About the Fourier transform of the logarithm function I want to calculate / simplify:
$$\mathcal{F} (\ln(|x|)\mathcal{F(f)}(x))=\mathcal{F} (\ln(|x|)) \star f$$
where $\mathcal{F}$ is the Fourier transform ($\mathcal[f](\xi)=\int_{\mathbb R}f(x)e^{ix\xi}\,dx$) and where $f$ is an even function.
Looking here: wiki, we find that 
$$\mathcal{F}[\log|x|](\xi)=-2\pi\gamma\delta(\xi)-\frac\pi{|\xi|},$$
so we should have:
$$\mathcal{F} (\ln(|x|)) \star f = (-2\pi\gamma\delta(x)-\frac\pi{|x|}) \star f(x) $$
$$ = -2\pi\gamma f(x)- \pi \int_{-\infty}^{\infty} \frac{f(t)}{|x-t|} dt $$
but the integral of the second term does not converge... whereas the term $\mathcal{F} (\ln(x)\mathcal{F(f)}(x))$ is well defined providing the function $f$ is of rapide decrease near zero and infinity. So where is the problem ? and what is finally the "simplified expression" of $\mathcal{F} (\ln(x)\mathcal{F(f)}(x))$ ? We cannot use this distribution in a convolution product with a function?
I already post this on Stackexchange but did not receive an answer.
 A: I thought that it might be instructive to present an approach to deriving the Fourier transform of $\log(|x|)$.  The result includes a distributional interpretation of $\frac1{|x|}$.  Finally, we show that the distributional interpretation of $\frac1{|x|}$ is non-unique and that it differs from other interpretations by a multiple of the Dirac Delta distribution.  With that introduction, we now proceed.
PRELIMARIES
Let $\psi(x)=\log(|x|)$ and let $\Psi$ denote its Fourier Transform .  Then, we write
$$\Psi(x)=\mathscr{F}\{\psi\}(x)\tag 1$$
where $(1)$ is interpreted as a Tempered Distribution.  That is, for any $\phi \in \mathbb{S}$, we can write
$$\langle \mathscr{F}\{\psi\}, \phi\rangle =\langle \psi, \mathscr{F}\{\phi\}\rangle$$
Now, let $\psi_\epsilon(k) =e^{-\varepsilon|k|}\log(|k|)$.  Therefore, $\psi(k)=\lim_{\varepsilon\to 0^+}\psi_\varepsilon(k)$ and we see that
$$\begin{align}
\lim_{\varepsilon\to 0^+}\langle \mathscr{F}\{\psi_\varepsilon\}, \phi\rangle&=\lim_{\varepsilon\to 0^+}\langle \psi_\varepsilon, \mathscr{F}\{\phi\}\rangle \\\\
&=\langle \psi,\mathscr{F}\{\phi\}\rangle\\\\
&=\langle \mathscr{F}\{\psi\}, \phi\rangle
\end{align}$$
Next, we evaluate the Fourier transform of $\psi_\varepsilon$.


EVALUATING THE FOURIER TRANSFORM OF $\displaystyle \psi_\varepsilon$
Denote by $\Psi_\epsilon$, the Fourier transform of $\psi_\varepsilon$.  Then, we have
$$\begin{align}
\Psi_\varepsilon(x)&=\mathscr{F}\{\psi_\epsilon\}(x)\\\\
&=\int_{-\infty}^\infty e^{-\varepsilon|k|}\log(|k|) e^{ikx}\,dk\\\\
&=2\text{Re}\left(\int_0^\infty e^{-(\varepsilon -ix)k}\log(k) \,dk\right)\\\\
&=-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma -\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2)-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)\\\\
&=\psi^{(1)}_\varepsilon(x)+\psi^{(2)}_\varepsilon(x)+\psi^{(3)}_\varepsilon(x)\tag2
\end{align}$$
where
$$\begin{align}
\psi^{(1)}_\varepsilon(x)&=-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma\\\\
\psi^{(2)}_\varepsilon(x)&=-\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2)\\\\
\psi^{(3)}_\varepsilon(x)&=-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)
\end{align}$$
Next, we will find the distributional limits of $\psi^{(1)}_\varepsilon$, $\psi^{(2)}_\varepsilon$, and $\psi^{(3)}_\varepsilon$.

DISTRIBUTIONAL LIMITS OF $\displaystyle \psi^{(1)}_\varepsilon$, $\displaystyle 
 \psi^{(2)}_\varepsilon$, and $\displaystyle  \psi^{(3)}_\varepsilon$
Again, let $\phi\in \mathbb{S}$.  Then,
$$\begin{align}
\lim_{\varepsilon\to 0^+}\langle \psi^{(1)}_\varepsilon,\phi \rangle &=\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \psi^{(1)}_\varepsilon(x)\phi(x)\,dx\\\\
&=\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \left(-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma \right)\phi(x)\,dx\\\\
&=-2\gamma\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\phi(\varepsilon x)}{x^2+1}\,dx\\\\
&=-2\pi \gamma \phi(0)\tag3
\end{align}$$

