Past exam of the mathematical methods II course of the University of Oxford / 2014 resit / 9 / b / Solution

For the specified change of variables,Thus the inverse-coordinate Jacobian determinant is $\partial (u,v)/\partial (x,y)=-2/u$ where $u>0$, and the area element uses the absolute value $dxdy=(u/2)dudv$.

Now $u=x+y$, $v=(x-y)/(x+y)$ and $xy=u^2(1-v^2)/4$. The positive quadrant maps to the half-strip $u\ge 0$, $-1\le v\le 1$. The original origin corresponds to the collapsed boundary $u=0$, where $v$ is not unique; this measure-zero degeneracy does not affect the integral. ThereforeThe second factor is the area of a unit semicircle. Independently, the original integrand factors into two identical gamma integrals, giving $\Gamma (3/2{)}^2=\pi /4$.