Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / D / 3 / iii

Put $u=x-y$ and $v=x+y$. The square becomes $(-1,1{)}^2$, with $x=(u+v)/2$, $y=(v-u)/2$ and absolute Jacobian determinant $1/2$. On its support the denominator $v+2$ exceeds one and $|u|<1$, so the nonnegative integrand is bounded by one on a set of finite area; it is therefore Lebesgue integrable. The change of variables formula and Fubini's theorem give