= Solution
If $\operatorname{Im}z>0$, the same upper-half-plane contour calculation gives
$$
G(z)=\frac{2\pi i^ne^{iz}}{(n-1)!}.
$$
If $\operatorname{Im}z<0$, the pole lies below the real integration contour. Closing upward encloses no pole, so <Jordan lemma> and the <Cauchy integral theorem> give
$$
G(z)=0.
$$
Hence at a real point $x$ the upper and lower boundary values satisfy
$$
\boxed{
G(x+i0)-G(x-i0)
=\frac{2\pi i^ne^{ix}}{(n-1)!}
}.
$$
Thus 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 function>[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 $2\pi i^ne^{iz}/(n-1)!$ below the axis. Therefore $G$ cannot be the analytic continuation of $F$.
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