Suppose $\Gamma$ is a smooth and closed curve in $\mathbb{R}^2$, the area is $D$. Denote point $x\in\mathbb{R}^2$ as $x=(x_1,x_2)$. $f$ is continuous on the boundary $\Gamma$, consider the following Neumann PDE: \begin{equation} \frac{\partial^2 u}{\partial x_{1}^{2}}+\frac{\partial^2 u}{\partial x_{2}^{2}}=0,\ x\in D \end{equation} \begin{equation} \frac{\partial u(x)}{\partial n}=f(x),\ x\in\Gamma \end{equation} There is another function $E(x)=\frac{1}{2\pi}\ln\sqrt{x_{1}^2+x_{2}^2}$, for $x\in D$ and a point $y\neq x$, show the following equation: \begin{equation} u(x)=\int_{\Gamma}\frac{\partial E(x-y)}{\partial n_{y}}u(y)\mathrm{d}r_{y}-\int_{\Gamma}E(x-y)f(y)\mathrm{d}r_{y} \end{equation}
My attempt was this: I recognized the right part of equation is in the form of Green's identity \begin{equation} RHS=\int_{\Gamma}\frac{\partial E(x-y)}{\partial n_{y}}u(y)\mathrm{d}r_{y}-\int_{\Gamma}E(x-y)\frac{\partial u(y)}{\partial n_{y}}\mathrm{d}r_{y}=\int_{D}(u(y)\Delta E(x-y)-E(x-y)\Delta u(y))\mathrm{d}S \end{equation} Since function $u$ is harmonic in $D$, so it only needs to show \begin{equation} u(x)=\int_{D}u(y)\Delta E(x-y)\mathrm{d}S \end{equation} But it's a bit weird that I calculated $\Delta E(x-y)$, it turns out to be zero. For more reason, I don't know how to calculate it. If there's any other problems in my attempt, your critics is welcomed.
According to the comment, the function $E(x)$ is Green's function. So it only needs to show $\Delta E(x)=\delta(x)$. For $x\neq0$, $\Delta E(x)=0$.
When $x=y$, the last integral becomes $\int_{D}u(x)\Delta E(0)\mathrm{d}S=\int_{D}u(x)\delta(x)\mathrm{d}S$, equals to LHS. Hence complete the proof.