Though it may sound like a physical problem, but the thing I will introduce is rather mathematical. For the Fourier transform of Coulomb potential $$ V(\vec{x})=\frac{1}{\vert x\vert} $$ I can calculate its fourier tranform through $$V_d(q)=\int d^{d}\vec{x}\frac{e^{-i\vec{q}\vec{x}}}{\vert x\vert}$$ if $d=2$ or $d=3$, we get $$ V_3(q)=\frac{4\pi}{q^2}, V_2(q)=\frac{2\pi}{q} $$ However, how to calculate it in one-dimension? The answer seems to be $-2(\gamma+\ln q)$ where $\gamma$ is Euler's const.
ref: see eq(2.9) in the arXiv paper.
I agree with the comment of @copper.hat. I think that the way I'm describing below, using tempered distributions, and what is called the "finite part", is what is used in that arxiv preprint.
For functions $\phi\in\mathcal{S}(R)$ (the Schwartz space), we define $\text{f.p.}\,\frac{1}{|x|}$ as $$ \langle\text{f.p.}\,\frac{1}{|x|},\phi\rangle=\lim_{\delta\to 0^+}\biggl(\int_{|x|>\delta}\frac{1}{|x|}\phi(x)\,dx-c_\delta\phi(0)\biggr). $$ Exercise to you: Show that this makes sense precicely when $c_\delta=2\log(1/\delta)$, and that $\text{f.p.}\,\frac{1}{|x|}$ becomes a tempered distribution.
For tempered distribution $u$, its Fourier transform is defined via $$ \langle \hat{u},\phi\rangle = \langle u,\hat{\phi}\rangle, $$ so we need to find out how $u$ acts on Fouriertransforms of Schwartz functions. We calculate bravely(?), and find that $$ \begin{aligned} \langle \text{f.p.}\,\frac{1}{|x|},\hat{\phi}\rangle &=\lim_{\delta\to 0^+}\biggl(\int_{|x|>\delta}\frac{1}{|x|}\hat{\phi}(x)\,dx-2\log(1/\delta)\hat{\phi}(0)\biggr)\\ &=\lim_{\delta\to 0^+}\biggl(\int_{|x|>\delta}\frac{1}{|x|}\int_{-\infty}^{+\infty}\phi(\xi)e^{-ix\xi}\,d\xi\,dx-2\log(1/\delta)\int_{-\infty}^{+\infty}\phi(\xi)\,d\xi\biggr)\\ &=\lim_{\delta\to0^+}\int_{-\infty}^{\infty}\phi(\xi)\biggl[\int_{|x|>\delta}\frac{1}{|x|}e^{-ix\xi}\,dx-2\log(1/\delta)\biggr]\,d\xi\\ &=\lim_{\delta\to0^+}\int_{-\infty}^{\infty}\phi(\xi)\biggl[2\int_\delta^{+\infty}\frac{\cos(x\xi)}{x}\,dx-2\log(1/\delta)\biggr]\,d\xi \end{aligned} $$ The limit of the expression inside square brackets is our Fourier transform. It is even in $\xi$, so we just need to consider $\xi>0$. We change variables in the integral, and put $t=x\xi$. Then the integral becomes $$ \int_\delta^{+\infty}\frac{\cos(x\xi)}{x}\,dx=\int_{\delta\xi}^{+\infty}\frac{\cos t}{t}\,dt $$ Using the general formula (see wiki) $$ \int_{a}^{+\infty}\frac{\cos t}{t}\,dt=-\gamma-\log a-\int_0^{a}\frac{\cos t-1}{t}\,dt, $$ we get (with $a=\delta\xi$) $$ \int_\delta^{+\infty}\frac{\cos(x\xi)}{x}\,dx=-\gamma-\log(\delta\xi)-\int_{0}^{\delta\xi}\frac{\cos t-1}{t}\,dt $$ So (the last integral tends to zero), $$ \lim_{\delta\to0^+}\biggl[2\int_\delta^{+\infty}\frac{\cos(x\xi)}{x}\,dx-2\log(1/\delta)\biggr]=-2(\gamma+\log|\xi|). $$
Comment In Grafakos book Classical Fourier Analysis, analytic continuation is used to make sense of distributions $c_z|x|^z$ for all $z\in\mathbb C$, and a certain constant $c_z$. Also, their Fourier transforms are calculated.