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.

• Assuming there is, indeed, a global holomorphic section of the vector bundle, your form is, up to a scalar, is $\partial \overline{\partial}(h(\xi,\xi)\Phi^{n-1})$ because $\Phi$ is closed. Let $u$ be the form inside: since $\overline{\partial}^2u=0$, the form is, up to a scalar, $d(\overline{\partial}u)$, so is exact so has integral zero. At least I hope it’s right. Commented Jun 14, 2020 at 21:40

\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.