Kernel of an integrable holomorphic dee-bar connection is a holomorphic vector bundle

Question

I'm looking for references for the following fact, to pass on to a grad student. I think I can prove it, but it is a bit sloppy and I'd rather have something which is already written up.

Let $X$ be a complex manifold. Let $E$ be a smooth, complex vector bundle. Let $C^{\infty}(E)$ be the sheaf of smooth sections of $E$. A dee-bar connection is a $\mathbb{C}$-linear map $\nabla : C^{\infty}(E) \to C^{\infty}(E) \otimes \Omega^{0,1}$, obeying the Liebnitz rule $$\nabla(f \sigma) = f \nabla(\sigma) + \bar{\partial}f \otimes \sigma$$ Just as with connection in the $C^{\infty}$ setting, we can define the curvature; it is a section of $\Omega^{0,2} \otimes \mathrm{End}(E)$. Suppose that the curvature is zero.

Let $\mathrm{Ker}(\nabla)$ be the kernel of $\nabla : C^{\infty}(E) \to C^{\infty}(E) \otimes \Omega^{0,1}$. Let $\mathcal{O}$ be the sheaf of homolomorphic functions. It is easy to see that $\mathrm{Ker}(\nabla)$ is a $\mathcal{O}$-module. I want a reference for the fact that:

$\mathrm{Ker}(\nabla)$ is a locally free $\mathcal{O}$-module of rank $\dim E$.

The analogous fact for smooth connections is (one direction of) the Riemann-Hilbert correspondence; see, for example, Theorem 2.6 here.

Answer 1

This is Theorem 2.1.53 in Donaldson--Kronheimer's Geometry of 4-manifolds. They prove it from scratch via the contraction mapping principle but there are arguments from the Newlander--Nirenberg theorem also.

Answer 2

Kobayashi, "Differential geometry of complex vector bundles", Proposition 3.7. He does it using the Frobenius theorem.