A given vector field \mathbf{F} with second derivatives can be decomposed as

(1)   \begin{equation*} \mathbf{F}=-\nabla\phi+\nabla\times \mathbf{A}, \end{equation*}

where \phi is a scalar field and \mathbf{A} is a vector field. Additionally,

    \begin{align*} \phi&=\frac{1}{4\pi}\iiint\frac{\nabla'\cdot\mathbf{F}(\mathbf{r'})}{R}d^3r'-\frac{1}{4\pi}\oint\frac{\mathbf{F}(\mathbf{r'})\cdot d\mathbf{S}}{R}\\ \mathbf{A}&=\frac{1}{4\pi}\iiint\frac{\nabla'\times \mathbf{F}(\mathbf{r'})}{R}d^3r'-\frac{1}{4\pi}\oint\frac{d\mathbf{S}\times\mathbf{F}(\mathbf{r'})}{R}, \end{align*}

where R=|\mathbf{r}-\mathbf{r'}|.

 

To prove Helmholtz theorm, we use the following indentities

(2)   \begin{equation*} \nabla(\mathbf{r}-\mathbf{r'})=-\frac{1}{4\pi}\nabla^2\left(\frac{1}{R}\right), \end{equation*}

Additionally if \mathbf{F}(\mathbf{r'}) is not a function of \mathbf{r} then

(3)   \begin{equation*} \nabla^2\left(\mathbf{F}\phi\right)=\mathbf{F}\nabla^2\phi, \end{equation*}

which may appear to be too obvious to explicitly mention since \mathbf{F} is not a function of \mathbf{r}. However, it should be noted that the LHS Laplacian operates on a vector field, while the RHS one operates on a scalar function. To proof the above identity, we use the definition of Laplacian of a vector:

(4)   \begin{equation*} \nabla^2\mathbf{G}=\nabla(\nabla\cdot\mathbf{G})-\nabla\times\times\mathbf{G}. \end{equation*}

Replacing \mathbf{G} by \mathbf{F}\phi one gets

(5)   \begin{align*} \nabla^2(\mathbf{F}\phi)&=\nabla(\nabla\cdot\mathbf{F}\phi)-\nabla\times\nabla\times(\mathbf{F}\phi)\\ &=\nabla(\phi\underbrace{\nabla\cdot\mathbf{F}}_0+\mathbf{F}\cdot\nabla\phi)-\nabla\times(\nabla\phi\times\mathbf{F}+\phi\underbrace{\nabla\times \mathbf{F}}_0)\\ &=\nabla(\mathbf{F}\cdot\nabla\phi)-\nabla\times(\nabla\phi\times\mathbf{F})\\ \end{align*}

Using the identity:

    \[\nabla(\mathbf{G}\cdot\mathbf{L})=(\mathbf{G}\cdot\nabla) \mathbf{L}+(\mathbf{L}\cdot\nabla) \mathbf{G}+\mathbf{G}\times(\nabla\times\mathbf{L})+\mathbf{L}\times(\nabla\times\mathbf{G}),\]

and noting that \mathbf{F} does not on \mathbf{r} and that \nabla\times\nabla\phi=0,

(6)   \begin{equation*} \nabla(\mathbf{F}\cdot\nabla\phi)=(\mathbf{F}\cdot\nabla)\nabla\phi. \end{equation*}

Furthermore we use the following identity to simplify the second term

    \[\nabla\times(\mathbf{G}\times\mathbf{L})=\mathbf{G}(\nabla\cdot\mathbf{L})-\mathbf{L}(\nabla\cdot\mathbf{G})+(\mathbf{L}\cdot\nabla)\mathbf{G}-(\mathbf{G}\cdot\nabla)\mathbf{L}.\]

Therefore,

(7)   \begin{equation*} \nabla\times(\nabla\phi\times\mathbf{F})=(\mathbf{F}\cdot)\nabla\phi-\mathbf{F}\nabla^2\phi. \end{equation*}

This means that

(8)   \begin{equation*} \nabla^2(\mathbf{F}\phi)=\mathbf{F}\nabla^2\phi. \end{equation*}

Now we are ready to start the prove of Helmholtz theorem.

