The Fourier transform of the Coulomb potential $1/\vert \mathbf r \vert$ of an electric charge doesn't converge because one obtains $$F(k)=\frac {4\pi}{k} \int_0^\infty \sin(kr) dr.$$

The standard way to obtain a sensible value is to multiple the integrand by $f(\alpha,r)=e^{-\alpha r}$ and after doing the integral, taking the limit $\alpha\to 0$ (which has a nice physical reason). So one gets $$F(k)=\frac{4\pi}{k^2}.$$

Would any other function $f(\alpha,r)$ that makes the integral converge and that satisfies $\lim_{\alpha\to\alpha_0}f(\alpha,r)=1$ give the same result? For example $$F(k)=\lim_{\alpha\to 0}\frac {4\pi}{k} \int_0^\infty \frac{\sin(kr)}{\Gamma(\alpha r)} dr\stackrel{?}{=}\frac{4\pi}{k^2}.$$

In this case, Cesàro integration gives the same result. What would be the sufficient condition for uniqueness of regularization (maybe the theory of tempered distributions can answer this).

This question was asked in Math StackExchange.

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    $\begingroup$ the distributions are continuous linear operators on some vector space. the key word is "continuous" : $h(x) = \lim_{y \to x} h(y)$, in the same way, $FT[f] = \lim_{\epsilon \to 0} FT[T_\epsilon(f)]$ where $T_\epsilon$ is a regularization operator, more generally a en.wikipedia.org/wiki/Mollifier . continuity is what is sufficient for the result doesn't depend on the chosen regularization, as far as everything is proven continuous. $\endgroup$ – reuns Apr 2 '16 at 22:51
  • $\begingroup$ For me Cesaro integration gives $$\int_0^\infty \sin x dx=1$$ $\endgroup$ – Anixx May 8 '18 at 14:22

Yes, the answer is unique. What this "regularization" is doing is computing the Fourier Transform in the sense of distributions.

Editing to add (see comment of Christian Remling below): For your precise question, the hypotheses you propose for the regularization are too weak: it's not enough that $f(\alpha,r)$ converge to the constant function pointwise. What is true that most "natural" choices will give the same answer, because reasonable regularizations would converge weakly (i.e. in the sense of tempered distributions).

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    $\begingroup$ This answers the question the OP should have asked (Q: How does one naturally define the FT of $1/r$? A: as a tempered distribution), not the one he/she actually asked, to which the answer is no, since one can easily cook up functions $f(\alpha, k)$ that will give a different result (certainly for fixed $k$). $\endgroup$ – Christian Remling Apr 3 '16 at 2:29

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