The Proposition somewhat goes like this:
$$\text{Let $\Omega$ be a bounded region in $\mathbb{C}$, with smooth oriented boundary $\partial\Omega$. If $u \in C^1(\bar{\Omega})$, then we have $u(a)=\dfrac{1}{2\pi i}\int_{\partial \Omega}\dfrac{u(\lambda)d\lambda}{\lambda-a}-\dfrac{1}{2\pi i}\int_{\Omega}\dfrac{\bar{D}u(\lambda)d\bar{\lambda}\wedge d\lambda}{\lambda-a} $}$$
Proof : Let $D_{\epsilon}=\{\lambda: |\lambda-a| \le \epsilon\}$, choose $\epsilon$ so small that $D_{\epsilon} \subset \Omega$, put $\Omega_{\epsilon}=\Omega-D_{\epsilon}$ and we apply Stoke's Theorem to $\beta=\dfrac{u(\lambda)d\lambda}{\lambda-a}$ in $\Omega_{\epsilon}$. Since $d\lambda$ occurs in $\beta$ and since $(\lambda-a)^{-1}$ is holomorphic in $\Omega_{\epsilon}$, we have $d\beta=\bar{\partial}\beta=(\lambda-a)^{-1}(\bar{D}u)(\lambda)d\bar{\lambda}\wedge d\lambda$ so that $$\int_{\partial \Omega}\beta-\int_{\partial D_{\epsilon}} \beta=\int_{\Omega_{\epsilon}}\frac{(\bar{D}u)(\lambda)}{\lambda-a}d\bar{\lambda}\wedge d\lambda$$
I expect that the result holds when $\epsilon$ goes to zero. Because If I look at $$\int_{\partial D_{\epsilon}}\frac{u(\lambda)d\lambda}{\lambda-a}=i\int_{0}^{2\pi}u(a+\epsilon e^{i\theta})d\theta$$, and by taking $\epsilon$ to zero, the function is inside the integral is finite and hence I can take the limit inside to get $i2\pi u(a)$
The only thing that remains to be shown is when $\epsilon$ is taken to $0$, the region $\Omega_{\epsilon}$ becomes the whole of $\Omega$.
Thanks for the help!!
Since $\bar Du$ is bounded (since $u \in C^1$) and $1/\lambda$ is $L^1_{loc}(d\bar\lambda \wedge d\lambda)$ (probably easiest to see via polar coordinates), $$ \lim_{\varepsilon \to 0} \int_{D_\varepsilon} \frac{\bar Du(\lambda)}{\lambda-a}\,d\bar\lambda \wedge d\lambda = 0. $$ (And of course $\int_\Omega = \int_{\Omega_\varepsilon} + \int_{D_\varepsilon}$.)