(9)   \begin{align*} \mathbf{F}(\mathbf{r})&=\iiint\mathbf{F}(\mathbf{r'})\delta(\mathbf{r}-\mathbf{r'})d^3r'\\ &=-\frac{1}{4\pi}\iiint\mathbf{F}(\mathbf{r')}\nabla^2\left(\frac{1}{R}\right)\\ &=-\frac{1}{4\pi}\nabla^2\iiint\frac{\mathbf{F}(\mathbf{r'})}{R}d^3r'\\ &=-\frac{1}{4\pi}\nabla\left(\nabla\cdot\iiint\frac{\mathbf{F}(\mathbf{r'})}{R}d^3r' \right)+\frac{1}{4\pi}\nabla\times\nabla\times\left( \iiint\frac{\mathbf{F}(\mathbf{r'})}{R} d^3r'\right)\\ &=-\frac{1}{4\pi}\nabla\left(\iiint\nabla\cdot\left(\frac{\mathbf{F}(\mathbf{r'})}{R}\right)d^3r' \right) + \frac{1}{4\pi}\nabla\times\left( \iiint\nabla\times\left(\frac{\mathbf{F}(\mathbf{r'})}{R} \right)d^3r'\right)\\ &=-\frac{1}{4\pi}\nabla\left(\underbrace{\iiint\mathbf{F}(\mathbf{r'}) \cdot\nabla\left(\frac{1}{R}\right)d^3r'}_{I_1} \right) - \frac{1}{4\pi}\nabla\times\left(\underbrace{\iiint\mathbf{F}(\mathbf{r'})\times\nabla\left(\frac{1}{R}\right)d^3r'}_{I_2} \right). \end{align*}

Note that

    \[\nabla\left(\frac{1}{R}\right)=-\nabla'\left(\frac{1}{R}\right).\]

Therefore,

    \begin{align*} I_1&=-\iiint \mathbf{F}(\mathbf{r'})\cdot\nabla'\left(\frac{1}{R} \right)d^3r'\\ &=-\iiint\nabla'\cdot\left(\frac{\mathbf{F}(\mathbf{r'})}{R} \right)d^3r'+\iiint\frac{\nabla'\cdot\mathbf{F}(\mathbf{r'})}{R}d^3r'\\ &=\iint\frac{\nabla'\cdot\mathbf{F}(\mathbf{r'})}{\mathbf{R}}d^3r'-\oint\frac{F(\mathbf{r'})}{R}\cdot d\mathbf{S} \end{align*}

and

    \begin{align*} I_2&=-\iiint \mathbf{F}(\mathbf{r'})\times\nabla'\left(\frac{1}{R}\right)d^3r'\\ &=\iiint\nabla'\left(\frac{1}{R}\right)\times\mathbf{F}(\mathbf{r'})d^3r'\\ &=\iiint\nabla'\times\left(\frac{\mathbf{F}(\mathbf{r'})}{R}\right)d^3r'-\iiint\frac{\nabla'\times\mathbf{F}(\mathbf{r'})}{R}d^3r'. \end{align*}

We know that

    \[\iiint \left(\nabla'\times\mathbf{G}\right)d^3r'=\oint d\mathbf{S}\times\mathbf{G},\]

therefore,

    \begin{equation*} I_2=-\iiint\frac{\nabla'\times\mathbf{F}(\mathbf{r'})}{R}d^3r'+\oint\frac{d\mathbf{S}\times\mathbf{F}(\mathbf{r'})}{R}. \end{equation*}

If we let

    \[\phi\triangleq \frac{1}{4\pi}I_1\]

and

    \[\mathbf{A}=-\frac{1}{4\pi}I_2,\]

we get

    \begin{equation*} \mathbf{F}(\mathbf{r})=-\nabla\phi+\mathbf{A}. \end{equation*}