Why is $\int_M i \partial\bar \partial u \wedge \Phi^{n-1}$ zero? I was studying vanishing theorems on holomorphic sections of holomorphic Hermitian vector bundles on Kähler manifolds. Here $(E,h)$ is a Hermitian holomorphic bundle on a compact Kähler manifold $M$. The book claims that the following integral is $0$.
$$
\int_{M}i\partial \bar{\partial}h(\xi,\xi) \wedge \Phi^{n-1} = 0. 
$$
Here $\Phi$ is the Kähler form on the manifold $M$ and $h$ is the Hermitian metric on $E$ and $\xi$ is a global holomorphic section of $E$.
I'm not able to see why this integral should be $0$.
I'll be happy to provide more details.

As suggested in the comment, one can write
\begin{align*}
\partial \bar{\partial}h(\xi,\xi) \wedge \Phi^{n-1} &= d \bar{\partial}h(\xi,\xi) \wedge \Phi^{n-1} \\
&= d(\bar{\partial}h(\xi,\xi) \wedge \Phi^{n-1})
\end{align*}
Since $\partial^2 = 0$ and $\Phi$ is closed. Thus the integrand is exact and Stokes theorem implies that it's zero.
