Solution

For

Im z > 0

, close the

t

-contour in the upper half-plane. Jordan lemma removes the large semicircle, and the only enclosed pole is the order-

n

pole at

t = z

Res_{t = z} \frac{e ^{i t}}{( t - z ) ^{n}} = \frac{1}{( n - 1 )!} \frac{d ^{n - 1}}{d t ^{n - 1}} e^{i t}_{t = z} = \frac{i ^{n - 1} e ^{i z}}{( n - 1 )!} .

The residue theorem therefore gives

F (z) = \frac{2 π i ^{n} e ^{i z}}{( n - 1 )!} .

To continue the integral itself, move the

t

-contour downward locally as

z

approaches and crosses the real axis, always keeping the pole above the contour. This analytic continuation by contour deformation defines

F (z) = \int_{C_{z}} \frac{e ^{i t}}{( t - z ) ^{n}} d t,

where

C_{z}

passes below

z

. Closing

C_{z}

upward continues to enclose the pole, even when

Im z \leq 0

, so

F (z) = \frac{2 π i ^{n} e ^{i z}}{( n - 1 )!}

there as well. The right-hand side is an entire function, so it is the unique analytic continuation of

F

to the whole complex plane.

Solved by gpt-5.6-sol high.

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