 Solution

If $\mathrm{Im}z>0$, the same upper-half-plane contour calculation givesIf $\mathrm{Im}z<0$, the pole lies below the real integration contour. Closing upward encloses no pole, so Jordan lemma and the Cauchy integral theorem giveHence at a real point $x$ the upper and lower boundary values satisfyThus crossing from the upper half-plane to the lower half-plane produces the negative of this jump.

An analytic continuation must be holomorphic, hence continuous, across every point where it is defined. The undeformed real-axis formula for $G$ has the nonzero jump above, while the continuation constructed in part (c) retains $2\pi i^n{e}^{iz}/(n-1)!$ below the axis. Therefore $G$ cannot be the analytic continuation of $F$.