TITLE: Integral $(1, 1)$ forms and holomorphic line bundles
QUESTION [12 upvotes]: Let $X$ be a complex manifold.
We say that a cohomology class in $H^2(X,\mathbb{C})$ is integral if it lies in the image of the natural morphism $j : H^2(X,\mathbb{Z}) \longrightarrow H^2(X,\mathbb{C})$.
Define the first Chern class of a holomorphic line bundle $L \in \text{Pic}(X)$ on $X$ as the image of $L$ under the boundary map $c_1: \text{Pic}(X) \cong H^1(X,\mathcal{O}^*_X) \longrightarrow H^2(X,\mathbb{Z})$.
Question: Given a $d$-closed differential form $\omega$ of type $(1,1)$ on $X$ with integral class $[\omega] \in H^2(X,\mathbb{C})$, does it follow automatically that there exists a holomorphic line bundle $L \in \text{Pic}(X)$ such that the image of $c_1(L)$ under $j$ is equal to $[\omega]$, up to some $2\pi i$ factor or so? 
I would like an answer without assuming that $X$ is compact Kähler, so that Lefschetz's theorem on $(1,1)$ classes does not apply.
Also, in case the answer is positive, can anyone point out a reference?

REPLY [3 votes]: We will use the following theorem which is stated in D. Huybrechts's Complex Geometry (Theorem 2.6.26) (A proof which uses the Newlander-Nirenberg theorem can be found here):

Theorem: Let $E$ be a complex vector bundle on a complex manifold $X$. A holomorphic structure is uniquely determined by a $\mathbb C$-linear operator $\bar \partial_E : \mathscr A^0 (E) \to \mathscr A^{0,1} (E)$ which satisfies the Leibniz's rule and $\bar \partial_E^2 = 0$. 

Now let $\omega$ be a $(1,1)$-form with integral class $[w]$. The second condition imply that there is a smooth complex line bundle $L$ so that $c_1(L) = [w]$. Let $D$ be any complex connection on $L$. Then $[D^2] = [w]$ and there is a one form $a$ so that $D^2 + da = \omega$. Now let $\bar\partial_L$ be the projection of $D+a$ to $\mathscr A^{0,1}(L)$. Then $\bar\partial^2_L = 0$ as $(D+a)^2 = D^2 + da = \omega$ has no $(0,2)$ part. Thus $L$ has a holomorphic structure defined by $\bar\partial_L$.