What is the Fourier transform of $1/|x|$? I looked it up in several tables and calculated it in Mathematica and Matlab. Some tables say that the answer is simply $$\frac{1}{|\omega|}$$ and in other table it is $$-\sqrt{\frac{2}{\pi}}\ln|\omega|$$
and in Mathematica and Matlab (mupad) it is
$$-\sqrt{\frac{2}{\pi}}\ln|\omega|-2\gamma$$
where $\gamma$ is the Euler–Mascheroni constant. Why are there so many answers? Are they all equivalent in some way or two (or all) of them are wrong?
 A: This can be computed rigorously in terms of distributions, see here. The transform of a distribution is defined as
$$(\mathcal F[f], \phi) = (f, \mathcal F[\phi]).$$
If $f$ is a regular distribution, induced by an ordinary function $f_o$, and the order of integration in the resulting double integral can be changed, we get
$$(\mathcal F[f], \phi) =
\int dx f_o(x) \mathcal F[\phi](x) = \\
\int dx f_o(x) \int d\xi \phi(\xi) e^{i x \xi} = \\
\int d\xi \left( \int dx f_o(x) e^{i \xi x} \right) \phi(\xi) = \\
\int d\xi F_o(\xi) \phi(\xi),$$
which means that $\mathcal F[f]$ is a regular distribution induced by $F_o$. Further, $F_o = (f, e^{i \xi x})$ (this can be thought of as just a shorthand notation for the integral, since $e^{i \xi x}$ is not in the space of test functions), which explains why it's valid to compute $\mathcal F[f]$ by directly applying $f$ to $e^{i \xi x}$:
$$\mathcal F\!\left[ |x|^{-1} \right] =
2\int_0^{1} dx \frac {\cos \xi x - 1} x +
 2\int_1^\infty dx \frac {\cos \xi x} x =
-2 \ln |\xi| - 2 \gamma.$$
A: Although it is too late, let me give an explanation in hope for future reference.
Let me first compute the distributional Fourier transform (FT) of the function $f(x) = \log|x|$. Since $f$ is locally integrable and has slow growth, $f$ defines a tempered distribution on $\mathbb{R}$. So its FT is also well-defined.
Before we proceed, let us fix the convention for FT. Here, we will use the version
$$ \mathcal{F}[f](\xi) = \hat{f}(\xi) = \int_{\mathbb{R}} e^{-2\pi i \xi x} f(x) \, dx. $$
Now with a bit of computation, you can check that for any Schwartz function $\varphi \in \mathcal{S}(\mathbb{R})$,
\begin{align*}
\langle \hat{f}, \varphi \rangle
 = \langle \hat{\varphi}, f \rangle
&= \int_{\mathbb{R}} \hat{\varphi}(x) \log|x| \, dx \\
&= -(\gamma + \log(2\pi))\varphi(0) - \int_{\mathbb{R}} \frac{\varphi(x) - \varphi(0)\mathbf{1}_{[-1,1]}(x)}{|x|} \, dx. \tag{1}
\end{align*}
This means that $\hat{f}$ does not reduce to a linear combination of locally integrable functions and Dirac delta. But if we restrict the domain of $\hat{f}$ onto the space $X=\{\varphi \in \mathcal{S}(\mathbb{R}) : \varphi(0)= 0\}$, then the FT has the following nice representation:
$$ \hat{f}(x) = -(\gamma + \log(2\pi))\delta_0(x) - \frac{1}{|x|} \quad \text{on } X. \tag{2}$$
Precisely in this sense, we may loosely say that the FT of $\frac{1}{|x|}$ is $- \log|\xi| - \gamma - \log(2\pi)$ in light of the inverse FT.
But wait! Since $\text{(21)}$ is being considered on the space $X$ where $\varphi(0) = 0$, taking Dirac delta gives no difference. So we can equally say that
$$ \hat{f}(x) = -\frac{1}{|x|} \quad \text{on } X $$
and hence we can also say that the FT of $\frac{1}{|x|}$ is simply $-\log |\xi|$. There is no contradiction on this, since we are considering this distribution on the subspace
$$ \mathcal{F}[X] = \left\{\varphi \in \mathcal{S}(\mathbb{R}) : \int_{\mathbb{R}} \varphi(x) \, dx = 0 \right\}$$
and thus adding constants makes no difference.

