Calculus/Helmholtz Decomposition Theorem

The Helmholtz Decomposition Theorem, regarded as the fundamental theorem of vector calculus, dictates that any vector field $$\mathbf{F}$$ can be expressed as the sum of a conservative vector field $$\mathbf{G}$$ and a divergence free vector field $$\mathbf{H}$$: $$\mathbf{F} = \mathbf{G} + \mathbf{H}$$.

Approach #1
Given a vector field $$\mathbf{F}$$, the vector field $$\mathbf{G}(\mathbf{q}) = \frac{1}{4\pi}\iiint_{\mathbf{q}' \in \R^3} \frac{(\nabla \cdot \mathbf{F})|_{\mathbf{q}'}(\mathbf{q} - \mathbf{q}')}{|\mathbf{q}-\mathbf{q}'|^3}dV'$$ has the same divergence as $$\mathbf{F}$$, and is also conservative: $$\nabla \cdot \mathbf{G} = \nabla \cdot \mathbf{F}$$ and $$\nabla \times \mathbf{G} = \mathbf{0}$$. The vector field $$\mathbf{H} = \mathbf{F} - \mathbf{G}$$ is divergence free.

Therefore $$\mathbf{F} = \mathbf{G} + \mathbf{H}$$ where $$\mathbf{G}(\mathbf{q}) = \frac{1}{4\pi}\iiint_{\mathbf{q}' \in \R^3} \frac{(\nabla \cdot \mathbf{F})|_{\mathbf{q}'}(\mathbf{q} - \mathbf{q}')}{|\mathbf{q}-\mathbf{q}'|^3}dV'$$ and $$\mathbf{H} = \mathbf{F} - \mathbf{G}$$. Vector field $$\mathbf{G}$$ is conservative and $$\mathbf{H}$$ is divergence free.

Approach #2
Given a vector field $$\mathbf{F}$$, the vector field $$\mathbf{H}(\mathbf{q}) = \frac{1}{4\pi}\iiint_{\mathbf{q}' \in \R^3} \frac{(\nabla \times \mathbf{F})|_{\mathbf{q}'} \times (\mathbf{q} - \mathbf{q}')}{|\mathbf{q}-\mathbf{q}'|^3}dV'$$ has the same curl as $$\mathbf{F}$$, and is also divergence free: $$\nabla \times \mathbf{H} = \nabla \times \mathbf{F}$$ and $$\nabla \cdot \mathbf{H} = 0$$. The vector field $$\mathbf{G} = \mathbf{F} - \mathbf{H}$$ is conservative.

Therefore $$\mathbf{F} = \mathbf{G} + \mathbf{H}$$ where $$\mathbf{H}(\mathbf{q}) = \frac{1}{4\pi}\iiint_{\mathbf{q}' \in \R^3} \frac{(\nabla \times \mathbf{F})|_{\mathbf{q}'} \times (\mathbf{q} - \mathbf{q}')}{|\mathbf{q}-\mathbf{q}'|^3}dV'$$ and $$\mathbf{G} = \mathbf{F} - \mathbf{H}$$. Vector field $$\mathbf{G}$$ is conservative and $$\mathbf{H}$$ is divergence free.

Approach #3
The Helmholtz decomposition can be derived as follows:

Given an arbitrary point $$\mathbf{q}'$$, the divergence of the vector field $$\frac{\mathbf{q}-\mathbf{q}'}{4\pi|\mathbf{q}-\mathbf{q}'|^3}$$ is $$\nabla_{\mathbf{q}} \cdot \frac{\mathbf{q}-\mathbf{q}'}{4\pi|\mathbf{q}-\mathbf{q}'|^3} = \delta(\mathbf{q};\mathbf{q}')$$ where $$\delta(\mathbf{q};\mathbf{q}')$$ is the Dirac delta function centered on $$\mathbf{q}'$$ (The subscript $$_\mathbf{q}$$ clarifies that $$\mathbf{q}$$ as opposed to $$\mathbf{q}'$$ is the parameter that the differential operator is being applied to). Since $$\nabla_\mathbf{q}(\frac{-1}{|\mathbf{q}-\mathbf{q}'|}) = \frac{\mathbf{q}-\mathbf{q}'}{|\mathbf{q}-\mathbf{q}'|^3}$$, it is the case that $$\nabla_{\mathbf{q}}^2 \frac{-1}{4\pi|\mathbf{q}-\mathbf{q}'|} = \nabla_{\mathbf{q}} \cdot \frac{\mathbf{q}-\mathbf{q}'}{4\pi|\mathbf{q}-\mathbf{q}'|^3} = \delta(\mathbf{q};\mathbf{q}')$$

