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](https://en.wikipedia.org/wiki/Fourier_transform#Distributions), 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.



About the Fourier transform of the logarithm function
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](https://math.stackexchange.com/a/138013/87355). 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.](http://mathworld.wolfram.com/CosineIntegral.html) So it works out.


