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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 $$\int_M d\omega = \int_{\partial M} \omega.$$

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  • $\begingroup$ How is it possible that $\partial M = \Bbb{R}_{\ge 0}$? One should have $\partial \partial M = \emptyset$, while $\partial \Bbb{R}_{\ge 0} = \{ 0 \} \neq \emptyset$. $\endgroup$
    – Crostul
    Nov 16, 2016 at 12:51
  • $\begingroup$ @Crostul Thanks for the comment, I edited it $\endgroup$
    – user389900
    Nov 16, 2016 at 12:54

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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

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  • $\begingroup$ I dont fully understand what $\alpha$ is exactly. According to me it is an element of $W^*$, if $W$ is a vector space and $M$ is the union of coordinate patches $A\rightarrow B$ with $A$ open in $W$ and $B$ open in $M$. Is that right? $\endgroup$
    – user389900
    Nov 16, 2016 at 13:05
  • $\begingroup$ @DannyRand No. $\Omega^0(M)=C^\infty(M)$, i.e. $\alpha$ is just a smooth function on $M$. $\endgroup$
    – user2520938
    Nov 16, 2016 at 13:08
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    $\begingroup$ Well you don't necessarily get a minus at $\int_{\partial M}\alpha$, but since the orientation of $\partial M$ is induced by the orientation of $M$ the signs of both $\alpha(0)$'s are equal. If you don't want the minus at $\int_{\partial M}\alpha$ then reverse the orientation on $M$. In that case though you'd get $$\int_Md\alpha=\int_{\infty}^0\alpha'(x)dx$$ so then here too the minus disappears in front of the $\alpha(0)$ $\endgroup$
    – user2520938
    Nov 16, 2016 at 13:18
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    $\begingroup$ compactly supported means that $\alpha=0$ everywhere except on a compact subset $V\subset M$. In the specific case of $M=\mathbb{R}_{\geq 0}$ this is equivalent with the statement that there is a $c\in\mathbb{R}$ s.t. $\alpha(x)=0$ for $x\geq c$. $\endgroup$
    – user2520938
    Nov 16, 2016 at 13:26
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    $\begingroup$ @user2520938 If $a(x)=2$ for $x \ge 0$, then does Stokes hold as well? $\endgroup$
    – Amontillado
    Nov 16, 2016 at 13:34

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