Question
Let $\alpha=\sum_{i=1}^m\alpha_i(x)dx^i\in\Omega^1(\mathbb{R}^m)$ be a closed 1-form on $\mathbb{R}^m$, and $f(x)=\sum_{i=1}^mx^i\int_0^1\alpha^i(ux)du$ define a smooth function satisfying $df=\alpha$.
Show that if $\alpha$ has compact support and $m>1$, then there exists a smooth function $g$ with compact support such that $dg=\alpha$. Show that this doesn't need to hold for $m=1$.
Attempt
Since $\alpha$ has a compact support in $\mathbb{R}^m$ with $m>1$, by the Poincaré lemma, there exists a function $g$ such that $dg=\alpha$.
Let’s construct $g$ so that it has compact support. Consider a bump function $\beta$ that is 1 on the support of $\alpha$, and smoothly decreases to zero outside a slightly larger compact set. Then $\beta g$ will have compact support and still satisfy:
$d(\beta g)=\beta dg=\beta \alpha=\alpha$.
(because $\alpha=0$ outside the support of $\alpha$, so is $d\alpha$).
Now, suppose $m=1$, then $\mathbb{R}$ is not simply connected on any bounded interval. So, the Poincaré lemma does not apply. For example, if $\alpha=dx$ on $\mathbb{R}$, then the function $g(x)=x$ is such that $dg=dx=\alpha$, but $g$ does not have compact support. In other words, if there is a function with compact support on $\mathbb{R}$ whose derivative is $dx$ everywhere, the integral of such function will have to be constant outside its support, which would contradict the fact that the derivative is $dx$.
Note
I am not sure I attempted this the right way, or at least there are a couple of things I left out in this proof.
Your help will be appreciated.