Compute $\int_{\mathbb R^n}\frac{e^{-2i\pi x\cdot \xi}}{|\xi|^{2s}}d\xi$. I'm trying to compute the following integral for a long time, but can't conclude : $$\int_{\mathbb R^n}\frac{e^{-2i\pi x\cdot \xi}}{|\xi|^{2s}}d\xi,$$
or at least $$\int_{\mathbb R}\frac{e^{-2i\pi x\xi}}{|\xi|^{2s}}d\xi,$$
with $s$ small enough for the integral converge. I know it's the Fourier transform of $\xi\longmapsto \frac{1}{|\xi|^{2s}},$ but I don't see anyway if we know this integral or not. I tried to use Wolframalfo, but unfortunately it's not conclusive. Any idea ?

For $\xi > 0$ and $Re(s) \in (0,1)$ $$\mathcal{F}[|x|^{-s}](\xi)=\int_{-\infty}^\infty e^{-2i \pi \xi x} |x|^{-s}dx = \int_{-\infty}^\infty e^{-2i \pi y} |y/\xi|^{-s}dy/\xi = \xi^{s-1}\mathcal{F}[|x|^{-s}](1)$$
For $a > 0$ it holds that $\int_0^\infty e^{-a x} |x|^{-s}dx = a^{s-1}\int_0^\infty e^{-t} t^{-s}dt=e^{(s-1) \log a} \Gamma(1-s)$. By analytic continuation it stays true for $Re(a)  > 0$ and by continuity for $a = 2i \pi$, thus
 $$\mathcal{F}[|x|^{-s}](1) = (2 \pi )^{s-1} \Gamma(1-s)(e^{(s-1) i \pi/2}-e^{-(s-1) i \pi/2}) = (2 \pi )^{s-1} \Gamma(1-s) 2 i \sin((s-1) \pi /2)$$ 
In $\mathbb{R}^n$ it works quite the way, except the  reflection formula for $\Gamma(s)$ coming into play implicitely in $\mathcal{F}[|x|^{-s}](1)$ 
