Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 7E / a / Solution

For $\mathrm{Re}z>0$, absolute convergence permits the double-integral calculationPut $r=s+t$ and $u=t/(s+t)$, so $s=r(1-u)$, $t=ru$, and the Jacobian has magnitude $r$. The first quadrant becomes $r>0$, $0<u<1$, and therefore