Suppose I have a distribution $F$ which I know on the derivative of an arbitrary test function. That is, I know $F(\phi')$ for all $\phi ∈ \mathcal{D}(ℝ)$. I also know that $F(\phi') = -F'(\phi)$. I want, however, to know $F(\phi)$. Is there a 'distributional antiderivative' that immediately tells me what this should be? If not, is it possible for such $F'$ that are induced by simpler, specific functions? (e.g. polynomials)
For $\Phi(x) := ∫\phi(x)dx$, simply putting $F(\phi) = -F'(\Phi)$ seems a little dangerous.
If there exists $\psi\in C^\infty(\mathbb R)$ such that $\phi=\psi'$ then we can just set $F(\phi) = -F'(\psi).$
But this is in general not the case. It is true precisely when $\int_{\infty}^{\infty}\phi(x)\,dx=0.$ Now fix $\rho\in C^\infty_c(\mathbb R)$ such that $\int_{\infty}^{\infty}\rho(x)\,dx=1$ and given $\phi\in C^\infty_c(\mathbb R)$ set $\tilde\phi=\phi-(\int_{\infty}^{\infty}\phi(x)\,dx) \, \rho.$ Then $\int_{\infty}^{\infty}\tilde\phi(x)\,dx=0$ so we can apply the first paragraph on $\tilde\phi.$ Now we note that $F\left((\int_{\infty}^{\infty}\phi(x)\,dx)\rho\right) = F(\rho)(\int_{\infty}^{\infty}\phi(x)\,dx)$ is just some constant $F(\rho)$ times the constant distribution $\phi\mapsto\int_{\infty}^{\infty}\phi(x)\,dx$ acting on $\phi.$
Thus, we can set $F(\phi) = F(\tilde\phi) + F(\rho)(\int_{\infty}^{\infty}\phi(x)\,dx).$