# Do we have $\int_M \mathcal L_X\omega=0$ for $m$-form $\omega$?

Let $$M$$ be an oriented $$m$$-manifold without boundary, $$X\in \mathfrak X(M)$$ be a vector field, and $$\omega \in \Omega^m_c(M)$$ be a compactly supported top-degree differential form.

Question: Do we always have $$\int_M \mathcal L_X\omega=0,$$ where $$\mathcal L_X$$ is the Lie derivative?

Motivation: The integration operation has a diffeomorphism invariance, i.e., if $$F:M\to M$$ is an orientation-preserving diffeomorphism, then $$\int_M \omega = \int_M F^*\omega.$$ Now, suppose that the diffeomorphism $$F$$ is infinitesimal, i.e., $$F$$ maps each $$p\in M$$ to a point that is infinitesimally separated from $$p$$. Such infinitesimal diffeomorphism can be heuristically represented by a vector field $$X\in \mathfrak X(M)$$, which is an 'arrow' from $$p$$ to $$F(p)$$.

Furthermore, the infinitesimal version of the pullback $$F^*\omega$$ is the Lie derivative $$\mathcal L_X\omega$$. Hence, I suspect that the infinitesimal statement of the diffeomorphism invariance of integral should be $$\int_M \mathcal L_X\omega=0.$$

$$\def\intprod{\mathbin{\lrcorner}}$$ Consider the Cartan magic formula $$\mathcal{L}_X\omega = X\intprod(d\omega) + d(X\intprod \omega),$$ where $$d$$ is the exterior derivative and $$X\intprod \omega$$ is the interior product of $$X$$ and $$\omega$$.
For a top-degree form $$d\omega=0$$, so we conclude by Stokes theorem that $$\int_M \mathcal{L}_X\omega = \int_M d(X\intprod \omega) = \int_{\partial M}X\intprod \omega = 0$$ since the manifold $$M$$ has no boundary.