About the Fourier transform of the logarithm function

Question

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.

Answer 1

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.

Answer 2

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 $\psi_\varepsilon$.

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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 $\psi^{(3)}_\varepsilon$.

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DISTRIBUTIONAL LIMITS OF $\displaystyle \psi^{(1)}_\varepsilon$, $\displaystyle \psi^{(2)}_\varepsilon$, and $\displaystyle \psi^{(3)}_\varepsilon$

Again, let $\phi\in \mathbb{S}$. Then,

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

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

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

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

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

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NOTE:

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

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