Hello I am trying to integrate $$ I:=\int_{a}^{b}\int_{a}^{b}xy\log\left(\,\left\vert\,x - y\,\right\vert\,\right) \,{\rm d}y\,{\rm d}x,\qquad 0 < a <b $$ for $x,y\in \mathbb{R}$. I added the bounds of integration to deal with any convergence problems as suggested. We can re-write this as $$ I=\frac{1}{2}\int_a^b \int_a^b xy\log \left((x-y)^2\right)\, dy\,dx. $$ From here we can re-write the log as $$ \frac{1}{2}\int_a^b \int_a^b xy \left(\int\frac{2}{x-y}dx\right)dy\, dx. $$ I am not sure how to approach this from here till end. Thanks.
Trying to Integrate$ \iint xy\log|x-y|\, dy\,dx $
398 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 2 best solutions below
On
First notice the integrand is symmetric in $x$ and $y$, so it suffices to double the $y > x$ portion; that is $$ I:=2\int_{a}^{b}\int_{x}^{b}xy\log\left(y - x \right) \,{\rm d}y\,{\rm d}x,\qquad 0 < a <b$$ Change variables to $u = y - x$ and $v = y + x$. The triangle of integration rotates to another triangle and we get $${1 \over 4}\int_{0}^{b - a}\int_{ 2a + u }^{2b - u}{v - u \over 2}{v + u \over 2}\log\left(u \right) \,{\rm d}v\,{\rm d}u,\qquad 0 < a <b$$ $$= {1 \over 16}\int_{0}^{b - a}\log\left(u \right)\int_{ 2a + u }^{2b - u}(v^2 - u^2) \,{\rm d}v\,{\rm d}u,\qquad 0 < a <b$$ $$ = {1 \over 16}\int_{0}^{b - a}\log\left(u \right)(-{4 \over 3})(2 a^3 + 3 a^2 u - (b - u)^2 (2 b + u))) \,du$$ The integral rather sucks but it can be done in an elementary fashion through an integration by parts.
If $x,y\in[a,b]\:\:$it converges, otherwise it does not. For simplicity lets say double integral is taken over $[0,1]\times[0,1]$.It has log singularity which is not that bad and integral converges $$ \int_{0}^{1}\int_{0}^{1} xy\log|x-y|\, dy\,dx=\int_{0}^{1}\int_{0}^{y}xy\log(y-x)\,dx\,dy+\int_{0}^{1}\int_{y}^{1}xy\log(x-y)\,dx\,dy $$ $$=\int_{0}^{1}y\int_{0}^{y}x\log(y-x)\,dx\,dy+\int_{0}^{1}y\int_{y}^{1}x\log(x-y)\,dx\,dy$$ $$=\int_{0}^{1}y\int_{0}^{y}(y-t)\log t\,dt\,dy+\int_{0}^{1}y\int_{0}^{1-y}(y+s)\log s\,ds\,dy$$ $$=\int_0^1y(\frac{1}{2}y^2\log y-\frac{3}{4}y^2)dy+\int_0^1y\big(\frac{1}{2}(1-y^2)\log(1- y)-\frac{1}{4}(1-y)(1+3y)\big)dy\approx -0.437.$$