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


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.

  • 2
    $\begingroup$ 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". $\endgroup$ Commented May 12, 2018 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. $\endgroup$ Commented May 12, 2018 at 11:24
  • $\begingroup$ @Mateusz Kwaśnicki, even this type of integral seems not to be convergent ? (are you sure about parenthesis ?) $\endgroup$
    – Bertrand
    Commented May 12, 2018 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
    Commented May 12, 2018 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.) $\endgroup$ Commented May 12, 2018 at 19:17

2 Answers 2


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.


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.


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

$$\begin{align} \Psi_\varepsilon(x)&=\mathscr{F}\{\psi_\epsilon\}(x)\\\\ &=\int_{-\infty}^\infty e^{-\varepsilon|k|}\log(|k|) e^{ikx}\,dk\\\\ &=2\text{Re}\left(\int_0^\infty e^{-(\varepsilon -ix)k}\log(k) \,dk\right)\\\\ &=-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma -\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2)-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)\\\\ &=\psi^{(1)}_\varepsilon(x)+\psi^{(2)}_\varepsilon(x)+\psi^{(3)}_\varepsilon(x)\tag2 \end{align}$$


$$\begin{align} \psi^{(1)}_\varepsilon(x)&=-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma\\\\ \psi^{(2)}_\varepsilon(x)&=-\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2)\\\\ \psi^{(3)}_\varepsilon(x)&=-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \end{align}$$

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,

$$\begin{align} \lim_{\varepsilon\to 0^+}\langle \psi^{(1)}_\varepsilon,\phi \rangle &=\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \psi^{(1)}_\varepsilon(x)\phi(x)\,dx\\\\ &=\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \left(-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma \right)\phi(x)\,dx\\\\ &=-2\gamma\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\phi(\varepsilon x)}{x^2+1}\,dx\\\\ &=-2\pi \gamma \phi(0)\tag3 \end{align}$$

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

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


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 $\phi\in \mathbb{S}$,

$$\int_{-\infty}^\infty \phi(x) \text{PV}\left(\frac1{|x|}\right)\,dx=\int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx$$


It was arbitrary to split the integration in $(5)$ into intervals $|x|\le 1$ 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 $\phi\in \mathbb{S}$,

$$\int_{-\infty}^\infty \phi(x) \text{PV}\left(\frac1{|x|}\right)\,dx=\int_{|x|\le \nu}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge \nu}\frac{\phi(x)}{|x|}\,dx$$

  • 1
    $\begingroup$ This question already had an accepted answer. Could you perhaps clarify what your new answer adds to what has gone before? $\endgroup$
    – Yemon Choi
    Commented Apr 20, 2021 at 4:05
  • 2
    $\begingroup$ Yes. It provides a rigorous development of the Fourier Transform on $\log(|x|)$ as a tempered distribution. And it clarifies the meaning of the object "$-\frac\pi{|\xi|}$," which is interpreted in the distributional sense. Finally, it provides a discussion of the arbitrariness of the definition of the principal value as defined herein. Does that help clarify? $\endgroup$
    – Mark Viola
    Commented Apr 20, 2021 at 4:11
  • $\begingroup$ @YemonChoi I am happy to delete if you don't believe that this is useful. $\endgroup$
    – Mark Viola
    Commented Apr 20, 2021 at 4:14
  • $\begingroup$ I don't believe this answer needs to be deleted; I was just hoping that you could add some of this extra context at the top of your new answer, so that people who see this question and your answer will have a clearer idea of what the new contribution is. $\endgroup$
    – Yemon Choi
    Commented Apr 20, 2021 at 4:19
  • $\begingroup$ @YemonChoi Thank you for the suggestion. I've modified accordingly. $\endgroup$
    – Mark Viola
    Commented Apr 20, 2021 at 4:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.