Let the $x\in \mathbb{R}^d, t\in \mathbb{R}$, i.e. $(x,t)\in {\mathbb{R^{d+1}}}$.
I already know the Fourier transform of $|x|^{-\alpha}$ is $|\xi|^{-d+\alpha}$.
How do I get the Fourier transform of $|(x,t)|^{-\alpha} = \left(\sqrt{|x|^2 +t^2}\right)^{-\alpha}$ with respect to $x$ ?
You are interested in obtaining an explicit expression for $$I(k,t) =\frac{1}{(2\pi)^{n/2}} \int\left(\sqrt{|x|^2 +t^2}\right)^{-\alpha} e^{i x\cdot k} d^n x\,.$$ Let us introduce spherical coordinates with the first coordinate of $x$ with component $r \cos \phi$ pointing along $k$. We thus reduce the integral to the form $$I(k,t) =\frac{S_{n-2}}{(2\pi)^{n/2}} \int_0^\infty \int_{0}^\pi r^{n-1}\left(\sqrt{r^2 +t^2}\right)^{-\alpha} e^{i r |k| \cos(\phi)} \sin^{n-2}(\phi)\,d\phi dr $$ where $S_n$ is the surface area of the $n$-sphere given by $$ S_{n-1} =\frac{n\pi^\frac{n}{2}}{\Gamma\left(\frac{n}{2}+1 \right)}.$$
The integral over $\phi$ can be easily executed with the result $$I(k,t) =\frac{S_{n-2}}{(2\pi)^{n/2}} \int_0^\infty \frac{\pi r^{n-1} J_{n/2-1}(r |k|)}{(r^2+t^2)^{\alpha/2}} dr\,. $$ I am not sure about the remaining integral. Maybe somebody has an idea...