> [!theorem] > > Let $X$ be a [[Riemannian Metric|Riemannian]] [[n-Manifold|n-manifold]], $h \in C^1$, $f \in C^2$ such that $h\grad{f}$ has [[Compactly Supported|compact support]], and $N$ be the outward normal [[Vector Field|vector field]] on $\partial X$, then > $ > \int_X h\Delta f + \angles{\grad{h}, \grad{f}}dV = \int_{\partial X}h\angles{N, f}dA > $ > and if both functions are compactly supported, then > $ > \int_X h \Delta f - f \Delta h dV = \int_{\partial X} h\angles{N, f} - f\angles{N, h}dA > $ > > *Proof*. Firstly, by the product rule, > $ > \div{h\grad{f}} = h\Delta f + \angles{\grad{h}, \grad{f}} > $ > By the [[Divergence Theorem|divergence theorem]], > $ > \int_X\div{h\grad{f}}dV = \int_{\partial X} \angles{hN, f}dA = \int_{\partial X} h\angles{N, f}dA > $ > Similarly, since > $ > \begin{align*} > \div{h\grad{f} - f\grad{h}} &= h\Delta f - f\Delta h + \angles{\grad{h}, \grad{f}} \\ > &- \angles{\grad{f}, \grad{h}} \\ > &= h\Delta f - f \Delta h > \end{align*} > $ > and $\angles{N, f} = \angles{N, \grad{f}}$. Applying the divergence theorem again yields the desired result.