Double gaussian integral with variable limits of integration.

$$\frac{1}{2\pi}\int\limits_{0}^{\infty}\int\limits_{-x}^{\infty}e^{-\frac{(x^2+y^2)}{2}}dydx$$

Is there a particularly nice way of working this to an exact value? The -x on the limits of integration makes this a little different from how I am used to solving these.

I was told to look at differentiating it this way, but I didn't see where this was heading:

$$=\frac{1}{2\pi}\int_{0}^{\infty}e^{-x^2/2}\int_{-x}^{\infty}e^{-y^2/2}dydx$$ Let $$g(x) =\int_{-x}^{\infty}e^{-y^2/2}dy$$

$$=\frac{1}{2\pi}\int_{0}^{\infty}e^{-x^2/2}g(x)dx$$

Let $$2\pi*F(t)=\int_{0}^{t}e^{-x^2/2}g(x)dx$$.

Thus $$2\pi*F'(t)=e^{-t^2/2}g(t)-g(0)=e^{-t^2/2}g(t)-\frac{\sqrt{2\pi}}{2}$$

At this point I don't see which move to make.

Use polar coordinates & the integral becomes $$\begin{eqnarray*} \int_0^{\frac{3 \pi}{4}} d \theta \int_0^{\infty} e^{r^2/2} r dr. \end{eqnarray*}$$
Hint: The $$x$$ and $$y$$ axes, together with the lines $$y=x$$ and $$y=-x$$, divide the plane into $$8$$ regions. Now use the rotational symmetry of the integrand, together with the fact that the integral over the whole plane is $$1$$.