electrostatic potential, analytic properties An electrostatic potential associated  with some delocalized  charge $\int \rho(\mathbf{r}) d{\mathbf{r}}$ is given by:
$$v_H(\mathbf{r}) = \int \frac{\rho(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|}d\mathbf{r'}$$
This potential is finite at $\mathbf{r}=0$. Since $\frac{1}{|\mathbf{r}-\mathbf{r'}|}$ is a solution to a singular Poisson's equation, we can show that:
$$\nabla^2v_H(\mathbf{r})=-4\pi\rho(\mathbf{r})$$
where $\rho$ is a smooth function being finite everywhere.
Would a function
$$v(\mathbf{r}) = \int \frac{\rho(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^n}d\mathbf{r'}$$
(where $n$ is some nonnegative integer)
be also finite at $\mathbf{r}=0$?
In this case, $\frac{1}{|\mathbf{r}-\mathbf{r'}|^n}$ does not represent a solution to Poisson's equation ($\mathbf{r} \in \mathbb{R}^3$) and the above analysis is not valid.
 A: Maybe too simple, and certainly not a general solution, but anyway here goes.  
The simplest case is a spherical charge distribution:  
$$\rho= \left\{ \begin{array}{cc} \rho_0 & r \le a \\ 0 & r>a \end{array} \right. $$
Then 
$$ V(0) = \rho_0 \int_0^a \frac{4 \pi r^2}{r^n} dr = 4 \pi \rho_0 \int_0^a r^{2-n} dr $$
Cases:
$$ 0<n<3 \qquad  V(0) = \frac{4 \pi \rho_0}{3-n} a^{3-n} \text{ , finite}$$
$$ n=3 \qquad  V(0) = 4 \pi \rho_0 \lim_{\epsilon\rightarrow 0} \ln \left(\frac{r}{\epsilon} \right)  \rightarrow \infty $$
$$ n>3 \qquad V(0) = \frac{4 \pi \rho_0}{n-3} \lim_{\epsilon\rightarrow 0} \left(\frac{1}{\epsilon^{n-3}} - \frac{1}{a^{n-3}} \right) \rightarrow \infty $$
In all these cases, the denominator of the original integral, $|r-r'|^n$, grows without bound for large $r$, and the charge is localized around the origin, so  
$$ \lim_{r \rightarrow \infty} V(r) =0 $$
So, $n=3$ divides the finite from the infinite cases.  (Technically, I suppose this calc constitutes a counter-example for $n \ge 3$ by exhibiting an unbounded result for those cases). 
