If $\frac{\partial u}{\partial \overline{z}}=f$ and $|u(z)|$ goes to $0$ as $z\to \infty$, then $u$ is of this form. The Problem:
Suppose $f$ is $C^{\infty}$ and of compact support. Let $u$ also be $C^{\infty}$ such that $\frac{\partial u}{\partial \overline{z}}=f$. Then show that $$u(w)=\frac{1}{2\pi i}\iint_{\mathbb{C}}\frac{f(z)}{z-w} dz\wedge d \overline{z}$$ for all $w\in \mathbb{C}$. 
My thoughts:
I believe the approach is to use a variant of Cauchy's Integral Formula as stated in Chapter 5, Section 2, Theorem 3 of Narasimhan's Complex Analysis textbook, which states the following:
Let $\Omega$ be open in $\mathbb{C}$, $K$ a compact subset of $\Omega$. Let $\alpha\in C_{0}^{\infty}(\Omega)$ such that $\alpha=1$ on a neighborhood of $K$. Then for any $f$ holomorphic on $\Omega$, we have $$f(z)=-\frac{1}{\pi}\iint_{\Omega}\frac{\partial \alpha}{\partial \overline{\zeta}}\cdot f(\zeta)\cdot \frac{1}{\zeta-z}d\xi d\eta,$$ where $\zeta=\xi+i\eta$. 
My issue is that I am not sure what our $K$ should be, nor do I see why our $u$ would actually be $1$ on a neighborhood. I am also not sure how one would rephrase this fact to yield the result we want.
 A: I don't have Narasimhan's book near me atm but this result is typically achieved using Stokes theorem. 
Note that \begin{align} \frac{1}{2\pi i} \int_{\mathbb{C}\setminus B_\epsilon(w)} \frac{1}{z-w}\frac{\partial f(z)}{\partial \overline{z}}\,dz \wedge \,d\overline{z} &= \frac{1}{2\pi i}\int_{\mathbb{C}\setminus B_{\epsilon}(0)}\frac{1}{z}\frac{\partial f(z+w)}{\partial \overline{z}}\,dz \wedge \,d\overline{z} \tag{1}\\&= \frac{1}{2\pi i} \int_{\partial B_{\epsilon}(0)} \frac{f(z+w)}{z}\,dz \tag{2}\\&= \frac{1}{2\pi}\int_{0}^{2\pi} f(w+\epsilon e^{i\theta})\,d\theta \underset{\epsilon \to 0^+}{\longrightarrow} f(w). \tag{3}\end{align} where, in line $(1)$ we used the change of variable $z \mapsto z + w$, in line $(2)$ we used the Stokes-Theorem ($f$ has compact support) and in line $(3)$ we made the change of variable $z \mapsto \epsilon e^{i\theta}$.
Now, by DCT we may interchange derivative and limit to write \begin{align*} \frac{\partial u}{\partial \overline{z}}(w) &= \frac{1}{2\pi i} \frac{\partial}{\partial \overline{z}} \lim\limits_{\epsilon \to 0^+} \int_{\mathbb{C}\setminus B_\epsilon(w)} \frac{f(z)}{z-w}\,dz \wedge \,d\overline{z} \\&= \frac{1}{2\pi i} \lim\limits_{\epsilon \to 0^+} \int_{\mathbb{C}\setminus B_\epsilon(w)} \frac{1}{z-w}\frac{\partial f(z)}{\partial \overline{z}}\,dz \wedge \,d\overline{z} = f(w). \end{align*}
To see that $u$ given above is the unique solution under the given condition $|u(z)| \to 0$ as $|z| \to \infty$ note that for any other solution $u'$ satisfying $\displaystyle \frac{\partial u'}{\partial \overline{z}} = f$ we have $$\frac{\partial }{\partial \overline{z}}(u' - u) = 0$$
i.e., $(u' - u)$ is is a bounded entire function vanishing at infinity. Thus, by Liouville's theorem $u' - u \equiv 0$.
