Fourier transform of $\frac{(x-iw)^\alpha}{(x+iw)^\alpha}$ Let $w>0$ and $\alpha>0$. I want to compute the Fourier transform of $\frac{(x-iw)^\alpha}{(x+iw)^\alpha}$ in the distribution sense, i.e., evaluate
$$\int_{-\infty}^\infty \frac{(x-iw)^\alpha}{(x+iw)^\alpha} e^{-ikx} dx.$$
By rescaling, one may set $w=1$. Mathematica cannot solve this integral. If $\alpha \in \mathbb Z$, then the standard contour integration gives the answer.
How can I compute the above integral?
 A: The result uses confluent hypergeometric functions. A related integral representation here is $$\int_\lambda t^{a-1}(1-t)^{b-a-1}e^{zt}\,dt=\frac{2\pi i}{\Gamma(1-a)}U(a,b,z)\qquad\big[a,b,z\in\mathbb{C},\ \Re z>0\big]$$ where all the powers take the principal values, and the contour $\lambda$ encircles the negative real axis $\mathbb{R}_{\leqslant 0}$ (which is the branch cut of $t^{a-1}$), not crossing the branch cut $\mathbb{R}_{\geqslant 1}$ of $(1-t)^{b-a-1}$.
Then for $I(a,b,k):=\int_{-\infty}^\infty(i+x)^{a-1}(i-x)^{b-a-1}e^{-ikx}\,dx$, where $k\in\mathbb{R}$ and $\Re b<1$, we get $$I(a,b,k)=\begin{cases}J(a,b,k),&k\geqslant 0\\J(b-a,b,-k),&k\leqslant 0\end{cases},\quad J(a,b,k)=-\frac{\pi(2i)^b e^{-k}}{\Gamma(1-a)}U(a,b,2k)$$ (the values at $k=0$ coincide, as expected); reduction to the contour integral above is done by considering the integrand, for $k>0$, at $x=i(2t-1)$ with the branch cuts as specified above, and shrinking the contour to encircle $\mathbb{R}_{\leqslant 0}$; for $k<0$, use $x=i(1-2t)$ instead.
The present case at $w=1$ is obtained (up to a constant factor) by taking $(1+d/dk)^2$, in the distributional sense, of $I(-1-\alpha,0,k)$. I'm going to fill in the details later if needed. It's also worth noting that the case $U(a,0,z)$ arising here is studied by H. Bateman.
