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The integral is: $$\int_{-\infty}^\infty dx\ dy \ |x-y|\ e^{-\frac{x^2}{2}} e^{-\frac{y^2}{2}} $$ I have come across an integral of this form before and I think a substitution of the form:

$u=\frac{(x-y)}{2}$ and $v=\frac{(x+y)}{2}$

should be used. But the above doesn't work

Here is how I did it:

With the above substitution:

$du=(dx-dy)/2$

and

$dv=(dx+dy)/2$

so that the integral becomes:

$$\int_{-\infty}^\infty (du+dv)\ (dv-du) \ |2u|\ e^{-u^2} e^{-v^2}$$

I don't know how to proceed after this.

EDIT:

In order to calculate the pre-factor to integrand after the change of variables I calculate the Jacobian as follows: $$J=\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial v}{\partial x}\frac{\partial u}{\partial y}=\frac{1}{4}-(-\frac{1}{4})=\frac{1}{2}$$

which transforms the integral to:

$$(1/2)\int_{-\infty}^\infty du\ dv \ |2u|\ e^{-u^2} e^{-v^2} = 2\sqrt\pi$$

which differs from the answer of Jack.

Where am I wrong?

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    $\begingroup$ Ah, I see what you did wrong. Alternatively, your integral is $2\pi\Bbb E|X-Y|$ for independent $N(0,\,1)$ variables $X-Y$. Defining $Z:=(X-Y)/\sqrt{2}\sim N(0,\,1)$, this is $2\pi\sqrt{2}\Bbb E|Z|$, which lets you avoid using double integrals. $\endgroup$
    – J.G.
    Jun 23 at 9:58
  • $\begingroup$ Thanks! Please check the edit $\endgroup$
    – Lost
    Jun 23 at 11:41
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    $\begingroup$ You have $2$ errors that obscure each other. For one, $dxdy=\frac{dudv}{u_xv_y-u_yv_x}$ as opposed to $dxdy=dudv(u_xv_y-u_yv_x)$, so where you halved you should have doubled, so you need another factor of $4$. The other error is an erroneous factor of $2$, which is why the correct answer is twice what you got. Where this factor of $2$ occurred is unclear unless you make more explicit your work on$$\int_{-\infty}^\infty|2u|e^{-u^2}du=\int_0^\infty4ue^{-u^2}du=2.$$ $\endgroup$
    – J.G.
    Jun 23 at 12:04

1 Answer 1

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By a rotation of $45^\circ$ and Fubini's theorem $$ \iint_{\mathbb{R}^2}|x-y|e^{-\frac{x^2+y^2}{2}}\,dx\,dy = \sqrt{2}\iint_{\mathbb{R}^2}|u|e^{-u^2/2}e^{-v^2/2}\,du\,dv=4\sqrt{\pi}\int_{0}^{+\infty}u e^{-u^2/2}\,du=\color{red}{4\sqrt{\pi}}. $$

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