Let $M$ be a compact smooth manifold, oriented, without boundary. Let $u$ be a smooth $\mathbb{R}$-valued function on $M$. I'm working through a computation which asserts that $$\int_M \Delta e^{-u} dV =0.$$ I can't seem to justify this, however. I think it's the divergence theorem but cannot argue it rigorously.
This is undoubtedly a simple problem, but it escapes me at the moment.
Yes, this is the divergence theorem, which may be applied by compactness of $M$. If $X$ is any vector field, then $({\rm div}\,X){\rm d}V = {\rm d}(\iota_X({\rm d}V))$ and the fact that $\partial M=\varnothing$ allows us to use Stokes' theorem to conclude that $$\int_M ({\rm div}\,X)\,{\rm d}V = 0.$$In particular, if $f\colon M \to \Bbb R$ is any smooth function, we have $$\int_M \triangle f\,{\rm d}V = \int_M {\rm div}(\nabla f)\,{\rm d}V = 0,$$by the above. For you, $f = e^{-u}$.