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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.

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    $\begingroup$ That is because the distribution $\pi/|x|$ needs to be understood as a compensated integral: $(\pi/|x|) * f(x)$ is equal to $\pi \int_{-\infty}^\infty (f(y) - f(x) \mathbf{1}_{(-1,1)}(y - x)) / |y - x| dx$. (With this definition I am not sure about the constant $2 \pi \gamma$, though). See, for example, the theorem on page 40 in Vladimirov's book "Methods of the Theory of Generalised Functions". $\endgroup$ – Mateusz Kwaśnicki May 12 '18 at 11:14
  • $\begingroup$ I am sure that Gerry Folland makes a parallel between the "compensated integral" Mateusz refers to and the normalizations made in quantum field theory. It's in his Real Analysis book I think. $\endgroup$ – Giuseppe Negro May 12 '18 at 11:24
  • $\begingroup$ @Mateusz Kwaśnicki, even this type of integral seems not to be convergent ? (are you sure about parenthesis ?) $\endgroup$ – Bertrand May 12 '18 at 11:37
  • $\begingroup$ @Mateusz Kwaśnicki, it seems page 40 of Vladimirov's book mention $|x|^{\alpha -1}$ with $\alpha>0$ (that can be prolonged to meromorphic function but still with pole for $\alpha=0$...) $\endgroup$ – Bertrand May 12 '18 at 11:42
  • $\begingroup$ @Bertrand: I meant to write the integral with respect to $y$, not $x$, sorry. The theorem in Vladimirov's book is not written in a very clear way, but formula (7.4) with $N = 1$ and $\alpha = 0$ is what you were looking for. (Carlo Beenakker gives much more details in his answer.) $\endgroup$ – Mateusz Kwaśnicki May 12 '18 at 19:17
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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.

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