Help computing a double integral on $|x-y|$

Question

If $f(x) = x e^{-x^2}$, how can you compute the double integral:

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |x-y|f(x)f(y)\,dx\,dy $$

I know that:

$$ \int x e^{-x^2} dx = -\frac{1}{2}e^{-x^2} + C$$

Answer 1

Observe that the integral over the 1st and 3rd quadrants are the same; so are those of the 2nd and the 4th quadrants. So, rewrite the integral as,

$$I=2\left(\int_{Q_1}+\int_{Q_2}\right) |x-y|f(x)f(y)\,dx\,dy $$

Furthermore, the 2nd-quadrant integral can be rewritten as an equivalent 1st quadrant one,

$$\int_{Q_2} |x-y|f(x)f(y)\,dx\,dy =-\int_{Q_1} (x+y)f(x)f(y)\,dx\,dy $$

Thus, the integral $I$ becomes, $$I=2\int_{Q_1} |x-y|xy e^{-(x^2+y^2)}\,dx\,dy -2\int_{Q_1} (x+y)xy e^{-(x^2+y^2)}\,dx\,dy $$

Then, rewrite above integrals in polar coordinates and recognize the symmetry around $\theta=\pi/4$ in the first integral,

$$I=4\int_0^{\pi/4} (\cos^2\theta \sin\theta - \cos\theta \sin^2\theta)d\theta \int_0^{\infty}r^4 e^{-r^2}dr$$ $$ -2\int_0^{\pi/2}(\cos^2\theta \sin\theta + \cos\theta \sin^2\theta) d\theta \int_0^{\infty}r^4 e^{-r^2}dr $$

Carrying out the integrals over $r$ and $\theta$ to obtain,

$$I = 4\times \frac 16 (2-\sqrt{2})\times \frac{3\sqrt{\pi}}{8}-2\times\frac 23\times \frac{3\sqrt{\pi}}{8} = -\frac{\sqrt{2\pi}}{4}$$

Answer 2

$$ I=\iint_{\mathbb{R}^2} xy|x-y| \exp\left[-(x^2+y^2)\right]\,dx\,dy $$ can be easily computed by switching to polar coordinates. It equals

$$ \int_{0}^{+\infty}\int_{0}^{2\pi}\rho^4 e^{-\rho^2}\cos\theta\sin\theta|\cos\theta-\sin\theta|\,d\theta\,d\rho $$ which by Fubini's theorem is simply the product between $$ \int_{0}^{+\infty}\rho^4 e^{-\rho^2}\,d\rho = \frac{1}{2}\int_{0}^{+\infty}x\sqrt{x}e^{-x}\,dx = \tfrac{1}{2}\Gamma\left(\tfrac{5}{2}\right)=\frac{3\sqrt{\pi}}{8} $$ and $$ \int_{0}^{\pi}\sin(2\theta)|\cos\theta-\sin\theta|\,d\theta $$ which can be computed by splitting $[0,\pi]$ into four congruent sub-intervals. The final outcome is $$ I = -\frac{3\sqrt{\pi}}{8}\cdot\frac{2\sqrt{2}}{3} = -\frac{\sqrt{2\pi}}{4}.$$

Answer 3

The substitution $x=u+v, y=u-v$ transforms the given integral into $$\iint_{\mathbb{R}^2}2|v|(u^2-v^2)e^{-2u^2-2v^2}(2\,du\,dv)=16(I_2 I_1-I_0 I_3),$$ where $I_k=\int_0^\infty x^k e^{-2x^2}\,dx$: $$I_0=\frac{\sqrt{\pi}}{2\sqrt{2}},\ I_1=\frac{1}{4},\ I_2=\frac{\sqrt{\pi}}{8\sqrt{2}},\ I_3=\frac{1}{8}$$ ($I_0$ is known, $I_1$ is evaluated like you noted, and the rest are done with integration by parts; alternatively, all four are evaluated using the $\Gamma$-function).

This is yet another "separating" substitution (suggested in my comment initially).

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[#ffe,15px]{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \verts{x - y}\pars{x\expo{-x^{2}}}\pars{y\expo{-y^{2}}}\dd x\,\dd y} \\[5mm] = &\ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \bracks{\vphantom{A^{A^{A}}}\verts{z}\delta\pars{z - x + y}} \pars{x\expo{-x^{2}}}\pars{y\expo{-y^{2}}}\dd x\,\dd y\,\dd z \\[5mm] = &\ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \bracks{\verts{z}\int_{-\infty}^{\infty}\expo{\ic k\pars{z - x + y}} {\dd k \over 2\pi}}\pars{x\expo{-x^{2}}} \pars{y\expo{-y^{2}}}\dd x\,\dd y\,\dd z \\[5mm] = &\ {1 \over 2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \expo{\ic kz}\verts{z} \verts{\int_{-\infty}^{\infty}x\expo{-x^{2} - \ic k x}\dd x}^{2} \dd k\,\dd z \\[5mm] = &\ {1 \over 2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \expo{\ic kz}\verts{z} \verts{\int_{-\infty}^{\infty}x\exp\pars{-\bracks{x + \ic\,{k \over 2}}^{2} - {k^{2} \over 4}}\dd x}^{\ 2} \dd k\,\dd z \\[5mm] = &\ {1 \over 2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \expo{-k^{2}/2 + \ic kz}\verts{z} \verts{\int_{-\infty + \ic k/2}^{\infty + \ic k/2} \pars{x - \ic\,{k \over 2}} \exp\pars{-x^{2}}\dd x}^{\ 2}\dd k\,\dd z \\[5mm] = &\ {1 \over 2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \expo{-k^{2}/2 + \ic kz}\verts{z}{\pi k^{2} \over 4}\,\dd k\,\dd z \\[5mm] = &\ {1 \over 8}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\pars{-\,{1 \over 2}\bracks{k - \ic z}^{2} - {z^{2} \over 2}}\verts{z}k^{2}\,\dd k\,\dd z \\[5mm] = &\ {1 \over 8}\int_{-\infty}^{\infty}\expo{-z^{2}/2}\verts{z} \int_{-\infty - \ic z}^{\infty - \ic z}\expo{-k^{2}/2} \pars{k^{2} + 2\ic kz - z^{2}}\,\dd k\,\dd z \\[5mm] = &\ {1 \over 8}\int_{-\infty}^{\infty}\expo{-z^{2}/2}\verts{z}\ \underbrace{\int_{-\infty}^{\infty}\expo{-k^{2}/2} \pars{k^{2} - z^{2}}\,\dd k} _{\ds{\root{2\pi}\pars{1 - z^{2}}}}\ \dd z \\[5mm] = &\ {\root{2\pi} \over 4}\ \underbrace{\int_{0}^{\infty}\expo{-z^{2}/2}z\pars{1 - z^{2}}\,\dd z} _{\ds{=\ -1}}\ =\ \bbox[15px,#ffd,border:1px solid navy]{-\,{\root{2\pi} \over 4}} \approx -0.6267 \end{align}