Fourier transform of a potential I need help computing the distributional inverse Fourier transform of the function $1/|x|^2$ in dimension two. The integral makes sense written as
\begin{align} 1/2\pi \int_{\mathbb{R}^2} e^{ix\xi} |\xi|^{-2} d\xi \end{align}
is there a way to compute this explicitly as a principal value? 
 A: This is not the answer you want, but something slightly unexpected happens: the function $|x|^{-2}$ is not locally integrable in $\mathbb R^2$ (the problem is around $x=0$), so it's not obvious which distribution you mean. One can extend $|x|^{-2}$ to a distribution, e.g. if $c>0$ we can define $f_c$ via 
$$\langle f_c,\phi\rangle=\int_{|x|>c}\phi(x)|x|^{-2}\,dx+\int_{|x|\leq c}(\phi(x)-\phi(0))|x|^{-2}\,dx$$
and we have $f_c-f_{c'}=\pi(c^2-c'^2)\delta$. There is no "best" choice of $c$ (and the limit $c\to0$ of $f_c$ doesn't exist). The moral is that as a distribution, $|x|^{-2}$ can be defined only up to a multiple of $\delta$. No principal value is going to save us.
As for its inverse Fourier transformation, it can be computed indirectly as follows: since $|x|^2f_c=1$ (for any $c$), we get $\Delta\mathcal F^{-1}(f_c)=-2\pi\delta$ ($\mathcal F^{-1}$ means inverse Fourier transform, using your normalization), so $\mathcal F^{-1}(f_c)=\log|x|+g_c$ where $g_c$ is harmonic. All the involved functions/distributions are rotation-invariant, so $g$ is a rotation-invariant harmonic function, i.e. a constant. From $f_c-f_{c'}=\pi(c^2-c'^2)\delta$ we can see that $g_c-g_{c'}=(c^2-c'^2)/2$. The value of (say) $g_1$ seems beyond my skills.
