In this question I will use Stokes' theorem with manifolds (well, domains in manifolds), rather than with chains.
Let $M$ be a smooth manifold of dimension $m$. A subset $D\subseteq M$ is called a regular domain if 1) $D$ is compact, 2) $D$ is the closure of its interior, 3) the topological boundary $\partial D$ is a smooth hypersurface in $M$ (everything should also be valid when $\partial M$ is a "manifold with corners", i.e. it is piecewise smooth instead of smooth, but I now ignore that case for simplicity).
Stokes' theorem then states that $$ \int_{\partial D}\omega=\int_D d\omega, $$where $\omega$ is a $C^1$ $m-1$-form and the pullback to $\partial D$ on the left hand side is implicit.
The compactness of $D$ as well as the regularity of $\omega$ ensures that both integrals are convergent.
In the textbook G. de Rham: Differentiable Manifolds, there is a counterexample that shows that Stokes' theorem need not be valid when $D$ is noncompact (and $\omega$ has no compact support), even if everything converges. As an example, let $D=(-\infty,0]$ and $f$ is a function such that $f(x)= -1$ for $x\le -1$ and $f(x)=0$ for $x\ge 0$. The boundary of $D$ is $\partial D=+\{0\}$, where the $+$ sign means that this point is assigned a positive oriantation, then $$ \int_D df=\int_{-\infty}^0 df=1, $$ but $$ \int_{\partial D}f=f(0)=0, $$ which shows that Stokes' theorem does not apply to this case.
The problem is that the integral $\int_D df$ is calculated as $\int_D df=\int_{-\infty}^0f^\prime(x)dx=f(0)-\lim_{r\rightarrow\infty}f(-r)$, i.e. it is an improper Riemann integral, and a version of Stokes' theorem is obeyed with $D$ having a "fictious" boundary piece $-\{-\infty\}$ (the outer sign denotes orientation). But of course as a manifold with boundary, $\partial D=+\{0\}$, and the "boundary piece at infinity" is missing here and this is why Stokes' theorem fails.
A similar situation is often encountered in theoretical physics, where one often integrates over the entirety of $\mathbb R^3$. As a manifold, $\partial\mathbb R^3=\varnothing$ and Stokes' theorem says that $\int_{\mathbb R^3}d\omega=0$ if $\omega$ has compact support and does not say anything otherwise.
However in physics, such and integral would be considered with $\mathbb R^3$ having a fictious "boundary at infinity". In fact, what happens (often implicitly) is that instead of integrating over $\mathbb R^3$, one integrates over the closed ball $\bar B_r\equiv \bar B_r(0)$ of radius $r$, which is a regular domain whose boundary is $S_r\equiv S_r(0)$ the $2$-sphere of radius $r$. One may then take apply Stokes' theorem to $\mathbb R^3$ itself by $$ \lim_{r\rightarrow\infty}\int_{\bar B_r}d\omega=\lim_{r\rightarrow\infty}\int_{S_r}\omega, $$ with the compact closed balls $\bar B_r$ providing an exhaustion of $\mathbb R^3$ with compact sets.
Now, unlike calculus texts where improper Riemann integrals are frequently treated, I have never ever seen a textbook on differential geometry, where integration and Stokes' theorem are treated in a way that improper integrals are incorporated.
Ideally there would be a formalism where the Stokes' theorem $\int_M d\omega=\int_{\partial M}\omega$ could be interpreted for any $m$-manifold (with or without boundary) $M$ and any $m-1$-form $\omega$ such that the left hand side converges and where $\partial M$ is allowed to have "fictious" pieces at infinity. Probably such a version of Stokes' theorem would be defined in terms of exhaustions by regular domains, but of course there are questions I don't know the answer to such as Does every $M$ admit such an exhaustion? or Is the value of the integral independent of the exhaustion?
I am looking for resources (textbooks, papers) where a coherent theory of "improper integrals" on manifolds (including Stokes' theorem!) is elaborated.