Is the regularization of a Fourier transform unique? 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).
 A: Probably not the answer you're looking for but here it is anyway :
You didn't specified the dimension of the space, so lets say the dimension is 3. I'll also assume that you're familiar with distribution theory. 
We note $q(x)=1/\|x\|_2$, in $\mathbb R^3$ this function is locally integrable and bounded outside of the (compact) unit ball. So the function $q$ can be considered as a tempered distribution, this is a good thing cause we can define $\mathcal F(q)$ without ambiguity : for any function $\varphi$ in the Schwartz class $\mathcal S(\mathbb R^3)$ we have  $\langle\mathcal F(q)|\varphi\rangle=\langle q | \mathcal F (\varphi)\rangle=\int q\mathcal F(\varphi)d \lambda$ (So $\mathcal F (q)$ is also a tempered distribution).
Now suppose we have a sequence $(q_n)_n$ of tempered distributions. We have 
$$\langle\mathcal F(q_n)|\varphi\rangle=\langle q_n | \mathcal F (\varphi)\rangle$$ and we want $$\lim_n \langle\mathcal F(q_n)|\varphi\rangle=\lim_n\langle q_n | \mathcal F (\varphi)\rangle=\langle\mathcal F(q)|\varphi\rangle=\langle q | \mathcal F (\varphi)\rangle$$
for all $\varphi \in \mathcal S(\mathbb R^3)$. In other terms we want $\mathcal F (q_n)$ to converge to $\mathcal F (q)$ in the sens of $S'(\mathbb R^3)$. Since $\mathcal F$ is an automorphism of $\mathcal S(\mathbb R^3)$ this is equivalent to 
$$\lim_n \langle q_n|\mathcal \varphi\rangle= \langle\mathcal q|\varphi\rangle \;\;\forall \varphi \in \mathcal S(\mathbb R^3) $$
which is the definition of $q_n\to q$ in $\mathcal S'(\mathbb R^3)$ (convergence in the sense of tempered distribution). And by definition $q_n \to q\in \mathcal S'(\mathbb R^3)$ if and only if $\langle q_n |\varphi\rangle\to\langle q |\varphi\rangle$ $\forall \varphi \in S(\mathbb R^3) $.
Now if we assume moreover that the $q_n$ are $L^1(\mathbb R^3)$ functions (like in your examples) then we have $\langle q_n |\varphi\rangle=\int q_n \varphi d \lambda$ and thus you can show the convergence in $S'(\mathbb R^3)$ using  the dominated convergence theorem for exemple.
Now i believe that what you really wanted was the pointwise convergence, but this question is probablya bit more tricky. 
