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).

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