Knowing a scalar field from its Laplacian and gradient

As a part of the "rabbit hole" I am descending in order to understand the meaning of the integrals of a not-so-rarely found derivation of Ampère's law, I am trying to understand how to see the validity of the following result (found here):

Let $F$ be a $C^2$ scalar field defined on a region of $\mathbb{R}^3$ satisfying opportune assumptions [which I do not know in the details, but I think that $F\in C^2(\mathring{A})$, with $\bar{V}\subset\mathring{A}$ satisfying the conditions of the divergence theorem, and $r\in\mathring{V}$ is what is needed]; then $$\frac{1}{4\pi}\int_{V}\nabla^2 F(r')\frac{1}{|r-r'|}dV(r') =\lim_{\epsilon\downarrow 0}\frac{1}{4\pi}\int_{V\setminus B_{\epsilon}(r)}\nabla^2F(r')\frac{1}{|r-r'|}dV(r')$$$$= \lim_{\epsilon\downarrow 0}\frac{1}{4\pi}\int_{\partial(V\setminus B_{\epsilon}(r))}\frac{1}{|r-r'|}\frac{\partial F}{\partial n}dS(r') = -F(r)+\frac{1}{4\pi}\int_{\partial V}\frac{1}{|r-r'|}\frac{\partial F}{\partial n}(r')dS(r').$$

I applied Green's second identity, by taking the identity $\forall r'\ne r\quad\nabla^2\left(\frac{1}{|r-r'|}\right)=0$ into account, to get (the differentiations are intended with respect to the components of $r'$; $\hat{N}_e$ is the external normal to the surface of $V\setminus B_{\epsilon}(r)$) $$\frac{1}{4\pi}\int_{V\setminus B_{\epsilon}(r)}\nabla^2F(r')\frac{1}{|r-r'|}-F(r')\nabla^2\left(\frac{1}{|r-r'|}\right)dV(r')$$$$=\frac{1}{4\pi}\int_{\partial(V\setminus B_{\epsilon}(r))}\frac{1}{|r-r'|}\frac{\partial F}{\partial n}+ F(r')\frac{r-r'}{|r-r'^3|}\cdot \hat{N}_e\,dS(r')$$$$=\frac{1}{4\pi}\int_{\partial V}\frac{1}{|r-r'|}\frac{\partial F}{\partial n}dS(r')+\frac{1}{4\pi}\int_{\partial V}F(r')\frac{r-r'}{|r-r'|^3}\cdot \hat{N}_e\,dS(r')$$$$+\frac{1}{4\pi}\int_{\partial B_{\epsilon}(r)}\frac{1}{|r-r'|}\frac{\partial F}{\partial n}dS(r')-\frac{1}{4\pi}\int_{\partial B_{\epsilon}(r)}\frac{F(r')}{\epsilon^2}dS(r')$$In this last member I see that, as $\epsilon\to 0$, the last addend approaches $-F(r)$, the last to one approaches $0$, but I do not understand why $\frac{1}{4\pi}\int_{\partial V}F(r')\frac{r-r'}{|r-r'^3|}\cdot \hat{N}_e\,dS(r')$ disappears. I thank both TrialAndError who told me about this identity and any other people willing to answer me.

Edit (warning): The user who first told me about this result corrected himself: its correct form is

$$\lim_{\epsilon\downarrow 0}\frac{1}{4\pi}\int_{V\setminus B_{\epsilon}(r)}\frac{\nabla^2F(r')}{|r-r'|}dV(r')$$$$=-F(r)+\frac{1}{4\pi}\int_{\partial V}\frac{1}{|r-r'|}\frac{\partial F}{\partial n'}(r')-F(r')\frac{\partial}{\partial n'}\frac{1}{|r-r'|}dS(r')$$

• The last term will go to $4\pi F$. Then, take $V$ as all of space. Under suitable conditions on $F$, the surface integrals at infinity will vanish. – Mark Viola Jan 30 '16 at 21:11
• @Dr.MV Thank you very much! Mmh.. well, I think that it would be enough for $F$ to be bounded, if I am understanding... Nevertheless, it seems quite unusual to me that no limit but $\lim_{\epsilon\downarrow 0}$ appears in the statement... – Self-teaching worker Jan 30 '16 at 21:24
• Yes, sometimes the authors take that ommission lightly. Bounded is sufficient. In fact, if the scalar is a physical quantity, we assume that it is zero outside a finite volume $V'$ that is inside $V$. – Mark Viola Jan 30 '16 at 21:29
• @Dr.MV Thank you so much! – Self-teaching worker Jan 31 '16 at 9:21
• You're welcome! My pleasure. - Mark – Mark Viola Jan 31 '16 at 17:14

If $F\in C^2(\mathring{A})$, $F:\mathring{A}\to\mathbb{R}$, $\bar{V}\subset \mathring{A}$, $r\in\mathring{V}$ and there exists a $\delta$ such that, for all $\epsilon\le \delta$, $\epsilon>0$, the region $V\setminus B_\epsilon(r)$ satisfies the condition of the divergence theorem, then $$F(r)=\frac{1}{4\pi}\int_{\partial V}\frac{1}{|r-r'|}\frac{\partial F}{\partial n'}(r')-F(r')\frac{\partial}{\partial n'}\frac{1}{|r-r'|}dS(r')$$$$-\lim_{\epsilon\downarrow 0}\frac{1}{4\pi}\int_{V\setminus B_{\epsilon}(r)}\frac{\nabla^2F(r')}{|r-r'|}dV(r')$$