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I have this relationship:

\begin{align} \frac{1}{|x|}=f(x)*f(x)\ , \end{align} where $*$ denotes the convolution. I want to solve for $f(x)$. My first instinct was to apply the convolution theorem:

\begin{align} \mathcal{F}\left\{\frac{1}{|x|}\right\}=\mathcal{F}\left\{f(x)\right\}^2\implies f(x)=\mathcal{F}^{-1}\left\{\sqrt{\mathcal{F}\left\{\frac{1}{|x|}\right\}}\right\}\ , \end{align}

where $\mathcal{F}\{\}$ is the Fourier transform. According to Mathematica, \begin{align} \sqrt{\mathcal{F}\left(\frac{1}{|x|}\right)}=\sqrt{\frac{-2 \log (\left| \omega\right| )-2 \gamma }{\sqrt{2 \pi }}}\ , \end{align} where $\omega$ is the Fourier conjugate variable to $x$ and $\gamma$ is Euler's constant. How should I compute the inverse Fourier transform of the above function? I was thinking of using contour integration, but I am not sure what contour I should choose.

Alternatively, is there an easier way to solve for $f(x)$, possibly using functional equation methods? I saw this question, which makes me concerned that there is not a unique $f(x)$ as solution.

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  • $\begingroup$ This is how I would do it. I'm not sure how to evaluate the inverse Fourier transform, but note that it is a distribution and you will likely have to get extremely creative. $\endgroup$
    – Cameron Williams
    Jun 19, 2018 at 20:09
  • $\begingroup$ The general idea looks good, but there are issues right at the beginning. I'm assuming you are in dimension $1$ (otherwise the FT of $|x|^{-1}$ is $c_d|\xi|^{1-d}$), and then $1/|x|$ isn't locally integrable, so doesn't define a distribution, and thus its FT isn't even defined in any obvious way. $\endgroup$
    – user138530
    Jun 20, 2018 at 22:35
  • $\begingroup$ Maybe one could get around this by looking for an $F$ with $F'=f$ and $f*F=\pm\log |x|$. $\endgroup$
    – user138530
    Jun 20, 2018 at 22:43
  • $\begingroup$ Right, I'm assuming 1 dimension, but it would be preferable if the Fourier transform would be able to work in 3 dimensions. $\endgroup$
    – user85503
    Jun 21, 2018 at 13:26
  • $\begingroup$ I'm a little uncertain by your suggestion of $\frac{dF}{dx}=f$. I want to see if I understand it. If I differentiate both sides of $f*F=\log|x|$, I get $\frac{1}{|x|}$ on the RHS, so I can I end of getting an equation that looks like $f*f+something=\frac{1}{|x|}$. I would then need to find the necessary $f$ such that $something=0$. Is this what you had in mind? $\endgroup$
    – user85503
    Jun 21, 2018 at 13:32

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