What is the correct substitution to integrate $\;\int_{-\infty}^\infty dp\ dp_o \ |p_o-p|\ e^{-\frac{p^2}{2}} e^{-\frac{p_0^2}{2}}\;$?

Question

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?

Answer 1

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