# function defined using an integral, finiteness

How to prove that a function

$$v_H(\mathbf{r}) = \int \frac{\rho(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|}d\mathbf{r'}$$

is finite at any $\textbf{r}$?

$\rho(\mathbf{r'})$ appearing inside the integral is some well-behaved function that is finite everywhere, decays exponentially and vanishes at infinity.

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 any $\mathbf{r}$?

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Do we assume $\textbf{r}\in\mathbb{R}^3$? – icurays1 Nov 21 '12 at 17:11
yes, we are considering 3d space. – molkee Nov 21 '12 at 21:42

Partial answer: If $\rho(\textbf{r}^\prime)\in C^2_c(\mathbb{R}^3)$ (twice continuously differentiable with compact support), then $v_H(\textbf{r})\in C^2(\mathbb{R}^3)$ and $v_H$ solves a Poisson equation:
$$\triangle v_H=4\pi\rho$$
This is Theorem 1, page 23-25 in L. Evans' book on Partial Differential Equations, paraphrased in some lecture notes here. Note of course that care must be taken near $\textbf{r}=\textbf{r}^\prime$ (this is addressed in the proof).