 Solution

In the common sector $0<\arg z<\pi /2$, rotate the positive real contour clockwise to the negative imaginary axis. The large quarter-circle contributes zero because $\mathrm{Re}(z\zeta )>0$ there. The rotated ray passes through the simple pole $\zeta =-i$, whose residue isTake a small semicircular indentation into the right half-plane, excluding the pole from the clockwise contour. The downward ray traverses this semicircle clockwise, contributing $-i\pi R$ in the limit. Thus the deformed downward integral is $\tilde{F}-i\pi R$, and it equals the original integral. ConsequentlyThis half-residue, rather than a full $2\pi i$ residue, results from a pole lying on the rotated integration ray.

Parametrizing $\zeta =-iu$ givesFor $\mathrm{Im}z>0$ its tail decays exponentially. Near $u=1$, subtract the pole coefficient before integrating; the remaining integrand is regular, and the principal-value pole term is explicit. This proves holomorphic dependence on $z$ throughout the upper half-plane, locally uniformly on compact subsets. Agreement in the overlap and the identity theorem therefore give the requested analytic continuation: