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I have been told that the Helmholtz decomposition theorem says that

every smooth vector field $\boldsymbol{F}$ [where I am not sure what precise assumptions are needed on $\boldsymbol{F}$] on an opportune region $V\subset\mathbb{R}^3$ [satisfying certain conditions for whose precisation I would be very grateful to any answerer] can be expressed as $$ \boldsymbol{F}(\boldsymbol{x})=-\nabla\left[\int_{V}\frac{\nabla'\cdot \boldsymbol{F}(\boldsymbol{x}')}{4\pi\|\boldsymbol{x}-\boldsymbol{x}'\|}dV'-\oint_{\partial V}\frac{\boldsymbol{F}(\boldsymbol{x}')\cdot\hat{\boldsymbol{n}}(\boldsymbol{x}')}{4\pi\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'\right]$$ $$+\nabla\times\left[\int_{V}\frac{\nabla'\times \boldsymbol{F}(\boldsymbol{x}')}{4\pi\|\boldsymbol{x}-\boldsymbol{x}'\|}dV'+\oint_{\partial V}\frac{\boldsymbol{F}(\boldsymbol{x}')\times\hat{\boldsymbol{n}}(\boldsymbol{x}')}{4\pi\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'\right]$$[where I suppose that the $\int_V$ integrals are intended as limits of Riemann integrals or Lebesgue integrals].

I think I have been able to prove it (below), interpretating the integrals as Lebesgue integrals with $dV'=d\mu'$ where $\mu'$ is the usual tridimensional Lebesgue measure, for a compactly supported $\boldsymbol{F}\in C^2(\mathbb{R}^3)$ with $\boldsymbol{x}\in \mathring{V}$ and $V$ satisfying the hypothesis of Gauss's divergence theorem.

Nevertheless, I am also interested in proofs of it under less strict assumptions on $\boldsymbol{F}$. What are the usual assumptions -I am particularly interested in the assumptions done in physics- on $\boldsymbol{F}$ and how can the theorem proved in that case? I heartily thank any answerer.

I think that it would be interesting to generalise it to some space containing $C_c^2(\mathbb{R}^3)$ whose functions have "smoothness" properties usually considered true in physics, where the Helmholtz decomposition is much used, but I am not able to find such a space and prove the desired generalisation.

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    $\begingroup$ I am not sure what your question is. A proof of the Helmholtz decomposition without Green functions? $\endgroup$ – mvw Jan 31 '16 at 14:30
  • $\begingroup$ @mvw Thank you for your comment! If I correctly understand, here it's implied that elementary multivariate calculus is enough to prove it and, since I don't even know what Green functions are (if I correctly understand this, they are studied within functional analysis in connection with Dirac's $\delta$, which I like better to avoid to prove this theorem in order to avoid to fall into problems [...] $\endgroup$ – Self-teaching worker Jan 31 '16 at 15:01
  • $\begingroup$ [...] like this), I hope they aren't indispensable to prove it. Anyhow I'm not allergic to the $\delta$, but, at my very low level, I'm not able to understand the nonchalant use that I see made of it in the proofs of the Helmholtz decomposition that I find on line (usually adressed to physics and engineering students)... $\endgroup$ – Self-teaching worker Jan 31 '16 at 15:01
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Let $\boldsymbol{F}:\mathbb{R}^3\to \mathbb{R}^3$, $\boldsymbol{F}\in C^2( \mathbb{R}^3)$ compactly supported and let $V\subset\mathbb{R}^3$ be a region satisfying the hypothesis of Gauss's divergence theorem, with $\boldsymbol{x}\in \mathring{V}$ contained in its interior.

By using this result with $f(z)=\|z\|^{-1}$ and $g=F_i$, $i=1,2,3$, and taking the validity of this for $\varphi\in C_c^2(\mathbb{R}^3)$ into account, we see that $$\boldsymbol{F}(\boldsymbol{x})=-\frac{1}{4\pi}\int_{\mathbb{R}^3} \frac{\nabla'^2\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'=-\frac{1}{4\pi}\nabla^2\int_{\mathbb{R}^3} \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$which, according to a known identity valid for $C^2$ vector fields, equates $$\frac{1}{4\pi} \nabla\times\left[\nabla\times\int_{\mathbb{R}^3} \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'\right]-\frac{1}{4\pi}\nabla\left[\nabla\cdot \int_{\mathbb{R}^3} \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'\right]$$which, according to this again, is$$\frac{1}{4\pi} \nabla\times\int_{\mathbb{R}^3} \frac{\nabla'\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu' -\frac{1}{4\pi}\nabla \int_{\mathbb{R}^3} \frac{\nabla'\cdot\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$ $$=\frac{1}{4\pi} \nabla\times\int_V \frac{\nabla'\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu' -\frac{1}{4\pi}\nabla \int_V \frac{\nabla'\cdot\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$ $$+\frac{1}{4\pi} \nabla\times\int_{\mathbb{R}^3\setminus V} \frac{\nabla'\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu' -\frac{1}{4\pi}\nabla \int_{\mathbb{R}^3\setminus V} \frac{\nabla'\cdot\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'\quad\quad (1)$$ where the existence of the curls and divergences of the last member of the equality is guaranteed for example by the following lemma:

Let $\varphi:V\subset\mathbb{R}^3\to\mathbb{R}$ be bounded and $\mu_{\boldsymbol{y}}$-measurable, with $\mu_{\boldsymbol{y}}$ as the usual $3$-dimensional Lebesgue measure, where $V$ is bounded and measurable (according to the same measure). Let us define, for all $\boldsymbol{x}\in\mathbb{R}^3$, $$\Phi(\boldsymbol{x}):=\int_V \frac{\varphi(\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|}d\mu_{\boldsymbol{y}}$$then $\Phi\in C^1(\mathbb{R}^3)$ and, for $k=1,2,3$, $$\forall\boldsymbol{x}\in\mathbb{R}^3\quad\quad\frac{\partial \Phi(\boldsymbol{x})}{\partial x_k}=\int_V\frac{\partial}{\partial x_k} \left[\frac{\varphi(\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|}\right]d\mu_{\boldsymbol{y}}=\int_V \varphi(\boldsymbol{y})\frac{y_k-x_k}{\|\boldsymbol{x}-\boldsymbol{y}\|^3}d\mu_{\boldsymbol{y}}$$

which is proved here. The integrand functions of the integrals above are$$ \frac{\nabla'\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}=\nabla'\times\left[ \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]-\nabla'\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] \times\boldsymbol{F}(\boldsymbol{x}')$$$$=\nabla'\times\left[ \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]+\nabla\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] \times\boldsymbol{F}(\boldsymbol{x}')$$$$=\nabla'\times\left[ \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]+\nabla\times\left[\frac{ \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]$$ and $$ \frac{\nabla'\cdot\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}=\nabla'\cdot\left[ \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]-\nabla'\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] \cdot\boldsymbol{F}(\boldsymbol{x}')$$$$=\nabla'\cdot\left[ \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]+\nabla\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] \cdot\boldsymbol{F}(\boldsymbol{x}')$$$$=\nabla'\cdot\left[ \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]+\nabla\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] $$ and therefore $$\frac{1}{4\pi} \nabla\times\int_{\mathbb{R}^3\setminus V} \frac{\nabla'\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu' -\frac{1}{4\pi}\nabla \int_{\mathbb{R}^3\setminus V} \frac{\nabla'\cdot\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$ $$=\frac{1}{4\pi} \nabla\times\int_{\mathbb{R}^3\setminus V} \nabla'\times\left[ \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] + \nabla\times\left[\frac{ \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] d\mu'$$$$-\frac{1}{4\pi}\nabla\int_{\mathbb{R}^3\setminus V} \nabla'\cdot\left[ \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] + \nabla\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] d\mu'.$$ Let us chose a parallelepiped $R$ or sphere such that $\boldsymbol{F}$ and its derivatives are null outside its interior and such that $\bar{V}\subset \mathring {R}$ and let us call $E:=\overline{R\setminus V}$. By using Leibniz's rule we see that $$\int_E \nabla\times\left[\nabla\times\left[\frac{ \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\right] d^3x'=\nabla\times\int_E \nabla\times\left[\frac{ \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] d^3x'$$$$= \nabla\times\int_{\mathbb{R}^3\setminus V} \nabla\times\left[\frac{ \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] d\mu'$$and $$\int_E \nabla\left[\nabla\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] \right]d^3x'=\nabla\int_E \nabla\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] d^3x'$$$$=\nabla\int_{\mathbb{R}^3\setminus V} \nabla\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] d\mu'$$and their difference is $$\nabla\times\int_{\mathbb{R}^3\setminus V} \nabla\times\left[\frac{ \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] d\mu'-\nabla\int_{\mathbb{R}^3\setminus V} \nabla\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] d\mu'$$$$=\int_E \nabla\times\left[\nabla\times\left[\frac{ \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\right] - \nabla\left[\nabla\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] \right]d^3x'$$$$=\int_E-\nabla^2\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] d^3x'=\mathbf{0}$$ and so $$\frac{1}{4\pi} \nabla\times\int_{\mathbb{R}^3\setminus V} \frac{\nabla'\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu' -\frac{1}{4\pi}\nabla \int_{\mathbb{R}^3\setminus V} \frac{\nabla'\cdot\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$ $$=\frac{1}{4\pi} \nabla\times\int_{\mathbb{R}^3\setminus V} \nabla'\times\left[ \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu' +\frac{1}{4\pi} \nabla\times\int_{\mathbb{R}^3\setminus V} \nabla\times\left[\frac{ \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] d\mu'$$$$-\frac{1}{4\pi}\nabla\int_{\mathbb{R}^3\setminus V} \nabla'\cdot\left[ \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] d\mu' -\frac{1}{4\pi} \nabla\int_{\mathbb{R}^3\setminus V} \nabla\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] d\mu'$$$$=\frac{1}{4\pi} \nabla\times\int_{\mathbb{R}^3\setminus V} \nabla'\times\left[ \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu' -\frac{1}{4\pi}\nabla\int_{\mathbb{R}^3\setminus V} \nabla'\cdot\left[ \frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right] d\mu' $$which, by using Gauss's theorem of divergence and the derived $\int_D\nabla\times\mathbf{F}d^3x=\int_{\partial D} \hat{\mathbf{n}}\times\mathbf{F}d\sigma$ identity, we can see to equate$$\frac{1}{4\pi} \nabla\times\int_{\partial E} \frac{\hat{\boldsymbol{n}}(\boldsymbol{x'})\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\sigma' -\frac{1}{4\pi}\nabla\int_{\partial E} \frac{\boldsymbol{F}(\boldsymbol{x}')\cdot\hat{\boldsymbol{n}}(\boldsymbol{x'})}{\|\boldsymbol{x}-\boldsymbol{x}'\|} d\sigma' $$$$=-\frac{1}{4\pi} \nabla\times\int_{\partial V} \frac{\hat{\boldsymbol{n}}(\boldsymbol{x'})\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\sigma' +\frac{1}{4\pi}\nabla\int_{\partial V} \frac{\boldsymbol{F}(\boldsymbol{x}')\cdot\hat{\boldsymbol{n}}(\boldsymbol{x'})}{\|\boldsymbol{x}-\boldsymbol{x}'\|} d\sigma' $$where the last member is derived by taking the fact that the normal on $\partial V$ is the opposite of the normal to the internal frontier surface of $E$ into account, as well as the nullity of $\boldsymbol{F}$ on the external surface. By substituting this in the expression $(1)$ we conclude that $$\boldsymbol{F}(\boldsymbol{x})=\frac{1}{4\pi} \nabla\times\int_V \frac{\nabla'\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu' -\frac{1}{4\pi}\nabla \int_V \frac{\nabla'\cdot\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$$$-\frac{1}{4\pi} \nabla\times\int_{\partial V} \frac{\hat{\boldsymbol{n}}(\boldsymbol{x'})\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\sigma' +\frac{1}{4\pi}\nabla\int_{\partial V} \frac{\boldsymbol{F}(\boldsymbol{x}')\cdot\hat{\boldsymbol{n}}(\boldsymbol{x'})}{\|\boldsymbol{x}-\boldsymbol{x}'\|} d\sigma' .$$I do not accept this answer of mine because, since many statements usually found in texts of physics do not specify particular assumptions, I suspect that much more relaxed assumptions on $\boldsymbol{F}$ might be intended.

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Thanks to a conversation I have had with user TrialAndError, whom I deeply thank, here, I think I have been able to understand a proof for the particular case where $\boldsymbol{F}:\mathring{A}\subset\mathbb{R}^3\to\mathbb{R}^3$ is such that $\exists \boldsymbol{G}\in C^3(\mathring{A}):\boldsymbol{F}=\nabla^2 \boldsymbol{G}$ and $V$, with $ \bar{V}\subset\mathring{A}$, satisfies the assumptions of the divergence theorem, with $\boldsymbol{x}\in\mathring{V}$.

