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.
Best Answer
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.