 Solution

The integral identity requires compactness of $M$, compact support of $X$, or suitable conditions eliminating flux at infinity. The printed question does not state such a hypothesis. We prove the Riemannian divergence theorem for a compact manifold, and more generally for a compactly supported smooth vector field on an oriented Riemannian manifold with boundary.

Write $j:\partial M\hookrightarrow M$ for the inclusion. Give $\partial M$ the outward-normal-first boundary orientation, and let $N$ be the outward unit normal vector. The Riemannian volume form of the induced Riemannian metric isIndeed, if $(e_1,\dots ,e_{n-1})$ is a positively oriented orthonormal basis tangent to the boundary, $(N,e_1,\dots ,e_{n-1})$ is a positively oriented orthonormal basis in $M$. Both sides therefore evaluate to one on the boundary basis.

At a boundary point decompose $X=g(X,N)N+X^{\top }$, where $X^{\top }$ is tangent to $\partial M$. Evaluating on $n-1$ tangent vectors, the term with $X^{\top }$ vanishes: all $n$ arguments of the alternating volume form then lie in the $(n-1)$-dimensional boundary tangent space. Thus the interior product of a differential form satisfiesThe general Stokes theorem for an oriented manifold with boundary states ${\int }_{M}d\eta ={\int }_{\partial M}{j}^{\ast }\eta$ for any smooth differential form $\eta$ of degree $n-1$ with compact support; compactness of $M$ makes the support condition automatic. Applying it to $\eta ={\iota }_X{\omega }_g$ and using the defining equation for the divergence of a Riemannian vector field proves

For a concrete failure without the extra hypothesis, take $M=[0,\infty )$ with Riemannian metric $d{x}^2$ and $X=(\arctan x){\partial }_x$. The divergence of a Riemannian vector field is $\mathrm{div}X=(1+x^2{)}^{-1}$, sobecause $X$ vanishes at the only boundary point. Both integrals are finite. The missing $\pi /2$ is the limiting flux at infinity, so mere integrability of the divergence cannot repair the unrestricted assertion.