I want to use the fact that Fourier transform of a radial function is a radial function. I'm trying to find the Fourier transform of $$f(x)=\frac{1}{|x|^2}, x\in R^3$$ I began like this: \begin{align*} \mathcal{F}(f)(\xi)=\int_{R^3}e^{-2\pi i\xi\cdot x}\frac{1}{|x|^2}dx=[\text{ spherical coordinates}]\\= \int_0^{\infty}\int_0^{\pi}\int_0^{2\pi}e^{-2\pi i\xi\cdot (r\sin\theta\cos\phi, r\sin\theta\sin\phi, r\cos\theta)}\frac{1}{r^2}r^2\sin(\phi) d\phi d\theta dr\\=\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi}e^{-2\pi i\xi\cdot (r\sin\theta\cos\phi, r\sin\theta\sin\phi, r\cos\theta)}\sin(\phi) d\phi d\theta dr \end{align*} but I don't know how to continue further. I wanted to find a rotation matrix (a composition of it) so I could use the property above, but without success. Any hints, please?
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$\begingroup$ Possible duplicate of math.stackexchange.com/questions/2799385/… $\endgroup$– Andy WallsJun 8, 2018 at 22:39
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2 Answers
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You may assume $\xi=(0,0,|\xi|)$. This will simplify the integral.
answered Jun 1, 2018 at 10:30
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$\begingroup$ Why can I assume that? Because the Fourier transform is radial? $\endgroup$– WaddlesJun 1, 2018 at 10:39
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$\begingroup$ Yes, since $\hat f(\xi)$ depends only on $|\xi|$. $\endgroup$ Jun 1, 2018 at 10:49
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$\begingroup$ This makes no sense because first integral to compute is $\int_{0}^{2\pi}sin(x)dx$, which is zero. $\endgroup$– MartinJun 7, 2018 at 15:31
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$\begingroup$ @Martin The integral would be $$\int_0^\infty\int_0^\pi\int_0^{2\pi}e^{-2\pi i|\xi|r\cos\theta}\sin\phi\,d\theta d\phi dr=2\int_0^\infty\int_0^{2\pi}e^{-2\pi i|\xi|r\cos\theta}\,d\theta dr$$ $\endgroup$ Jun 10, 2018 at 13:32
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Even though $|x|^{-2}$ is locally integrable in $\mathbb R^3$, the Fourier integral doesn't exist. You have to work with distributions or introduce a regularization of the Fourier integral in some other way in order to define the transform rigorously. In terms of distributions, $$(|x|^{-2}, e^{-2 \pi i w \cdot x}) = \pi |w|^{-1}.$$ Certain results will still hold in terms of ordinary functions as well. For instance, Plancherel's theorem for the inner product holds provided that the other function is sufficiently well-behaved.