I am not good at physics, but I guess we can give a physical interpretation of it. Notice that $\text{(1)}$ has a term that cancels out the singularity of $\frac{1}{|x|}$. So in a very loose sense, it has the form
$$ (\log |x|)^{\wedge}(\xi) = - \frac{1}{|\xi|} + \infty \cdot \delta_0(\xi)
\qquad \Leftrightarrow \quad
\left( \frac{1}{|x|}\right)^{\wedge}(x) = \infty - \log |\xi|. $$
Thus in order that we can ignore the cancellation part, we should work on the situation where the addition of constant does not affect the meaning of the FT of $\frac{1}{|x|}$. Probably this is the case for potentials, where only the difference in values has actual physical meaning.

Addendum. (Derivation of $\text{(1)}$) The key trick is the following Frullani-type integral
$$ \forall a > 0 \ : \quad \int_{0}^{\infty} \frac{e^{-2\pi s} - e^{-2\pi as}}{s} \, ds = \log a. $$
From this, we find that
\begin{align*}
\int_{\mathbb{R}} \hat{\varphi}(x) \log|x| \, dx
&= \int_{\mathbb{R}} \hat{\varphi}(x) \left( \int_{0}^{\infty} \frac{e^{-2\pi s} - e^{-2\pi |x|s}}{s} \, ds \right) \, dx \\
&= \int_{0}^{\infty} \frac{1}{s} \left( \int_{\mathbb{R}} \hat{\varphi}(x) (e^{-2\pi s} - e^{-2\pi |x|s}) \, dx \right) \, ds \tag{i} \\
&= \int_{0}^{\infty} \frac{1}{s} \left( e^{-2\pi s} \varphi(0) -  \int_{\mathbb{R}} \hat{\varphi}(x) e^{-2\pi |x|s} \, dx \right) \, ds \tag{ii} \\
&= \int_{0}^{\infty} \left( \frac{e^{-2\pi s}}{s}  \varphi(0) -  \int_{\mathbb{R}} \frac{\varphi(x)}{\pi(s^2 + x^2)}  \, dx \right) \, ds \tag{iii}
\end{align*}
Here, 
(i) We interchanged the order of integrations by Fubini's theorem. It is not hard to check that the integrability condition is indeed met.
(ii) We utilized Fourier inversion $\varphi(\xi) = \int_{\mathbb{R}} \hat{\varphi}(x) e^{2\pi i \xi x} \, dx$.
(iii) We utilized the fact that $\langle \hat{\varphi}, \psi \rangle = \langle \varphi, \hat{\psi} \rangle$ for any integrable functions $\varphi, \psi$, together with the fact that the FT of $e^{-2\pi s |\cdot|}$ is $\frac{1}{\pi(s^2 + (\cdot)^2)}$.
Now we add and subtract the term 
$$ \int_{\mathbb{R}} \frac{\varphi(0) \mathbf{1}_{[-1,1]}(x)}{\pi(s^2 + x^2)} \, dx = \frac{2\arctan(1/s)}{\pi s} \, \varphi(0)$$
inside the outer integral. Then
\begin{align*}
\int_{\mathbb{R}} \hat{\varphi}(x) \log|x| \, dx
&= \varphi(0) \int_{0}^{\infty} \left( \frac{e^{-2\pi s}}{s} - \frac{2\arctan(1/s)}{\pi s} \right) \, ds \\
&\qquad + \int_{0}^{\infty} \int_{\mathbb{R}} \left( \frac{\varphi(x) - \varphi(0) \mathbf{1}_{[-1,1]}(x)}{\pi(s^2 + x^2)} \right) \, dx \, ds
\end{align*}
Now the first term is just a constant multiple of $\varphi(0)$, and this constant turns out to have the value
$$ \int_{0}^{\infty} \left( \frac{e^{-2\pi s}}{s} - \frac{2\arctan(1/s)}{\pi s} \right) \, ds = -\gamma - \log(2\pi). $$
For the second term, interchanging the order of integrations using Fubini's theorem gives
$$ \int_{\mathbb{R}} \frac{\varphi(x) - \varphi(0) \mathbf{1}_{[-1,1]}(x)}{|x|} \, dx $$
Therefore the claim follows.
A: To be fully rigorous, we should concede that $1/|x|$ is not directly a distn on $\mathbb R$. But it does arise as an even tempered distn $u$ such that $x\cdot u=sign(x)$.
Up to a constant, Fourier transform gives ${d\over dx}\hat{u}=PV{1\over x}$. This equation has at least solution $\hat{u}=\log|x|$. The associated homog eqn has only multiples of $1$ as solns, so the eqn’s solns are $\hat{u}=\log|x|+c$ for arbitrary constants $c$ (and with the normalization constant).
The constant $c$ be determined by evaluation against $e^{-\pi x^2}$.
