Compute $E(4XY(X^{2} + Y^{2} + 1)^{-1})$ for $X$, $Y$ independent and uniform on $\left[0,1\right]$

Question

Calculate $$ E\left(4XY \over X^{2} + Y^{2} + 1\right). $$ Right now I have gotten to $$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{4xy \over x^{2} + y^{2} + 1}\,{\rm d}y\,{\rm d}x. $$

Answer 1

To compute $$ A=E\left(\frac{4XY}{X^2+Y^2+1}\right), $$ one first uses the fact that the density of $(X,Y)$ is $1$ on the square $(0,1)^2$ and $0$ elsewhere (and not $1$ on the whole plane $\mathbb R^2$, which would not make for the density of a probability distribution). Second, one uses the change of variables $(x,y)=(u^2,v^2)$. This yields $$ A=\int_0^1\int_0^1\frac{4xy}{x^2+y^2+1} \mathrm dy\mathrm dx=\int_0^1\int_0^1\frac1{u+v+1} \mathrm dv\mathrm du. $$ Integrating with respect to $v$, one sees that the RHS is $$ \int_0^1\log(u+v+1){\Large\mid}_{v=0}^{v=1}\,\mathrm du=\int_0^1(\log(u+2)-\log(u+1))\mathrm du. $$ Integrating with respect to $u$, one sees that the (new) RHS is $$ (u+2)\log(u+2)-(u+1)\log(u+1){\Large\mid}_{u=0}^{u=1}=3\log3-4\log2, $$ hence $$ A=3\log3-4\log2=\log\left(\frac{27}{16}\right).$$