Truncated gaussian integral I wish to calculate $$\int_0^\infty \int_{-x}^\infty \frac{e^{-x^2/2}}{\sqrt {2\pi}} \frac{e^{-y^2/2}}{\sqrt {2\pi}}\, dy\, dx = \int_0^\infty \int_0^\infty \frac{e^{-x^2/2}}{\sqrt {2\pi}} \frac{e^{-y^2/2}}{\sqrt {2\pi}}\, dy\, dx + \int_0^\infty \int_{-x}^0 \frac{e^{-x^2/2}}{\sqrt {2\pi}} \frac{e^{-y^2/2}}{\sqrt {2\pi}}\, dy\, dx $$
$$= \frac{1}{4} + \int_0^\infty \int_{-x}^0 \frac{e^{-x^2/2}}{\sqrt {2\pi}} \frac{e^{-y^2/2}}{\sqrt {2\pi}}\, dy\, dx. $$
However, I'm not sure how to evaluate the second term. Since the inner integral is truncated, I don't know if converting to polar coordinates will help. 
 A: By symmetry, note that:
$$\int_0^\infty \int_0^x f(x,y)\,dy\,dx = \int_0^\infty \int_y^\infty f(x,y)\,dx \,dy=\int_0^\infty \int_x^\infty f(y,x)\,dy\,dx,$$
so if $f$ is symmetric:
$$\int_0^\infty \int_0^x f(x,y)\,dy\,dx = \frac{1}{2}\int_0^\infty \int_0^\infty f(x,y)\,dy\,dx.$$
Can you take it from here?
A: If we don't want to go deep into the details, I would consider the question as the training excercise for two classical methods:
1) integration by parts,
2) use of the property of the CDF of a standard normal distribution (denote it by $\Phi(x)$) that $\Phi(-x)=1-\Phi(x)$ for any $x$.
Let $\phi(x)=\Phi'(x)$. Since $\Phi(\infty)=1$, 
then what we are interested in is the value of
$$
\int_0^\infty dx
\phi(x)
\int_{-x}^\infty d\Phi(y)
=
\int_0^\infty dx
\phi(x)
(1-\Phi(-x))
$$
Since $\Phi(-x)=1-\Phi(x)$ we have 
$$
\int_0^\infty dx
\phi(x)
\int_{-x}^\infty d\Phi(y)
=
\int_0^\infty dx
\phi(x)
\Phi(x).
$$
The integral on the right is computed using the method 1),
which yields the required solution $0.375$.
