# 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.

• @Argon do you mean like: $x^2+y^2=r$ and $dydx=rd\phi dr$? I did try, but there is also an $x$ there. – Badshah Jan 14 '13 at 17:10
• Have you tried switching the order of integration? – Michael Biro Jan 14 '13 at 17:11
• This looks like substitution of $e^{something}$ – John Dvorak Jan 14 '13 at 17:13
• @MichaelBiro I also wanted to do that, but I was confused how to change it. I dont know the 'rules' exactly'. – Badshah Jan 14 '13 at 17:15

Interchange the order of the variables: $$\int_{-\infty}^\infty \int_{-\infty}^x x e^{-x^2} e^{-y^2}\ dy\ dx = \int_{-\infty}^\infty e^{-y^2} \int_{y}^\infty x e^{-x^2}\ dx\ dy$$
You can integrate by parts: $$\begin{eqnarray} \int_{-\infty}^{\infty} x e^{-x^2} \left(\int_{-\infty}^{x} e^{-y^2} dy\right) dx &=& -\frac{1}{2}\int_{-\infty}^{\infty} \left(\int_{-\infty}^{x} e^{-y^2} dy\right)d(e^{-x^2}) \\ &=&\frac{1}{2}\int_{-\infty}^{\infty}e^{-x^2}d\left(\int_{-\infty}^{x} e^{-y^2} dy\right) \\ &=&\frac{1}{2}\int_{-\infty}^{\infty}e^{-x^2}\left(e^{-x^2}dx\right) \\ &=&\frac{1}{2}\int_{-\infty}^{\infty}e^{-2x^2}dx \\ &=&\frac{1}{2}\sqrt{\frac{\pi}{2}}. \end{eqnarray}$$
$$\int_{-\infty}^\infty xe^{-x^2} \int_{-\infty}^x e^{-y^2}\, dy \, dx = \frac{\sqrt \pi}{2} \int_{-\infty}^\infty xe^{-x^2} \left[\operatorname{erf}x \right]_{-\infty}^x \, dx = \frac{\sqrt \pi}{2}\int_{-\infty}^\infty (\operatorname{erf} x+1)x e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2\sqrt 2}$$
and the last integral can be done by parts, because $\frac{d}{dx}\operatorname{erf} x = \frac{2e^{-x^2}}{\sqrt{\pi}}$