Past exam of the mathematical methods II course of the University of Oxford / 2017 resit / 2 / Solution

Invert the change of variables: $x=(v-u)/2$, $y=(v+u)/2$. The absolute Jacobian determinant is $1/2$, and the transformed triangle is $0\le v\le 3$, $-v\le u\le v$. HenceAt the single vertex $v=0$, the printed integrand is undefined. This causes no integral divergence: throughout the remaining triangle $-1\le u/v\le 1$, so the integrand is bounded between $e^{-1}$ and $e$. Giving the vertex any finite value, or taking a limiting integral with $v>0$, gives the same answer.