Let $\epsilon>0$ be a positive infinitesimal constant. What is the Fourier transform of $f(x) = [-i(x+i\epsilon)]^{-\alpha}$ for $\alpha>0$ (we always consider the principal branch); in other words, what is $$\lim_{\epsilon\to 0^+} \int_{-\infty}^\infty f(x) e^{-ikx} dx?$$ The integral is understood as the improper Riemann integral. Mathematica says that the answer is $$\frac{2\pi}{\Gamma(\alpha)} k^{\alpha-1} \theta(k),$$ where $\Gamma$ is the Gamma function and $\theta$ is the Heaviside function. How can I obtain this analytically?
Note: (1) The integral is easily seen to be zero for $k<0$, by using the residue theorem; closing the contour on the upper half plane, the function $f$ does not have any poles or branch cuts. (2) If $\alpha$ is an integer, then the result for $k>0$ is again obtained from the residue theorem. For fractional $\alpha$, due to the branch cut of $f$ on the lower half plane, the residue theorem is not straightforward.
The function $f(z)=\frac1{(-i(z+i\varepsilon))^\alpha}$, with $\alpha>0$ and $\varepsilon>0$, is multivalued with a branch point at $z=-i\alpha$.
We choose to cut the plane from $z=-i\alpha$ along a ray on the imaginary axis. Then, for $k>0$, we close the contour in the upper-half plane and find that
$$\int_{-\infty}^\infty \frac{e^{-ikx}}{(-i(x+i\varepsilon))^\alpha}\,dx=0\tag1$$
For $k<0$, we close the contour in the lower-half plane, deforming around the branch cut and find that for $\nu>0$
$$\int_{-\infty}^\infty \frac{e^{-ikx}}{(-i(x+i\varepsilon))^\alpha}\,dx=2\sin(\alpha \pi)e^{-ik\varepsilon}\int_{\nu}^\infty \frac{e^{-kt}}{t^\alpha}\,dy-e^{-k\varepsilon}\nu^{1-\alpha}\int_{-\pi}^\pi e^{\nu ke^{i\phi}}e^{i(1-\alpha)\phi}\,d\phi\tag2$$
Next, we let $\nu\to 0$. To do this, we repeatedly integrate by parts ($\lfloor \alpha \rfloor$ times) the first integral on the right-hand side of $(2)$. This is left as an exercise for the reader.