Let $f$ be a smooth function such on a compact kahler manifold $(M, w)$, and the component of $w$ is denoted by $g_{ij}$, assume there is a constant $s$ such that $sf = -g^{ij}\sqrt{-1}\partial_{j}\bar\partial_{i}f$, we have:
$s \nabla_j f = -\nabla_j \nabla_p \nabla_{\bar p} f$ suppresing the mertic.
Then how do you prove the following Bochner-Weitzenbock formula:
$(\int_{U}-\nabla_j \nabla_p \nabla_{\bar p} f \nabla_\bar j f)w^n = (\int_{U} -\nabla_{\bar p} \nabla_p \nabla_{ j} f \nabla_{\bar j} f + R^{q \bar j} \nabla_{q} f \nabla_{\bar j}f )w^n$
Thoughts: the left hand side is simply $\partial_j sf = \partial_j-g^{ik}\sqrt{-1}\partial_{j}\bar\partial_{k}f = - \partial_j g^{ik} \sqrt{-1}\partial_{j}\bar\partial_{k}f -\partial_j\sqrt{-1}\partial_{j}\bar\partial_{k}f g^{ik}$
How do you proceed?
Update: My definition for the Ricci curvature is the following:
Lowering the index of the curvature tensor we have $R_{i \bar j k \bar l} = -\partial_{k} \partial_{l} g_{i \bar j} + g^{p \bar q}(\partial_k g_{i \bar q})(\partial_{\bar l} g_{p \bar j})$, then let $R_{i\bar j} = g^{k \bar l } R_{i \bar j k \bar l}$.
It is not obvious how to go from this definition.
Your identity is only using the Ricci identity; that is, the formula for how to commute derivatives in a Kähler manifold. It requires neither integration nor the assumption $-\Delta f = sf$.
The derivation of your identity requires three observations. First, since the connection is torsion-free, it holds that $\nabla_{\bar p}\nabla_q u = \nabla_q\nabla_{\bar p}u$ for all functions $u$. Second, since $g$ is Kähler, it holds that $$ \nabla_j\nabla_p\nabla_{\bar q}f = \nabla_p\nabla_j\nabla_{\bar q}f . $$ Third, by definition of the Ricci curvature, $$ \nabla_p\nabla_{\bar p}\nabla_j f = \nabla_{\bar p}\nabla_p\nabla_j f - R_{j\bar p}\nabla_p f ; $$ here I use Einstein summation notation as in the problem statement. Putting all of this together, we compute that $$ -\nabla_j\nabla_p\nabla_{\bar p} f = -\nabla_p\nabla_j\nabla_{\bar p}f = -\nabla_p\nabla_{\bar p}\nabla_j f = -\nabla_{\bar p}\nabla_p\nabla_j f + R_{j\bar p}\nabla_p f . $$ Multiplying by $\nabla_{\bar j}f$ and integrating yields the display in question. You can find proofs of these facts in Besse's book "Einstein manifolds", for example.