Well James Munkres in the text Analysis on Manifolds prove the general Stoke's theorem for $k$-form when $k>1$ and then he proves it for $k=1$ only when the bounary of the Manifold is not empty and he leaves as exercise to show that if $f$ is a scalar function defined over a compact $1$-manifold $M$ then
$$
\int_M df=0
$$
So let's start to try to prove it and precisely we will do to scalar function whose support is covered by a single coordinate chart becasue as Munkres showed in the general proof this is sufficent. Indeed if $\Phi:\{\phi_i:i=1,…,l\}$ is a partition of unity dominated by the coordinate patches of $M$ then the support of the forn $\phi_i\omega$ is contained in a single coordinate patch for each $i=1,..,l$ and thus
$$
\int_Md\omega=\int_M0+\int_Md\omega=\int_M0\curlywedge\omega+\int_M\Biggl(\sum_{i=1}^l\phi_i\Biggl)\curlywedge d\omega=
\\
\int_Md1\curlywedge\omega+\int_M\Biggl(\sum_{i=1}^l\phi_i\curlywedge d\omega\Bigg)=
\\
\int_Md\Biggl(\sum_{i=1}^l\phi_i\Biggl)\curlywedge\omega+\sum_{i=1}^l\Biggl(\int_M\phi_i\curlywedge d\omega\Biggl)=
\\
\sum_{i=1}^l\Biggl(\int_Md\phi_i\curlywedge\omega\Biggl)+\sum_{i=1}^l\Biggl(\int_M(-1)^0\phi_i\curlywedge d\omega\Biggl)=\sum_{i=1}^l\Biggl(\int_Md\phi_i\curlywedge\omega+\int_M(-1)^0\phi_i\curlywedge d\omega\Biggl)
\\
\sum_{i=1}^l\Biggl(\int_Md\phi_i\curlywedge\omega+(-1)^0\phi_i\curlywedge d\omega\Biggl)=\sum_{i=1}^l\int_Md(\phi_i\omega)=0
$$
Well if $M$ is a compact $1$-manifold without boundary it is possible to prove that for any $p\in M$ there exist a coordinate patch $\alpha$ defined in the unitary positive interval $(0,1)$ and thus if $f:M\rightarrow\Bbb R$ is a scalar function defined over M whose support $S_f$ is covered by a single coordinate patch then
$$
\int_Mdf:=\int_{(0,1)}\alpha^*(df)=\int_{(0,1)}d(\alpha^*f)=\int_{(0,1)}d(\alpha\circ f)
$$
So Unfortunately I do not able to prove that
$$
\int_{(0,1)}d(\alpha\circ f)=0
$$
and thus I ask to do it. So could someone help me, please?
Best Answer
You say you know that it is sufficient to prove this for a scalar function $f$ with support $K$ contained in a coordinate open $U$, then $df$ also has support in $U$, say in $K'$. Now, let $\phi:(a,b)\to U$ be a parametrization. The integral in this case is defined by $$ \int_a^b\phi^*(df)=\int_a^bd(\phi^*f). $$ I think you also reached this conclusion. Write $g=\phi^*f$. Then $dg=g'(x)dx$. So, we have $$ \int_a^b\phi^*(df)=\int_a^bd(\phi^*f)=\int_a^bg'(x)dx=g(b)-g(a) $$ by the fundamental theorem of calculus. Now, as $K\subsetneq U$, $\phi^{-1}(K)$ is a proper subset of $(a,b)$ and hence does not contain $a$ or $b$. So, $g(b)=g(a)=0$. So $$ \int_a^b\phi^*(df)=\int_a^bd(\phi^*f)=\int_a^bg'(x)dx=g(b)-g(a)=0. $$