I have a simple question regarding complex geometry: **is there an analog for the Stokes Theorem for the Dolbeault Operator $\bar{\partial}$?** For instance, suppose that $M$ is a closed complex manifold and I am looking for some identity like


$$\int\_M \bar{\partial}(\cdots)=\cdots$$


For it to make sense, I suppose that $(\cdots)$ should be a $(0,n-1)$-form, and that we have a complex volume form... But anyway, which is the best identity one may get?


In case there is no such analog, could someone find a 'deep' reason ensuring this non-existence (a very crucial difference...)?


**EDIT** (just for enlarging the scope of the question)


Is there a possibility that something may be said for $(0,n-1)$-forms provided that there is some complex (or 'anticomplex') volume form, that is, a $(0,n)$-form which is locally $\theta=dz\_1\wedge\cdots\wedge dz\_n$ coming from (say) a $SU(n)$ structure? I know that this point of view is far away of the initial scope of the question, but... it is a soft question indeed.



Stokes-like Theorem for Dolbeault Operator
If $M$ is an $n$-dimensional complex manifold, then $M$ is a $2n$-dimensional smooth manifold, so you should integrate a $2n$-form (note that $\bar{\partial}$ of a $(0, n-1)$-form is an $n$-form). 


Consider the expression $\int\_M\bar{\partial}\alpha$ where $\alpha$ is a complex $(2n-1)$-form. As $\mathcal{E}^{2n-1}(X, \mathbb{C}) = \mathcal{E}^{n,n-1}(X)\oplus\mathcal{E}^{n-1,n}(X)$, we can write $\alpha = \alpha^{n,n-1} + \alpha^{n-1,n}$ where the superscripts denote the bidegree. Note that 


$$\bar{\partial}\alpha = \bar{\partial}\alpha^{n,n-1} + \bar{\partial}\alpha^{n-1, n} = \bar{\partial}\alpha^{n, n-1}.$$ 


As $\partial\alpha^{n,n-1} = 0$, $\bar{\partial}\alpha^{n,n-1} = d\alpha^{n,n-1}$ so 


$$\int\_M\bar{\partial}\alpha = \int\_M d\alpha^{n,n-1} = \int\_{\partial M}\alpha^{n,n-1} = 0.$$


