x}\cdot\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\cdot\frac{\partial v}{\partial x}\right|\,dx\,dy = 1\,dx\,dy.
$$
Hence your iterated integral becomes
$$
\int_{-\infty}^\infty\int_{-\infty}^0 \frac 1{2\pi} e^{-(u^2+v^2)/2}\,du\,dv
= \int_{-\infty}^\infty \int_{-\infty}^0 \left\{\frac 1{2\pi} e^{-v^2/2}\right\} e^{-u^2/2}\,du\,dv
$$
The part in $\{\text{braces}\}$ does not depend on $u$, so this is
$$
\int_{-\infty}^\infty\left( \frac 1{2\pi} e^{-v^2/2} \int_{-\infty}^0 e^{-u^2/2}\,du \right)\,dv.
$$
Now the inside integral does not depend on $v$, so it pulls out:
$$
\int_{-\infty}^\infty e^{-v^2/2}\,dv \cdot \frac1{2\pi} \int_{-\infty}^0 e^{-u^2/2}\,du
$$
and this is of course
$$
\int_{-\infty}^\infty \frac1{\sqrt{2\pi}} e^{-v^2/2}\,dv \cdot \int_{-\infty}^0 \frac1{\sqrt{2\pi}} e^{-u^2/2}\,du.
$$
The first integral comes to $1$ and the second, by a simple symmetry argument, is $1/2$.
A: 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.
A: Your integral is the probability:
$$\mathbb{P}[X\leq Y]$$
where $X$ and $Y$ are two independent normal variables $N(0,1)$, 
hence the value of the integral is just $\frac{1}{2}$, since:
$$\mathbb{P}[X\leq Y]=\mathbb{P}[Y\leq X],\qquad \mathbb{P}[X\leq Y]+\mathbb{P}[Y\leq X]=1.$$
A: Ah, I realise just now I'd misread to start with. Algebraically, we can solve it by noting that the value of the integral is the same under the change of variables $u=-x$, and since summing the two resulting integrals results in the $[-\infty,\infty] \times [-\infty,\infty]$ case, the answer is $1/2$.
i.e.:
$$\int_{- \infty}^\infty \int_{-\infty}^y \frac{1}{2 \pi} e^{-1/2 (x^2+y^2)} \,\mathrm{d}x\,\mathrm{d}y = \int_{- \infty}^\infty \int_{-y}^\infty \frac{1}{2 \pi} e^{-1/2 \left( x^2+y^2 \right)}\,\mathrm{d}x\,\mathrm{d}y.$$
$LHS+RHS=\int_{- \infty}^\infty \int_{-\infty}^\infty \frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y=1$, and $LHS=RHS$, so your integral is $1/2$.
EDIT: I guess I was a little bit short. There are several ways of showing $LHS+RHS$ is what I quote, but here is a simple, uninformative one.
$LHS+RHS=\int_{- \infty}^\infty \int_{-\infty}^\infty \frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y-\int_{- \infty}^\infty \int_{-y}^y\frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y$.
Now, Let $I=\int_{- \infty}^\infty \int_{-y}^y\frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y$. Use substitution $u=-y$. Then
$I=-\int_{\infty}^{-\infty} \int_{u}^{-u}\frac{1}{2 \pi} e^{-1/2 (x^2+u^2)}\,\mathrm{d}x\,\mathrm{d}u=\int_{-\infty}^{\infty} \int_{u}^{-u}\frac{1}{2 \pi} e^{-1/2 (x^2+u^2)}\,\mathrm{d}x\,\mathrm{d}u=-I$, hence as $I=-I$, $I=0$.
I think it could also be done by splitting the integration range up. (e.g. looking at the integrals on $(-\infty,0]$ and $[0,\infty)$)
