Evaluating the integral $\int_{-\infty}^{\infty} \frac{(1-ix)^{n-s_1}}{(1+ix)^{n+s_2}} dx$. How can one show that for $s_1,s_2 \in \mathbb{C}$ $$
\begin{aligned}
\int_{-\infty}^{\infty} \frac{(1-ix)^{n-s_1}}{(1+ix)^{n+s_2}}  dx = &   2\sin(\pi(n+s_2))2^{1-s_1-s_2} \Gamma(1-n-s_2)\Gamma(s_1+s_2-1)/ \Gamma(s_1-n)  & \\ = & 2^{2-s_1-s_2}\pi \Gamma(s_1+s_2-1)/[\Gamma(n+s_2)\Gamma(s_1-n)]. 
\end{aligned} 
$$
If I recall correctly this could be done with complex analysis noting that the integral is equal to
$$
\int_{C} \frac{(1-ix)^{n-s_1}}{(1+ix)^{n+s_2}}  dx,
$$
where $C$ starts from $+i \infty$ along the imaginary axis and goes around $i$ once in a small circle (positive direction) and then goes back to $+i\infty$.
However this doesn't help me since I don't know how we can deform the contour to $C$ and to evaluate this contour integral. 
 A: One considers the complex contour integral
$$\oint_C dz \frac{(1-i z)^{n-s_1}}{(1+i z)^{n+s_2}} $$
where $C$ is a semicircle of radius $R$ in the upper half plane, with a detour along either side of the imaginary axis and around the branch point $z=i$ with a circle of radius $\epsilon$.  We take as a branch cut the ray $\operatorname{Re}{z} =0$, $\operatorname{Im}{z} \gt 1$.  The contour integral is zero by Cauchy's theorem.  Writing the contour integral out, we get
$$\int_{-R}^R dx \frac{(1-i x)^{n-s_1}}{(1+i x)^{n+s_2}} + i R \int_0^{\pi/2} d\theta \, e^{i \theta} \frac{(1-i R e^{i \theta})^{n-s_1}}{(1+i R e^{i \theta})^{n+s_2}} \\ + i \int_R^{1+\epsilon} dy \, \frac{(1+y)^{n-s_1}}{(1-y)_+^{n+s_2}} + i \epsilon \int_{\pi/2}^{-3 \pi/2} d\phi \, e^{i \phi} \frac{(2-i \epsilon e^{i \phi})^{n-s_1}}{(i \epsilon e^{i \phi})^{n+s_2}} \\+ i \int_{1+\epsilon}^R dy \, \frac{(1+y)^{n-s_1}}{(1-y)_-^{n+s_2}} + i R \int_{\pi/2}^{\pi} d\theta \, e^{i \theta} \frac{(1-i R e^{i \theta})^{n-s_1}}{(1+i R e^{i \theta})^{n+s_2}} = 0 $$
In the third and fifth intgerals, I use the subscripts $\pm$ to denote the effect of the different sides of the branch cut on the quantity that subscript modifies.  That is, $(1-y)_+^{n+s_2} = e^{i \pi (n+s_2)} (y-1)^{n+s_2}$ and $(1-y)_-^{n+s_2} = e^{-i \pi (n+s_2)} (y-1)^{n+s_2}$.
We take the limits as $R \to \infty$ and $\epsilon \to 0$.  Thus, we assume that $s_1$ and $s_2$ are defined such that the second and sixth integrals vanish in this limit, i.e., $\operatorname{Re}{(s_1+s_2)} \gt 1$.  Also, the fourth integral vanishes when $\operatorname{Re}{(n+s_2)} \lt 1$, which we will assume to be the case.  Thus, we have,
$$\begin{align}\int_{-\infty}^{\infty} dx \frac{(1-i x)^{n-s_1}}{(1+i x)^{n+s_2}} &= -i \int_{\infty}^1 dy \frac{(y+1)^{n-s_1}}{(y-1)^{n+s_2} e^{i \pi (n+s_2)}} -i \int_1^{\infty} dy \frac{(y+1)^{n-s_1}}{(y-1)^{n+s_2} e^{-i \pi (n+s_2)}}\\ &= i (-2 i \sin{[\pi (n+s_2)]})  \int_1^{\infty} dy \frac{(y+1)^{n-s_1}}{(y-1)^{n+s_2}}\\ &=  2 \sin{[\pi (n+s_2)]} \int_0^{\infty} dy \, y^{-(n+s_2)} (y+2)^{n-s_1}\\ &= 2^{2-s_1-s_2} \sin{[\pi (n+s_2)]} \int_0^{\infty} dy \, y^{-(n+s_2)} (y+1)^{n-s_1}\\ &= 2^{2-s_1-s_2} \sin{[\pi (n+s_2)]} \int_1^{\infty} dy \, (y-1)^{-(n+s_2)} y^{n-s_1}\\&= 2^{2-s_1-s_2} \sin{[\pi (n+s_2)]} \int_0^1 du \, u^{s_1+s_2-2} (1-u)^{-(n+s_2)} \end{align}$$
We recognize that last integral as a beta integral.  Thus, we have
$$\int_{-\infty}^{\infty} dx \frac{(1-i x)^{n-s_1}}{(1+i x)^{n+s_2}} = 2^{2-s_1-s_2} \sin{[\pi (n+s_2)]} \frac{\Gamma(s_1+s_2-1) \Gamma(1-n-s_2)}{\Gamma(s_1-n)} $$
Finally, using the reflection formula, i.e., $\pi/\sin{(\pi z)} = \Gamma(z) \Gamma(1-z)$, we get the result we seek in final form:

$$\int_{-\infty}^{\infty} dx \frac{(1-i x)^{n-s_1}}{(1+i x)^{n+s_2}} = \frac{4 \pi \, \Gamma(s_1+s_2-1)}{2^{s_1+s_2} \Gamma(s_1-n) \Gamma(n+s_2)} $$

This is, as mentioned above, subject to the constraints $\operatorname{Re}{(s_1+s_2)} \gt 1$ and $\operatorname{Re}{(n+s_2)} \lt 1$.
ADDENUDUM
Here is a problem that is both a special case and a little bit of a generalization at the same time.  Note that there are two branch points rather than the single one here.
ADDENUDUM II
I made an error in the integral about the small circle about the branch point; there, $z=i+\epsilon \, e^{i \phi}$.  That error has been corrected and there is now an additional constraint that must be satisfied by the parameters $n$ and $s_2$ for the posted result to remain valid.
A: The substitution $\;x=\tan\theta\;$ makes the integral \begin{align}\int_{-\pi/2}^{\pi/2}\Big(\frac{\cos\theta-i\sin\theta}{\cos\theta}\Big)^{n-s_1}\Big(\frac{\cos\theta}{\cos\theta+i\sin\theta}\Big)^{n+s_2}\sec^2\theta\;d\theta&=\int_{-\pi/2}^{\pi/2}e^{-i(n-s_1)\theta}e^{-i(n+s_2)\theta}\cos^{s_1+s_2-2}\theta\;d\theta\\
&=\int_{-\pi/2}^{\pi/2}e^{-i(2n+s_2-s_1)\theta}\cos^{s_1+s_2-2}\theta\;d\theta\tag{$\star$}
\end{align}
This link shows how to solve $(\star)$.
