Fourier transform of a a function in the space Which is the Fourier transform (in the sense of distributions) of the function $f(x)=x/\|x\|^n $, where $x$ belongs to the Euclidean space $ R^n$?
 A: Contour integration yields
$$
\int_0^\infty\cos(\lambda r)\frac{\mathrm{d}r}{r^{1-\alpha}}
=\Gamma(\alpha)|\lambda|^{-\alpha}\cos(\pi\alpha/2)\tag{1}
$$
Euler's Reflection Formula, proven in this answer, implies
$$
\cos(\pi\alpha/2)=\frac\pi{\Gamma\left(\frac{1-\alpha}{2}\right)\Gamma\left(\frac{1+\alpha}{2}\right)}\tag{2}
$$
the Duplication Formula shows
$$
\Gamma\left(\frac\alpha2\right)\Gamma\left(\frac{\alpha+1}2\right)=\sqrt\pi\,2^{1-\alpha}\Gamma(\alpha)\tag{3}
$$
and this answer gives
$$
\omega_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}\tag{4}
$$
Putting these together, we get
$$
\begin{align}
&\int_{\mathbb{R}^n}\frac1{|x|^{n-\alpha}}e^{-2\pi ix\cdot\xi}\,\mathrm{d}x\\
&=\int_0^\infty\int_{-1}^1\frac1{r^{1-\alpha}}\cos(2\pi|\xi|rt)\omega_{n-2}\sqrt{1-t^2}^{\raise{2pt}{\,n-3}}\mathrm{d}t\,\mathrm{d}r\\
&=2\Gamma(\alpha)\cos(\pi\alpha/2)\omega_{n-2}(2\pi|\xi|)^{-\alpha}\int_0^1t^{-\alpha}\sqrt{1-t^2}^{\raise{2pt}{\,n-3}}\mathrm{d}t\\
&=\Gamma(\alpha)\cos(\pi\alpha/2)\omega_{n-2}(2\pi|\xi|)^{-\alpha}\int_0^1t^{-(\alpha+1)/2}(1-t)^{(n-3)/2}\mathrm{d}t\\
&=\Gamma(\alpha)\cos(\pi\alpha/2)\omega_{n-2}(2\pi|\xi|)^{-\alpha}\frac{\Gamma\left(\frac{1-\alpha}{2}\right)\Gamma\left(\frac{n-1}{2}\right)}{\Gamma\left(\frac{n-\alpha}{2}\right)}\\
&=\Gamma(\alpha)\cos(\pi\alpha/2)2\pi^{(n-1)/2}(2\pi)^{-\alpha}\frac{\Gamma\left(\frac{1-\alpha}{2}\right)}{\Gamma\left(\frac{n-\alpha}{2}\right)}|\xi|^{-\alpha}\\
&=\Gamma(\alpha)\frac{2\pi^{(n+1)/2}(2\pi)^{-\alpha}}{\Gamma\left(\frac{1+\alpha}{2}\right)\Gamma\left(\frac{n-\alpha}{2}\right)}|\xi|^{-\alpha}\\
&=\frac{\Gamma\left(\frac\alpha2\right)\pi^{(n-\alpha)/2}}{\Gamma\left(\frac{n-\alpha}{2}\right)\pi^{\alpha/2}}|\xi|^{-\alpha}\tag{5}
\end{align}
$$
Taking the gradient of $(5)$ and letting $\alpha\to0$ yields
$$
\int_{\mathbb{R}^n}\frac{x}{|x|^n}e^{-2\pi ix\cdot\xi}\,\mathrm{d}x
=\frac{-i\pi^{(n-2)/2}}{\Gamma\left(\frac{n}{2}\right)}\frac{\xi}{|\xi|^2}\tag{6}
$$
