I just wanted to know whether the integral of a closed n-form is invariant if we integrate it over homotopy equivalent spaces. This seems like a generalization of "Homotopy invariance of line integral on manifolds". Am I right?
Something like: $$ \int_{\Sigma} \omega = \int_{\Sigma'} \omega$$ for homotopy equivalent $\Sigma\equiv \Sigma'$
Edit: Ok, I found an obvious counterexample by setting $\Sigma=$n-dimensional ball and $\Sigma'=\{0\}$ embedded in $\mathbb{R}^{n+1}$ with $\omega= dx_1\wedge...\wedge dx_n$. However, I'm still curious as to the correct generalization of "Homotopy invariance of line integral on manifolds" since it is mentioned in page 2 here http://arxiv.org/pdf/hep-th/9706092v1.pdf
Your expression doesn't make sense. If $\omega$ is a form on $\Sigma$, then it's not a form on $\Sigma'$! You need a map $f:\Sigma'\to \Sigma$ to compare them. So a better version of your claim would be $\int_{\Sigma'} f^*\omega=\int_\Sigma\omega$ when $f$ is a homotopy equivalence. This is now false, rather than undefined.
What's missing from your conjecture is closedness. For closed forms, it's nearly true as I just stated it, and the reason is Stokes' theorem, as has been mentioned. For suppose $f:\Sigma'\to \Sigma$ is a homotopy equivalence with homotopy inverse $g$, so $F:\Sigma \times I\to \Sigma$ is a homotopy with $F_0=\text{id}_\Sigma$ and $F_1=fg.$ Then $\int_\Sigma \omega-\int_{\Sigma} (fg)^*\omega=\int_{\Sigma\times I} d(F^*\omega)=\int_{\Sigma\times I} F^*(d\omega)=0$, since by hypothesis $d\omega=0$. But also $\int_{\Sigma} (fg)^*\omega=\int_{\Sigma'} f^*\omega,$ as was to be shown.
Remember I said nearly true earlier? This is because the integral of $f^*\omega$ is really only defined on $\Sigma'$ if $\omega$ has the same degree as the dimension of $\Sigma'$, so one should also make that restriction. One more observation: this is actually a special case of a more general theorem, which is the most valuable one to remember, and which has the same proof. Namely,