Then, this result shows, with an application of Leibniz's rule for differentiation under the integral sign, that $$\boldsymbol{F}(\boldsymbol{x})=-\frac{1}{4\pi}\nabla^2\int_V\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$where $\mu'$ is the usual $3$-dimensional Lebesgue measure. The same result shows that the integral in the expression above belongs to $C^2(\mathbb{R}^3)$ and therefore a known identity for the curl of the curl means that$$\boldsymbol{F}(\boldsymbol{x})=\frac{1}{4\pi}\nabla\times\left[\nabla\times\int_V\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'\right]-\frac{1}{4\pi}\nabla\left[\nabla\cdot\int_V\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'\right]$$where the following

Lemma. Let $\varphi:V\subset\mathbb{R}^3\to\mathbb{R}$ be bounded and $\mu'$-measurable, with $\mu'$ as the usual $3$-dimensional Lebesgue measure, where $V$ is bounded and measurable (according to the same measure). Let us define, for all $\boldsymbol{x}\in\mathbb{R}^3$, $$\Phi(\boldsymbol{x}):=\int_V \frac{\varphi(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$then $\Phi\in C^1(\mathbb{R}^3)$ and, for $k=1,2,3$, $$\forall\boldsymbol{x}\in\mathbb{R}^3\quad\quad\frac{\partial \Phi(\boldsymbol{x})}{\partial x_k}=\int_V\frac{\partial}{\partial x_k} \left[\frac{\varphi(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'=\int_V \varphi(\boldsymbol{x}')\frac{x_k'-x_k}{\|\boldsymbol{x}-\boldsymbol{x}'\|^3}d\mu'$$

which is proved here allows the commutation between derivative and integral sings to get$$\boldsymbol{F}(\boldsymbol{x})=\frac{1}{4\pi}\nabla\times\int_V\nabla\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'-\frac{1}{4\pi}\nabla\int_V\nabla\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'$$$$=\frac{1}{4\pi}\nabla\times\int_V\nabla\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\times\boldsymbol{F}(\boldsymbol{x}')d\mu'-\frac{1}{4\pi}\nabla\int_V\nabla\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\cdot\boldsymbol{F}(\boldsymbol{x}')d\mu'$$$$=-\frac{1}{4\pi}\nabla\times\int_V\nabla'\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\times\boldsymbol{F}(\boldsymbol{x}')d\mu'+\frac{1}{4\pi}\nabla\int_V\nabla'\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\cdot\boldsymbol{F}(\boldsymbol{x}')d\mu'$$$$=-\frac{1}{4\pi}\nabla\times\int_V\nabla'\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]-\frac{\nabla'\times \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$$$+\frac{1}{4\pi}\nabla\int_V\nabla'\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]-\frac{\nabla'\cdot \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$$$=-\frac{1}{4\pi}\nabla\times\int_V\nabla'\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'+\frac{1}{4\pi}\nabla\times\int_V\frac{\nabla'\times \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$$$+\frac{1}{4\pi}\nabla\int_V\nabla'\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'-\frac{1}{4\pi}\nabla\int_V\frac{\nabla'\cdot \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$where an application of the divergence theorem gives

$$\int_V\nabla'\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'=\lim_{\delta\to 0}\int_{V\setminus B(\boldsymbol{x},\delta)}\nabla'\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]dx_1'dx_2'dx_3'$$$$=\lim_{\delta\to 0}\int_{\partial(V\setminus B(\boldsymbol{x},\delta))}\frac{\hat{\boldsymbol{n}}(\boldsymbol{x}')\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'=\int_{\partial V}\frac{\hat{\boldsymbol{n}}(\boldsymbol{x}')\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'$$ because the integral on the internal surface is $\int_{\partial B(\boldsymbol{x},\delta)}\frac{-\hat{\boldsymbol{n}}(\boldsymbol{x}')\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'\xrightarrow{\delta\to 0} \mathbf{0}$ and analogously$$\int_V\nabla'\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'=\int_{\partial V}\frac{\boldsymbol{F}(\boldsymbol{x}')\cdot \hat{\boldsymbol{n}}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'$$

and therefore $$\boldsymbol{F}(\boldsymbol{x})=-\frac{1}{4\pi}\nabla\times\int_{\partial V}\frac{\hat{\boldsymbol{n}}(\boldsymbol{x}')\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'+\frac{1}{4\pi}\nabla\times\int_V\frac{\nabla'\times \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$$$+\frac{1}{4\pi}\nabla\int_{\partial V}\frac{\boldsymbol{F}(\boldsymbol{x}')\cdot \hat{\boldsymbol{n}}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'-\frac{1}{4\pi}\nabla\int_V\frac{\nabla'\cdot \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$as we wished to prove.

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