Distributional "Antiderivative" Suppose that $\mathbb{R}^3$, fix $f \in L^1(\mathbb{R}^3;\mathbb{R})$ and let $g \in L^1(\mathbb{R}^3;\mathbb{R}^3)$ satisfies
$$
div(g)=\delta_a -f.
$$
Then what is $g$?  It what is the distributional anti-derivative of this thing?
 A: In dimension 3, it is classical that 
$$ \mathrm{div} \left( \frac{\vec x}{\| \vec x \|^3} \right) = 4 \pi \delta_0$$
See for instance $\nabla \cdot \big(\frac{\hat{r}}{r^{2}}\big)$ and Dirac Delta Function
So you can take
$$g = \frac{1}{4\pi}\left(  \frac{\vec x}{\| \vec x \|^3} -  \frac{\vec x-\vec a}{\| \vec x - \vec a\|^3} \right)$$
A: Let $\vec{G}(\vec{x}) := \frac{1}{4\pi} \frac{\vec x}{\| \vec x \|^3}$. Then $\operatorname{div} \vec{G} = \delta_0$. Now set $\vec{g}_p := \vec{G}*(\delta_{\vec{a}}-f)$ i.e.
$$
\vec{g}_p(\vec{x}) = \int \frac{1}{4\pi} \frac{\vec{x}-\vec{x}'}{\| \vec{x}-\vec{x}' \|^3} (\delta_{\vec{a}}(\vec{x}') - f(\vec{x}')) \, d^3x' \\
= \frac{1}{4\pi} \frac{\vec{x}-\vec{a}}{\| \vec{x}-\vec{a} \|^3}
- \int \frac{1}{4\pi} \frac{\vec{x}-\vec{x}'}{\| \vec{x}-\vec{x}' \|^3} f(\vec{x}') \, d^3x'.
$$
Then
$$
\operatorname{div} \vec{g}_p 
= \operatorname{div}(\vec{G}*(\delta_{\vec{a}}-f))
= (\operatorname{div}\vec{G})*(\delta_{\vec{a}}-f)
= \delta_0 * (\delta_{\vec{a}}-f)
= \delta_{\vec{a}}-f.
$$
But $\vec{g}_p$ is not the only solution. We can to it add any $\vec{g}_h$ such that $\operatorname{div}\vec{h}_h=0.$ All solutions are thus given by $\vec{g} = \vec{g}_p + \vec{g}_h.$
