Top Dolbeault cohomology group vs top De Rham cohomology group of a compact complex manifold

Question

I cannot reconcile two simple facts for a compact complex manifold $X$ with $dim_{\mathbb{C}}=n$. Let $A^{p,q}(X)$ be differential forms of degree $(p,q)$ on $X$.

One the one hand, $$H^{2n}_{DR} (X)= A^{n,n}(X)/d\left(A^{n-1,n}(X) + A^{n,n-1}(X)\right) = A^{n,n}(X)/\left(\partial(A^{n-1,n}(X)) + \bar{\partial}(A^{n,n-1})\right)$$ and this group is isomorphic to $\mathbb{C}$ via integration of forms and Stokes' theorem.

On the other hand, $$H^n(X,K_X)\stackrel{Dolbeault}{====} A^{n,n}(X)/ \bar{\partial}(A^{n,n-1}(X))$$ is also said to be isomorphic to $\mathbb{C}$ via integration and before Serre's duality is established (p. 135 of Voisin's "Hodge Theory and Complex Algebraic Geometry I"; in fact, the above fact is used to prove Serre's duality in the book.)

Without the $\partial\bar{\partial}$-lemma (that requires $X$ to be Kahler) I don't see how to show that $$\partial \omega=d \omega, \quad \omega \in A^{n-1, n}(X)$$ is zero in Dolbeault cohomology (since it clearly integrates to zero by Stokes' theorem). And yet it should be true by Serre's duality that does not require the Kahler condition...

Answer 1

One can look at the dual statement: since $H^{2n}(X)\cong H^0(X)$ by Poincaré duality and $H^{n,n}(X)\cong H^{0,0}(X)$ by Hodge star. Then the fact that $H^0(X)\cong H^{0,0}(X)$ is reflected by $df=0\Leftrightarrow \bar\partial f=0$ on a compact complex manifold. However, I do not know how to explicitly show $d$-exact is equivalent to $\bar\partial$-exact in top forms.