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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.

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@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 $e^{something}$ –  Jan 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

4 Answers 4

up vote 4 down vote accepted

Interchange the order of the variables: $$ \int_{-\infty}^\infty \int_{-\infty}^x x e^{-x^2} e^{-y^2}\ dy\ dx = \int_{-\infty}^\infty e^{-y^2} \int_{y}^\infty x e^{-x^2}\ dx\ dy $$

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You can integrate by parts: $$ \begin{eqnarray} \int_{-\infty}^{\infty} x e^{-x^2} \left(\int_{-\infty}^{x} e^{-y^2} dy\right) dx &=& -\frac{1}{2}\int_{-\infty}^{\infty} \left(\int_{-\infty}^{x} e^{-y^2} dy\right)d(e^{-x^2}) \\ &=&\frac{1}{2}\int_{-\infty}^{\infty}e^{-x^2}d\left(\int_{-\infty}^{x} e^{-y^2} dy\right) \\ &=&\frac{1}{2}\int_{-\infty}^{\infty}e^{-x^2}\left(e^{-x^2}dx\right) \\ &=&\frac{1}{2}\int_{-\infty}^{\infty}e^{-2x^2}dx \\ &=&\frac{1}{2}\sqrt{\frac{\pi}{2}}. \end{eqnarray} $$

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Hint: Change the order of integration. It turns into two known integrals.

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$$\int_{-\infty}^\infty xe^{-x^2} \int_{-\infty}^x e^{-y^2}\, dy \, dx = \frac{\sqrt \pi}{2} \int_{-\infty}^\infty xe^{-x^2} \left[\operatorname{erf}x \right]_{-\infty}^x \, dx = \frac{\sqrt \pi}{2}\int_{-\infty}^\infty (\operatorname{erf} x+1)x e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2\sqrt 2}$$

and the last integral can be done by parts, because $\frac{d}{dx}\operatorname{erf} x = \frac{2e^{-x^2}}{\sqrt{\pi}}$

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I cant follow this one, a little too fast. –  Badshah Jan 14 '13 at 17:16

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