Suppose I have a $d$ dimensional manifold $\mathcal{M}$ on which I want to perform the integral of a certain function $\mathcal{f}: \mathcal{M} \longrightarrow \mathbb{R}$ I will have then $$\int_{\mathcal{M}} f \ dx$$ where $dx$ is a formal integration measure defined on $\mathcal{M}$.
Now consider an application $$\psi : \mathcal{M} \longrightarrow \mathcal{M'}\\ \psi(x) \longmapsto x+\epsilon(x)$$ where $\epsilon(x)$ is a smooth function such that: $\|\epsilon\|_{\mathcal{M}}<<1$. The transformation above is supposed to be a "generalized translation". To my understanding $\mathcal{M'}$ is point to point infinitesimally different from $\mathcal{M}$ and is of the same dimension as $\mathcal{M}$. The integral then will become: $$\int_{\mathcal{M}+\epsilon(\mathcal{M})}f\circ\psi \ |det(d\psi)|\ dx'$$ where the jacobian is defined as $$J^{\mu}_{\nu}=\delta^{\mu}_{\nu}+\partial_{\nu}\epsilon^{\mu} \\ \mu,\nu=1......d$$
So $$det(d\psi)=e^{Tr\log(\delta^{\mu}_{\nu}+\partial_{\nu}\epsilon^{\mu})}$$
Let's define $A_{\mu \nu}=\partial_\nu \epsilon_{\mu}$ then $$log(\mathbb{1}+ A(\epsilon))=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} (A)^n=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} (\partial_\nu\epsilon_{\mu})^n$$ It is straightforward now to see that: $$det(d\psi)=exp \left(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} Tr(\partial_\nu\epsilon_{\mu})^n\right)\\ Tr A_{\mu\nu}=A^{\nu}_{\nu}=\partial^{\nu}\epsilon_{\nu}$$ So $det(d\psi) \approx 1+\partial^{\mu}\epsilon_{\mu}+O(\epsilon^2)$ then the integral is $$\int_{\mathcal{M}+\epsilon(\mathcal{M})}f(x+\epsilon(x)) \cdot (1+\partial^{\mu}\epsilon_{\mu})\ dx = \int_{\mathcal{M}+\epsilon(\mathcal{M})}( f(x)+\delta f(x)) \cdot (1+\partial^{\mu}\epsilon_{\mu})\ dx $$ keeping only $O(\epsilon)$ terms we get: $$\int_{\mathcal{M}+\epsilon(\mathcal{M})}( f(x)+\delta f(x)+f(x)\partial^{\mu}\epsilon_{\mu})\ dx=\int_{\mathcal{M}}( f(x)+\delta f(x)+f(x)\partial^{\mu}\epsilon_{\mu})\ dx+\int_{\mathcal{\epsilon(M)}} f(x)\ dx$$ having used the linearity of the integral and the fact that $|\int_{\mathcal{\epsilon(M)}}| \backsim |\epsilon|^d$
In order to ask my question I'll have to make an example: Let's say that $\mathcal{M} = \{ [a,b]\subset \mathbb{R}$ } so if we have $$\int_{\mathcal{M}}f=\int_a^b f \rightarrow \int_{a+\epsilon(a)}^{b+\epsilon(b)}f=\int_{a}^{b}f +\int_{b}^{b+\epsilon(b)}f -\int_a^{a+\epsilon(a)} f= \int_{\mathcal{M}}f+\int_{\epsilon(\mathcal{M})}f$$ but $$\Delta I=\int_{b}^{b+\epsilon(b)}f -\int_a^{a+\epsilon(a)} f =f(b)\epsilon(b)-f(a)\epsilon(a)$$ which can be written as $$\Delta I= \int_{a}^{b}d(\epsilon f)$$ and if we assume that $\epsilon(b)=\epsilon(a)=\epsilon$ (the function $\epsilon(x)$ is arbitrary!) we have that $$\Delta I= \epsilon (f(b)-f(a))=\epsilon \int_{a}^{b} df$$
I was wondering if there was a generalization of this relation in the general case I sketched above and if it's related in some way to Stoke's theorem.
EDIT: solution attempt, after some thinking I wrote this can anyone tell me if this qualifies as a proof or if there are mistakes/there's another one?
By definition we have that $M(\epsilon)=M'-M$ now let's take a step back and recall the definition of boundary: it is the complement of the internal part of the manifold, i.e. the set of all points which have neighborhoods homeomorphic to an open subset of $\mathbb{R}^d$. Ok now let's give a more precise meaning to the expression $M'-M$; we can see that we want all points which are in the "difference between the two manifolds" i.e. $$M(\epsilon)=\overline{M'\cap M}$$ Now that I have given a more useful definition for $M(\epsilon)$ let's explore it's meaning in a more precise way. Consider $x' \in M'$ and $x \in M$ then we will have two open neighborhoods $A_{x'}\subset M'$ and $B_x \subset M$ for which exist two Homeomorphisms: $$\phi_{M' }:A_{x'}\subset M'\longrightarrow \alpha \subset \mathbb{R}^d \\ \phi_{M }:B_{x}\subset M\longrightarrow \beta \subset \mathbb{R}^d$$ Let $\psi$ be the infinitesimal diffeomorphism defined above,such that :$$\psi:A_{x'}\subset M'\longrightarrow B'_x \subset M$$ The point $x'$ won't land exactly in the same neighborhood as the original $x$ due to the infinitesimal displacement given by $\psi$ defined in the question above. Let's consider the following composition, $\phi_{M'}\circ\psi^{-1}:B'_x\subset M\longrightarrow \alpha\subset \mathbb{R}^d$ this map allows us to compare $M$ and $M'$ so let's define: $$\chi\equiv(\phi_{M'}\circ\psi^{-1})\ \cap \ \phi_{M}: (B_x\cap B'_x) \subset M \longrightarrow (\alpha \cap \beta)\subset \mathbb{R}^d$$ where all intersections exists because the map $\psi$ is taken to be an infintesimal displacement between $M$ and $M'$ and for the definition of $M'$ itself given in the question. Now if we take the complement of this relation we find that $$\overline{\chi}:\overline{B \cap B'}\subset M \longrightarrow \overline{\alpha \cap \beta}\subset \mathbb{R}^d$$ but since $\alpha$ and $\beta$ (which remember are infintesimaly close) are just (multidimensional) intervals in $\mathbb{R}^d$ we know that $\overline{\alpha \cap \beta}$ for $\epsilon \rightarrow 0$ is precisely the boundary of the original neighborhood $\alpha$, just like in the example in the question above. Thus we get: $\Gamma = \overline{(\alpha \cap \beta)} \cap \partial \mathbb{R}^d\neq 0$ then since $\chi$ and $\overline{\chi}$ are Homeomorphisms we have that $\overline{ B \cap B'}$ must be in $\partial M$. We have "demonstrated" that the complement of the intersection between $M$ and $M'$ when considering infinitesmial maps is just the boundary of $M$. This should make clear that the formal relation defined above for $M(\epsilon)$ can be rewritten as: $$M(\epsilon) = \overline{M'}\cup \overline{M}=\partial M'\cup \partial M \approx \partial M$$ since $M' \approx M $.
In the end then the integral becomes : $$\int_{M(\epsilon)} f(x) dx \rightarrow \int_{\partial M} f(x)\epsilon^{\mu}d\Sigma^{\mu} = \int_{M} \partial_{\mu}(\epsilon^{\mu}f(x)) \ dx $$