Consider the usual concept of integral on a smooth manifold (the one built using partitions of unity). When applied to the usual smooth structure of $\mathbb{R}^n$, does it coincide with the concept of improper integral? I assume so, but I am not sure.
A concrete example: is the function $\mathbb{e} ^{\mathbb{i} \|x\|^2}$ integrable (in the sense of integration on manifolds, not in the sense of improper integrals) on $\mathbb{R^n}$? (I know that it isn't Riemann or Lebesgue integrable.)
As @Yubal says, the thing works as in the usual case: some improper integrals can be computed, some not. The clearest example is the volume, which can be computed for any manifold, by integrating the volume form $\varOmega$, which of course has not compact support unless the manifold $M$ is compact. There you pick one exhaustion of $M$ by compact sets $K_n\subset$Int$(K_{n+1})$ and bump functions $\varphi_n$ which are $\equiv 1$ on $K_n$ and $\equiv 0$ off $K_{n+1}$. Then you have integrals with compact support $I_n=\int_M \varphi_n\varOmega$ and show that $I=\lim_n I_n$ exists (including it is $+\infty$). Also that the limit doesn't depend on the compacts sets nor the bump functions chosen. By definition, $I=\int_M\varOmega$.