Distributional Fourier Transform of $f(x)=\frac{e^{ix\cdot \xi_0}}{\vert x-x_0\vert}$ where $x\in\mathbb R^2$ and $\xi_0,x_0\in\mathbb R^2$ fixed Problem: Find the Fourier transform of $f:\mathbb R^2\to\mathbb C$ given by
$$f(x)=\frac{e^{ix\cdot \xi_0}}{\vert x-x_0\vert}\quad\text{where }\xi_0,x_0\in\mathbb R^2\text{ are fixed.}$$
My Attempt: Note that $f\notin L^1(\mathbb R^2)$ and also $f\notin L^2(\mathbb R^2)$. Therefore, we compute the Fourier transform of $f$ in the distributional sense since $f\in L^1_{\operatorname{loc}}(\mathbb R^2)$. Let $\psi\in\mathcal C_c^\infty(\mathbb R^2)$. Then
\begin{align*}
\widehat{f}(\psi)&=f(\widehat{\psi})=\int_{\mathbb R^2}f(x)\widehat\psi(x)\,dx\\
&=\frac{1}{2\pi}\int_{\mathbb R^2}\frac{e^{ix\cdot \xi_0}}{\vert x-x_0\vert}\int_{\mathbb R^2}\psi(\xi)e^{-ix\cdot\xi}\,d\xi\,\,dx\\
&=\frac{1}{2\pi}\int_{\mathbb R^2}\psi(\xi)\int_{\mathbb R^2}\frac{e^{ix\cdot(\xi_0-\xi)}}{\vert x-x_0\vert}\,dx\,d\xi.
\end{align*}
Therefore, it suffices to evaluate the integral
$$I=\frac{1}{2\pi}\int_{\mathbb R^2}\frac{e^{ix\cdot(\xi_0-\xi)}}{\vert x-x_0\vert}\,dx.$$
The one thing that is confusing me about the problem is the multidimensionality, if on the other hand we were in $\mathbb R$, then some manipulations with the Heaviside function and the delta distribution would most likely do the trick. Also, the integral above is not absolutely convergent and the integrand is not nonnegative, whence we cannot even apply polar coordinates, as stated in Folland's book.

Therefore, my question is: how does one deal with the integral above, or is there a different and more efficient way to go about this problem?
Thank you for your time and appreciate any feedback.
 A: Here's an evaluation that should provide some hints as to what the correct answer is.
Consider the Fourier integral
$$F_{\epsilon}(\xi)=\int_{\mathbb{R}^2}d^2x~\frac{e^{-i(\xi-\xi_0)x}}{|x-x_0|}e^{-\epsilon |x-x_0|}$$
where $\epsilon$ is a positive number (can be readily generalized to a complex number with $\text{Re}( \epsilon)>0$). This regulated Fourier transform does converge absolutely, in  contrast to it's unregulated counterpart. This allows for an easy evaluation of the integral as follows. It suffices to consider the simpler integral
$$I(p)=\int_{\mathbb{R}^2}d^2x~\frac{e^{-ipx}}{|x|}e^{-\epsilon |x|}$$
since of course
$$F_{\epsilon}(\xi)=e^{-i(\xi-\xi_0)x_0}I(\xi-\xi_0)$$
One can evaluate $I$ in spherical coordinates in various ways. Here's one of them
$$I(p)=\int_{0}^{2\pi}d\theta\int_{0}^{\infty}dr~e^{-\epsilon r-i|p|r\cos\theta}=\int_{0}^{2\pi}\frac{1}{\epsilon+i|p|\cos\theta}d\theta=\frac{2\pi}{\sqrt{|p|^2+\epsilon^2}}$$
where the last integral can be easily performed with a standard contour integration trick. It would appear that one can easily take the limit $\epsilon\to 0^+$ to get that the requested FT is given by
$$\hat{f}(\xi)=\frac{e^{-i(\xi-\xi_0)x_0}}{|\xi-\xi_0|}$$
and that the limit is quite regular, despite the fact that actually if it wasn't for the cutoff parameter we wouldn't have been able to evaluate the double integral in the order we did. One may worry that this procedure is misleading, since the integral diverges if evaluated with $\epsilon=0$ directly step by step as above. However it turns out that the cutoff and distributional considerations are not needed if the right order of integration is employed since
$$I_0(p)=\int_{0}^{\infty}dr\int_{0}^{2\pi}d\theta~e^{-i|p|r\cos\theta}=2\pi\int_0^{\infty}J_0(|p|r)dr=\frac{2\pi}{|p|}$$
where the last step can be derived from the Laplace transform of a Bessel function
$$\int_{0}^{\infty}J_0(px)e^{-sx}dx=\frac{1}{\sqrt{p^2+s^2}}$$
In summary, this integral exists and is well defined but poses computational difficulties due to the lack of absolute covergence, which also precludes the free use of Fubini's theorem. If the integral exists, evaluating with a regulator is enough to calculate it's value if it makes the evaluation easier, but eventually one has to prove its existence somehow. In this example there is a way to do that without resorting to distributional results. Let us note, however, that this integral can also be shown to have the value above by using the following distributional identity(calculation is left to the reader)
$$\int_{0}^{\infty}dx~e^{-ipx\cos\theta}=\pi\delta(|p|\cos\theta)-i ~PV\left(\frac{1}{p\cos\theta}\right)$$