$$\begin{align}
\langle \psi^{(2)}_\varepsilon,\phi \rangle &=\int_{-\infty}^\infty \left(-\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2) \right)\phi(x)\,dx\\\\
&=-2\log(\varepsilon)\int_{-\infty}^\infty \frac{\phi(\varepsilon x)}{x^2+1}\,dx-\int_{-\infty}^\infty \frac{\log(1+x^2)}{1+x^2}\phi(\varepsilon x)\,dx\\\\
&= -2\pi \log(\varepsilon)\phi(0)-2\pi \log(2) \phi(0)+o(\varepsilon)
\end{align}\tag4$$

$$\begin{align}
\langle \psi^{(3)}_\varepsilon,\phi \rangle &=\int_{-\infty}^\infty \left(-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)\right)\phi(x)\,dx\\\\
&-\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx-\int_{|x|\ge 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx\\\\
&=-\phi(0)\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \,dx\\\\
&-\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) (\phi(x)-\phi(0))\,dx-\int_{|x|\ge 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx\\\\
&= \left(2\pi \log(\varepsilon) +2\pi \log(2)\right)\phi(0)+o(\varepsilon)\\\\
&-\pi \int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx-\pi \int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx\tag5
\end{align}$$


FINAL RESULTS
Substituting $(3)$, $(4)$, and $(5)$ into $(2)$, we find that
$$\begin{align}
\lim_{\varepsilon\to 0^+}\langle \mathscr{F}\{\psi_\varepsilon\},\phi\rangle =-2\pi \gamma \phi(0)-\pi \int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx-\pi\int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx\\\\
\end{align}$$
from which we assert that in distribution
$$\mathscr{F}\{\psi\}(x)=-2\pi \gamma \delta(x)-\pi \text{PV}\left(\frac1{|x|}\right)$$
where we interpret $\text{PV}\left(\frac1{|x|}\right)$ to mean that for any $\phi\in \mathbb{S}$,
$$\int_{-\infty}^\infty \phi(x) \text{PV}\left(\frac1{|x|}\right)\,dx=\int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx$$


NOTE:
It was arbitrary to split the integration in $(5)$ into inervals $|x|\le 1$ and $|x|\ge 1$.  Had we chosen instead the intervals $|x|\le \nu$ and $|x|\ge \nu$ for any $\nu>0$, we would have obtained
$$\mathscr{F}\{\psi\}(x)=-2\pi (\gamma+\log(\nu)) \delta(x)-\pi \text{PV}\left(\frac1{|x|}\right)$$
where we interpret $\text{PV}\left(\frac1{|x|}\right)$ to mean that for any $\phi\in \mathbb{S}$,
$$\int_{-\infty}^\infty \phi(x) \text{PV}\left(\frac1{|x|}\right)\,dx=\int_{|x|\le \nu}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge \nu}\frac{\phi(x)}{|x|}\,dx$$
A: As mentioned by Mateusz Kwaśnicki, the $1/|x|$ in the Fourier transform of the logarithm should be regularised in a "principal value" type of way, as explained for example in this MSE posting. To check that everything works out, it helps to walk through a specific example:
definition of Fourier transform and convolution theorem:
$${\cal F}_f(\xi)=\int_{-\infty}^\infty f(x)e^{i x\xi}dx,\;\;{\cal F}_{f\cdot g}(\xi)=\frac{1}{2\pi}\int_{-\infty}^\infty {\cal F}_f(\xi-t){\cal F}_g(t)dt$$
choose two functions $f$ and $g$, with their respective Fourier transforms,
$$f(x)=\delta(x-a),\;\;{\cal F}_f(\xi)=e^{ia\xi}$$
$$g(x)=\ln|x|,\;\;{\cal F}_g(\xi)=-2\pi\gamma\delta(\xi)-{\cal P}\frac{\pi}{|\xi|}$$
the ${\cal P}$ is there to remind us of the need to regularise $1/|\xi|$:
$$\int_{-\infty}^\infty h(\xi){\cal P}\frac{1}{|\xi|}\,d\xi=\int_{-\infty}^\infty \left[h(\xi)-\theta(1-|\xi|)h(0)\right]\frac{1}{|\xi|}\,d\xi$$
with $\theta(\xi)$ the unit step function.
Now first we Fourier transform the product of $f$ and $g$,
$${\cal F}_{f\cdot g}=\int_{-\infty}^\infty \ln|x|\delta(x-a)e^{ ix\xi}dx=e^{ia\xi}\ln|a|$$
and then we want to check that the convolution theorem gives the same answer:
$$\frac{1}{2\pi}\int_{-\infty}^\infty {\cal F}_f(\xi-t){\cal F}_g(t)dt=\int_{-\infty}^\infty\left(-\gamma\delta(t)-\tfrac{1}{2}{\cal P}\frac{1}{|t|}\right)e^{ia(\xi-t)}dt$$
$$=- e^{ia\xi}\left(\gamma+\int_{1}^\infty\frac{1}{t}\cos at\,dt+\int_{0}^1\frac{1}{t}(\cos at-1)\,dt\right)$$
$$=- e^{ia\xi}\left(\gamma-{\rm Ci}(|a|)-\gamma+{\rm Ci}(|a|)-\ln|a|\right)=e^{ia\xi}\ln|a|$$
with Ci the cosine integral. So it works out.
