evaluating this difficult integral I have to evaluate this integral:
$\int^{\infty}_{-\infty}\int^{x}_{-\infty}xe^{-x^2}e^{-y^2}dydx$
I know how to evaluate the next integral:
$\int^{\infty}_{-\infty}e^{-x^2}dx$
But I just cant figure out how to do first one. I need hints.
 A: Interchange the order of the variables:
$$ \int_{-\infty}^\infty \int_{-\infty}^x x e^{-x^2} e^{-y^2}\ dy\ dx
= \int_{-\infty}^\infty e^{-y^2} \int_{y}^\infty x e^{-x^2}\ dx\ dy $$
A: You can integrate by parts:
$$
\begin{eqnarray}
\int_{-\infty}^{\infty} x e^{-x^2} \left(\int_{-\infty}^{x} e^{-y^2} dy\right) dx
&=& -\frac{1}{2}\int_{-\infty}^{\infty} \left(\int_{-\infty}^{x} e^{-y^2} dy\right)d(e^{-x^2}) \\
&=&\frac{1}{2}\int_{-\infty}^{\infty}e^{-x^2}d\left(\int_{-\infty}^{x} e^{-y^2} dy\right) \\
&=&\frac{1}{2}\int_{-\infty}^{\infty}e^{-x^2}\left(e^{-x^2}dx\right) \\
&=&\frac{1}{2}\int_{-\infty}^{\infty}e^{-2x^2}dx \\ &=&\frac{1}{2}\sqrt{\frac{\pi}{2}}.
\end{eqnarray}
$$
A: $$\int_{-\infty}^\infty xe^{-x^2} \int_{-\infty}^x e^{-y^2}\, dy \, dx =
 \frac{\sqrt \pi}{2} \int_{-\infty}^\infty xe^{-x^2} \left[\operatorname{erf}x \right]_{-\infty}^x \, dx =
 \frac{\sqrt \pi}{2}\int_{-\infty}^\infty (\operatorname{erf} x+1)x e^{-x^2} \, dx =
\frac{\sqrt{\pi}}{2\sqrt 2}$$
and the last integral can be done by parts, because $\frac{d}{dx}\operatorname{erf} x = \frac{2e^{-x^2}}{\sqrt{\pi}}$
A: Hint: Change the order of integration. It turns into two known integrals.
