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).
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.