Alongside the identities $$\nabla \cdot (f\mathbf{G}) = (\nabla f) \cdot \mathbf{G} + f(\nabla \cdot \mathbf{G})$$, and $$\nabla \times (f\mathbf{G}) = (\nabla f) \times \mathbf{G} + f(\nabla \times \mathbf{G})$$, and most importantly $$\nabla \times (\nabla \times \mathbf{F}) = \nabla(\nabla \cdot \mathbf{F}) - \nabla^2\mathbf{F}$$, the following can be derived:

$$\mathbf{F}(\mathbf{q}) = \iiint_{\mathbf{q}' \in \R^3} \delta(\mathbf{q};\mathbf{q}')\mathbf{F}(\mathbf{q}')dV'$$ $$ = \iiint_{\mathbf{q}' \in \R^3} (\nabla_{\mathbf{q}}^2 \frac{-1}{4\pi|\mathbf{q}-\mathbf{q}'|})\mathbf{F}(\mathbf{q}')dV'$$ $$ = \iiint_{\mathbf{q}' \in \R^3} (\nabla_{\mathbf{q}}^2 \frac{-\mathbf{F}(\mathbf{q}')}{4\pi|\mathbf{q}-\mathbf{q}'|})dV'$$

$$ = \iiint_{\mathbf{q}' \in \R^3} (\nabla_{\mathbf{q}}(\nabla_{\mathbf{q}} \cdot \frac{-\mathbf{F}(\mathbf{q}')}{4\pi|\mathbf{q}-\mathbf{q}'|}) - \nabla_{\mathbf{q}} \times (\nabla_{\mathbf{q}} \times \frac{-\mathbf{F}(\mathbf{q}')}{4\pi|\mathbf{q}-\mathbf{q}'|}))dV'$$

$$ = \iiint_{\mathbf{q}' \in \R^3} (\nabla_{\mathbf{q}}(\frac{(\mathbf{q}-\mathbf{q}') \cdot \mathbf{F}(\mathbf{q}')}{4\pi|\mathbf{q}-\mathbf{q}'|^3}) - \nabla_{\mathbf{q}} \times (\frac{(\mathbf{q}-\mathbf{q}') \times \mathbf{F}(\mathbf{q}')}{4\pi|\mathbf{q}-\mathbf{q}'|^3}))dV'$$

$$ = \nabla_{\mathbf{q}} \iiint_{\mathbf{q}' \in \R^3} \frac{\mathbf{F}(\mathbf{q}') \cdot (\mathbf{q}-\mathbf{q}')}{4\pi|\mathbf{q}-\mathbf{q}'|^3}dV' + \nabla_{\mathbf{q}} \times \iiint_{\mathbf{q}' \in \R^3} \frac{\mathbf{F}(\mathbf{q}') \times (\mathbf{q}-\mathbf{q}')}{4\pi|\mathbf{q}-\mathbf{q}'|^3}dV'$$

$$\mathbf{G}(\mathbf{q}) = \nabla_{\mathbf{q}} \iiint_{\mathbf{q}' \in \R^3} \frac{\mathbf{F}(\mathbf{q}') \cdot (\mathbf{q}-\mathbf{q}')}{4\pi|\mathbf{q}-\mathbf{q}'|^3}dV'$$ is the gradient of a scalar field, and so is conservative.

$$\mathbf{H}(\mathbf{q}) = \nabla_{\mathbf{q}} \times \iiint_{\mathbf{q}' \in \R^3} \frac{\mathbf{F}(\mathbf{q}') \times (\mathbf{q}-\mathbf{q}')}{4\pi|\mathbf{q}-\mathbf{q}'|^3}dV'$$ is the curl of a vector field, and so is divergence free.

In summary, $$\mathbf{F} = \mathbf{G} + \mathbf{H}$$ where $$\mathbf{G}(\mathbf{q}) = \nabla_{\mathbf{q}} \iiint_{\mathbf{q}' \in \R^3} \frac{\mathbf{F}(\mathbf{q}') \cdot (\mathbf{q}-\mathbf{q}')}{4\pi|\mathbf{q}-\mathbf{q}'|^3}dV'$$ is conservative and $$\mathbf{H}(\mathbf{q}) = \nabla_{\mathbf{q}} \times \iiint_{\mathbf{q}' \in \R^3} \frac{\mathbf{F}(\mathbf{q}') \times (\mathbf{q}-\mathbf{q}')}{4\pi|\mathbf{q}-\mathbf{q}'|^3}dV'$$ is divergence free.