Integration by Parts with Laplacian over a Manifold I am studying a text on differential geometry where it states during a proof that integration by parts is used to prove the following integral over a closed manifold:
$$
\begin{split}
 &\int_M -\Delta u \log u \: dV=  \int_M \frac{|\nabla u|^2}{u} \: dV\\
\end{split}
$$
I am not sure what detail I am missing here, because if we apply the formula we get a mess as the integral of the log gives a messy term which does not take us closer to the result.  If I try swapping the order of the product and then doing the integration by parts, I just get $0$ since the integral of the Laplacian of a function over a closed manifold vanishes. 
 A: The following general identity holds on any Riemannian manifold, where $f$ and $X$ are a differentiable function and vector field, respectively:
$\nabla \cdot (fX) = \nabla f \cdot X + f \nabla \cdot X; \tag 1$
this may be easily verified by working in a local coordinate system; if we set
$f = \ln u \tag 2$
and
$X = \nabla u \tag 3$
then via (1) we have
$\nabla \cdot (\ln u \nabla u) = \nabla \ln u \cdot \nabla u + \ln u \nabla^2 u; \tag 4$
we have
$\nabla \ln u = \dfrac{\nabla u}{u}, \tag 5$
whence
$\nabla \ln u \cdot \nabla u = \dfrac{\nabla u \cdot \nabla u}{u} = \dfrac{\vert \nabla u \vert^2}{u}; \tag 6$
combining (4) and (6) yields
$\nabla \cdot (\ln u \nabla u) = \dfrac{\vert \nabla u \vert^2}{u} + \ln u \nabla^2 u; \tag 7$
if $M$ has a boundary $\partial M$ then by the divergence theorem
$\displaystyle \int_{\partial M} \ln u \nabla u \; dS =  \int_M \nabla \cdot (\ln u \nabla u) \; dV$
$= \displaystyle \int_M \left (\dfrac{\vert \nabla u \vert^2}{u} + \ln u \nabla^2 u \right ) \; dV = \int_M \dfrac{\vert \nabla u \vert^2}{u} \; dV + \int_M \ln u \nabla^2 u\; dV, \tag 8$
where of course we make the necessary assumptions on $u$ and $M$ so that these integrals are all well-defined.  In the event that
$\partial M = \emptyset, \tag 9$
then (8) reduces to
$\displaystyle \int_M \dfrac{\vert \nabla u \vert^2}{u} \; dV = - \int_M \ln u \nabla^2u \; dV, \tag{10}$
$OE\Delta$.
