Homotopy invariance of line integral on manifolds Consider a 1-form: $\omega\in\Gamma(\mathrm{T}^*M)$
and two differentiable curves: $\gamma,\tilde{\gamma}:[a,b]\to M:\gamma(a)=\tilde{\gamma}(a),\gamma(b)=\tilde{\gamma}(b)$
together with a differentiable homotopy: $\Phi:[0,1]\times[a,b]:\Phi(0,t)\equiv\gamma(t),\Phi(1,t)\equiv\tilde{\gamma}(t)$
Prove or disprove the homotopy invariance of the line integral: $\int_\gamma\omega=\int_\tilde{\gamma}\omega$
Note that the homotopy invariance fails for merely continuous functions...
 A: The integral of closed 1-forms is invariant under path-homotopy (Theorem 16.26, Lee SM).
Moreover, the integral over a not necessarily simply connected manifold with boundary of either compactly supported 1-forms with non vanishing non negative exterior derivative or equivalently 1-forms of non vanishing support on the boundary (or sth like that) certainly not invariant under path-homotopy.
A: If you mean, by the integral, $\int_\gamma f(s(t)) \dot{s}(t) dt$ and everything is in a subset of $\mathbb{R}^n$ (otherwise the definition of your integral is not clear), then such invariance can be hardly satisfied by a larger class of functions then constants. For nonconstant $f$, it holds only if $$\int_{\gamma-\tilde{\gamma}} f=0.$$
This means, for any vector $v\in\mathbb{R}^n$, $$\int_{\gamma-\tilde{\gamma}} (fv)\,\overrightarrow{ds}=0.$$ Assume that $\Omega$ is a $2$-disc bounded by $\gamma-\tilde{\gamma}$. The expression $fv$ can be considered to be a $1$-form and by Stokes theorem, 
$$\int_{\gamma-\tilde{\gamma}} (fv)=\int_{\Omega} d(fv)=0.$$ This should hold for any $\Omega$ and any $v$, so $d(fv)=0$, i.e. $v_i \partial_j f - v_j \partial_i f=0$. Taking for $v$ the basis vectors in $\mathbb{R}^n$, we get that $\partial_j f=0$ for all $j$, so $f$ is constant.
There must be some simpler explanation, I just don't see it at the moment :(
