Integration by parts and Levi form 
Let $\Omega $ be an open set in $\mathbb{C}^n$, $u$ be a smooth function on $\Omega$ and let $\Phi$ be a non negative test function on $\Omega$. Prove that for all $b=(b_1,..., b_n)\in \mathbb{C^n}$ $\int_\Omega u(z)\sum_{j,k}^n\frac{\partial^2\Phi}{\partial z_j\partial \bar{z_k}}b_j\bar{b_k} d\lambda(z)=\int_\Omega \Phi(z)\sum_{j,k}^n\frac{\partial^2 u}{\partial z_j\partial \bar{z_k}}b_j\bar{b_k} d\lambda(z)$. 

$\sum_{j,k}^n\frac{\partial^2\Phi}{\partial z_j\partial \bar{z_k}}b_j\bar{b_k}$ is called the Levi form of $\Phi$. As $\Phi\in C_0^\infty$, it is zero on $\partial\Omega$. What I wanted to prove is $\int_\Omega u(z)\frac{\partial^2\Phi}{\partial z_j\partial \bar{z_k}} d\lambda(z)=\int_\Omega \Phi(z)\frac{\partial^2u}{\partial z_j\partial \bar{z_k}} d\lambda(z)$. But how? Any help is appreciated.
 A: I guess you need the stoke theorem and product rule for differential forms: if $\alpha$ and $\beta$ are differential forms, then
$$
d(\alpha\wedge \beta)= d\alpha\wedge \beta+(-1)^{|\alpha|}\alpha\wedge d\beta.
$$
You want to use it twice. 
Write $\alpha=\overline{\partial}\Phi, \beta=u \omega$, where $\omega$ is a $d$-closed $(n-1,n-1)$-form so that $u\partial\overline{\partial}\Phi\wedge \omega$ is the top form that you want to integrate. We want to show that
$$
\int_\Omega u\partial\overline{\partial}\Phi\wedge \omega
= \int_\Omega \Phi\partial\overline{\partial}u\wedge \omega.
$$
Firstly, we have
$$
\int_\Omega u\partial\overline{\partial}\Phi\wedge \omega
=\int_\Omega d\overline{\partial}\Phi\wedge(u \omega) 
=\int_{\partial\Omega} \overline{\partial}\Phi\wedge (u\omega)+\int_\Omega\overline{\partial}\Phi\wedge d(u\omega)
=\int_\Omega\overline{\partial}\Phi\wedge d(u\omega)
$$
Note that $d(u\omega)=du\wedge \omega$, the RHS in above is equal to
$$
\int_\Omega\overline{\partial}\Phi\wedge du\wedge\omega
=\int_\Omega\overline{\partial}\Phi\wedge \partial u\wedge\omega
=\int_\Omega d\Phi\wedge \partial u\wedge\omega
$$
This time take $\alpha=\Phi,\beta=\partial u\wedge \omega$. Then
$$
\int_\Omega d\Phi\wedge \partial u\wedge\omega
=\int_{\partial\Omega}\Phi\wedge \partial u\wedge \omega-\int_\Omega \Phi\wedge d(\partial u\wedge \omega)
=-\int_\Omega \Phi\wedge d\partial u \wedge \omega,
$$
where we have used $d\omega=0$. So finally
$$
-\int_\Omega \Phi\wedge d\partial u \wedge \omega
=-\int_\Omega \Phi\wedge \overline{\partial}\partial u \wedge \omega
=\int_\Omega \Phi\wedge \partial\overline{\partial} u \wedge \omega.
$$
