Why does Stokes's theorem for manifolds without boundary not work for $\int_{a<x<b}f'(x)\,\mathrm{d}x$

Question

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?

Answer 1

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.

Answer 2

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.