Let $M$ be the manifold with boundary $ M=\mathbb{R}_{\geq0}$ and $\omega\in\Omega^0(M)$ a 0-form.
Suppose both $\int_M d\omega$ and $\int_{\partial M}\omega$ are finite.
Does Stokes theorem hold? Or does $\int_M d\omega=\int_{\partial M}\omega$ hold?
I have no idea how to prove of disprove this and would appreciate any hints to get me in the right direction. Also I use this version of Stokes:
Stokes' Theorem: Let $M$ be a smooth, oriented $n$-manifold with boundary, and let $\omega$ be a compactly supported smooth $(n-1)$-form on $M$. Then $$\int_M d\omega = \int_{\partial M} \omega.$$
Yes it does, if $\alpha$ is indeed compactly supported. say $\alpha(x)=0$ for $x\geq c$. In that case: $$\int_{\partial M}\alpha = -\alpha(0)$$ $$ \begin{align} \int_Md\alpha &= \int_0^\infty \alpha'(x)dx\\ &=\int_{0}^c\alpha'(0)dx\\ &=\alpha(c)-\alpha(0)=-\alpha(0) \end{align} $$
considering that you assume $\int_Md\alpha$ to exist, it might also be true if $\alpha$ is not compactly supported. I leave the similar proof to you