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In an attempt to solve a PDE for $f(\mathbf{x})$ with $\mathbf{x} \in \mathbb{R}^3$, I have arrived at an expression of form $\hat{f}(\mathbf{k}) = \ln|\mathbf{k}| \hat{g}(\mathbf{k})$. I would like to use the convolution theorem to write $f(\mathbf{x}) = T * g(\mathbf{x})$, where $T$ is a tempered distribution satisfying $\hat{T}(\mathbf{k}) = \ln|\mathbf{k}|$.

To compute $\mathcal{F}^{-1}(\ln|\mathbf{k}|)$, I unsuccessfully tried adapting the solution to the one-dimensional case given in this stack exchange post. A key ingredient in the answer is the representation of $\gamma + \ln|x|$ as an integral that, when treated as a distribution, amounts to taking the Fourier transform of the test function. I do not see any way to adapt this idea to three dimensions, but perhaps I am overlooking something.

Is an explicit formula for $\mathcal{F}^{-1}(\ln|\mathbf{k}|)$ known in the three-dimensional case? And if so, how would I compute it? Bonus points for the $n$-dimensional case.

user11553332
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    Have you considered combining it with something like https://www.math.arizona.edu/~faris/methodsweb/hankel.pdf , since $\hat{T}$ should be spherically symmetric? – Steven Stadnicki Apr 13 '21 at 22:44
  • Hi @StevenStadnicki, thank you for the suggestion--I did try the Hankel transform, but the integral doesn't converge in the sense of functions, and I am not sure how to proceed in the sense of distributions. I found a paper on a distributional Hankel transform but can't see clearly whether the relationship with the Fourier trasnform still holds. – user11553332 Apr 13 '21 at 23:00
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    Did you try differentiating wrt $a\in (0,1)$ the Fourier transform of $|x|^{-a}$ and letting $a\to 0^+$ – reuns Apr 14 '21 at 00:19
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    I get $\mathcal{F}^{-1}(\ln|\mathbf{k}|) = A |\mathbf{x}|^{-3} + B \delta(\mathbf{x})$ for some constants $A,B$ which I don't know how to find. – md2perpe Apr 14 '21 at 18:12
  • Thanks for the idea @reuns--I had looked into a similar strategy previously but realize now that I mishandled the limit. I am also a bit rusty with my analysis and will have to convince myself that the limit converges correctly in the sense of distributions. If you happen to know why the limit expression is valid and can spare a second to explain why, I'd be greatly appreciative. – user11553332 Apr 14 '21 at 18:19
  • @md2perpe--I think I 'agree' but think some care is required to explain what is meant by $|x|^{-3}$ because that function is not locally integrable. – user11553332 Apr 14 '21 at 18:21
  • One can do the reverse of the calculation I did, in 2d, in https://mathoverflow.net/questions/172916/free-boson-correlator-langle-xzxw-rangle-ln-z-w/172994#172994 – Abdelmalek Abdesselam Apr 14 '21 at 21:04
  • See this answer. If $$\mathcal F\phi = \int_{\mathbb R^3} \phi(x) e^{i w \cdot x} dx, \ \phi_S(r) = \frac 1 {4 \pi r^2} \int_{|x| = r} \phi(x) , dS, \ (|x|^{-3}, \phi) = \int_0^\infty \frac {\phi_S(r) - \phi(0) [r < 1]} r , dr$$ for a test function $\phi$, then $$\mathcal F \big x \mapsto \ln |x| \big = -8 \pi^3 (|w|^{-3} - (1 - \gamma) \delta(w)).$$ – Maxim May 04 '21 at 12:38

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As follows from the proof in my answer here by taking the inverse Fourier transform, in $\mathbb{R}^d$, if I define the Fourier transform with the convention that $\mathcal{F}(f)(x) = \int_{\mathbb{R}^d} f(y) \,e^{-2iπ\,x\cdot y}\,\mathrm{d}y$ and take its extension to distributions, the result is $$ \mathcal{F}\!\left(\ln|x|\right) = \frac{\psi(d/2)-\gamma}{2} - \ln(π) - \mathrm{pf}\!\left(\frac{1}{\omega_d\,|x|^d}\right) $$ where $\omega_d = \frac{2\,\pi^{d/2}}{\Gamma(d/2)}$ is the size of the unit sphere (so $\omega_3 = 4\pi$), $\gamma$ is Euler-Mascheroni constant and $\psi$ is the digamma function. The notation $\mathrm{pf}$ denotes Hadamard's finite part, which for the function here can be defined by $$ \mathrm{pf}\!\left(\frac{1}{|x|^d}\right) := \mathrm{div}\!\left(\frac{x\ln(|x|)}{|x|^d}\right) $$ where the derivative is taken in the sense of distributions, or equivalently, by its action on smooth and compactly supported test functions $\varphi$ $$ \left\langle \mathrm{pf}\!\left(\tfrac{1}{|x|^d}\right),\varphi\right\rangle = \int_{|x|\leq 1} \frac{\varphi(x)-\varphi(0)\,\mathbf{1}_{|x|\leq 1}}{|x|^d}\,\mathrm{d}x. $$ Since $\psi(3/2) = 2-\gamma-\ln(4)$, it implies that in dimension $3$ $$ \mathcal{F}\!\left(\ln|x|\right) = 1-\gamma -\ln(2π) - \mathrm{pf}\!\left(\frac{1}{4\pi\,|x|^3}\right). $$

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