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According to Wikipedia, the fundamental theorem of calculus $$\int_a^bf'(x)\,\mathrm{d}x=\int_{[a,b]}\mathrm df=\int_{\partial[a,b]}f=f(b)-f(a)$$ is an example of Stokes's theorem. It is known that $$\int_{a\leq x\leq b}f'(x)\,\mathrm{d}x=\int_{a<x<b}f'(x)\,\mathrm{d}x,$$ however, one would expect $$\int_{a<x<b}f'(x)\,\mathrm{d}x=\int_{(a,b)}\mathrm df=\int_{\partial(a,b)}f=\int_{\emptyset}f=0$$ by Stokes's theorem for manifolds without boundary.

Where's my mistake?

Filippo
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    It just doesn’t. The boundary is an important part of Stokes’ Theorem. It’s kinda like asking why you can’t prove $4$ is prime. – AJY Jul 15 '21 at 07:01
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    If you check the proper theorem statement in wikipedia it will assume that $\Omega$ has boundary – Calvin Khor Jul 15 '21 at 07:05
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    $\partial(a,b)$ is still ${a,b}$, not $\emptyset$. The boundary operator takes the surroundings of the set into account. – Vercassivelaunos Jul 15 '21 at 07:11
  • @Vercassivelaunos I am surprised: In Lee's introduction to smooth manifolds, the boundary $\partial M$ of a manifold $M$ is a subset of $M$: $\partial M\subset M$ (page 25). – Filippo Jul 15 '21 at 07:23
  • That's only true for a "manifold with boundary", not for regular manifolds. For Stokes' theorem we usually take a regular submanifold of some larger manifold. Then we take its topological boundary as a subspace of the larger manifold, and add that to the submanifold. Now it's a "manifold with boundary", and that's the object to which we can apply the theorem. – Vercassivelaunos Jul 15 '21 at 09:13

2 Answers2

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Here is the precise statement of the Stokes theorem:

Let $M$ be an $n$-dimensional oriented manifold (not necessarily compact) and $\omega$ an $n-1$-form with compact support. Then $$\int_M\mathrm{d}\omega=0$$ if $\partial M=\emptyset$.

In particular, the Stokes theorem does not say that the integral of all differentiable functions on $(a,b)$ is zero. It says that the integral of all such functions with compact support is zero. For a function $f$ satisfying these requirements we have that$$f(a):=\lim_{x\to a}f(x)=0$$ and $$f(b):=\lim_{x\to b}f(x)=0$$such that$$\int_a^bf'(x)\,\mathrm{d}x=f(b)-f(a)=0-0=0=\int_{(a,b)}\mathrm{d}f$$ and we get consistent results.

Filippo
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This question strikes to me as a very interesting question about definitions. Your argument that $$\int_{(a,b)}f'(x)dx=0,$$ using the Stokes theorem sounds right to me, because after all $(a,b)$ is a $1$-manifold with empty boundary and $f'(x)dx$ is a exact $1$-form on that manifold. In the other hand $$\int_{(a,b)}f'(x)dx= \int_a^b f'(x)dx,$$ where the integral on the RHS is the usual Riemann integral. But obviously is not true that
$$\int_a^b f'(x)dx=0 $$ for every function $f$ defined over $(a,b)$. So what is wrong ? What is wrong here is the fact that the integral over $(a,b)$ (as a manifold) is not well defined for every $1$-form, more specifically over a open manifold we only can integrate forms with compact support, so the integral $$\int_{(a,b)}f'(x)dx$$ only makes sense if the form $f'(x)dx$ has compact support, and as oberved by Fillipo equation $$\int_{(a,b)}f'(x)dx=0,$$ only makes sense (in the sense of using the stokes theorem) if the form $f(x)dx$ have compact suport, but in that case near the extremes of the interval $(a,b)$ the function $f'(x)$ is zero, what does not contradicts the equation $$\int_{(a,b)}f'(x)dx= \int_a^b f'(x)dx = f(b)-f(a).$$ So your argument sounds ok to me, you only have to restrict yourself to function $f$ with compact support.

LucLuzzi
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    Thank you very much! However, I still have a doubt: Let's assume $f'|{(a,b)}$ has compact support. For instance, let's say $f'$ is supported on $[a',b']$ with $a<a'<b'<b$. I'd expect $\int{(a,b)}\mathrm{d}f=\int_{a'}^{b'}f'(x),\mathrm{d}x$. However, I don't see why $f(a')=f(b')$. – Filippo Jul 15 '21 at 10:36
  • I guess $f$ would be constant on $[a,a']$ and $[b',b]$, but I don't see why $f$ should have the same value on both intervals. – Filippo Jul 15 '21 at 10:39
  • I think I just "solved" the problem: It doesn't suffice that $f'$ has compact support. $f$ needs to have compact support (which implies that $f'$ has compact support). – Filippo Jul 15 '21 at 10:51
  • As far as I understand, you are assuming that $\int_{a'}^{b'}f'(x)dx=0$, which would imply $f(a')=f(b')$ by the fundamental theorem of calculus. But $\int_{a'}^{b'}f'(x)dx=0$ is not necessarely true, even if the support of $f'$ is equal to $[a',b']$. However, the fundamental theorem of calculus says that $\int_{a'}^{b'}f'(x)dx=0$ if the support of $f$ is equal to $[a',b']$. I looked into two books on differential geometry (Lee's introduction to smooth manifolds and Hamilton's mathematical gauge theory) and in both books it is required that $f$ has compact support. – Filippo Jul 15 '21 at 11:31
  • Which solves my problem, since in that case $\int_{(a,b)}df=\int_a^bf'(x)dx=0$. – Filippo Jul 15 '21 at 11:31
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    You are right, for all to work well we need that $f$ has compact support, thank you by the observation i will edit the original answer to look more complete. – LucLuzzi Jul 15 '21 at 12:03
  • Thank you! I am happy that we now agree! – Filippo Jul 15 '21 at 12:19