= Solution
Choose $z=e^{it}$ on the right semicircle, with the continuous branch $\log z=it$. Then
$$
I=\frac1i\int_{-i}^{i}\frac{z^u}{z+iA}\,dz.
$$
Here the integral is along that semicircle. Since $|A|>1$, the pole lies outside the unit disc and the <geometric series> converges uniformly on the contour.
Put
$$
H_u(w)={}_2F_1(1,u+1;u+2;w).
$$
The <Euler integral for the hypergeometric function>, initially for $\operatorname{Re}u>-1$, gives
$$
H_u(w)=(u+1)\int_0^1\frac{s^u}{1-ws}\,ds
=(u+1)\sum_{n\ge0}\frac{w^n}{u+n+1},\qquad |w|<1.
$$
The prefactor follows from $\Gamma(u+2)/\Gamma(u+1)=u+1$. Hence an antiderivative of the contour integrand before multiplication by $1/i$ is
$$
\frac{z^{u+1}}{iA(u+1)}H_u\!\left(-\frac z{iA}\right).
$$
Evaluating at the endpoints with the chosen branch yields
$$
\boxed{I(u,A)=
-\frac{i}{A(u+1)}
\left[
e^{i\pi u/2}{}_2F_1(1,u+1;u+2;-1/A)
+e^{-i\pi u/2}{}_2F_1(1,u+1;u+2;1/A)
\right].}
$$
This is first obtained where the individual expressions are defined. The original finite-interval integral is entire in $u$, so <analytic continuation> gives the result elsewhere, taking removable limits at the apparent exceptional negative integers. That convention matters: the two separate hypergeometric terms may be singular even though their combination is finite.
Back to article page
