For $x = (x_1, x_2) \in \mathbb{R}^2$, let
$$ E(x) = \frac{1}{2\pi} \log(|x|), $$
where $|x| = \sqrt{x_1^2 + x_2^2}$. In fact $E$ is the fundamental solution for the Laplace equation.
Let $\Gamma$ be a simple curve that is closed and smooth, and $\mu$ a continuous function defined on $\Gamma$. Then, define
$$ u(x)=\int_{\Gamma} \frac{\partial E}{\partial n_y}(x-y) \mu(y) d \sigma_y, \quad x \in \mathbb{R}^2 \backslash \Gamma. $$
It could be verified $u$ is harmonic in its domain.
The question is, for $x_0 \in \Gamma$ and $\overrightarrow{n_0}$ the exterior normal at $x_0$ of $\Gamma$, find the limit
$$ \lim_{h\to +0} u(x_0 + h \overrightarrow{n_0}) - u(x_0 - h \overrightarrow{n_0}). $$
I find $u$ looks like a part in Green's representation formula, and tried to find the desired limit using the techniques in the proof of Green's representation formula, but failed. In fact, $\mu$ in the problem is only defined on $\Gamma$, not meeting the conditions of Green's 1st and 2nd formulas. Thanks for your help!
Now I somehow get an answer (while I'm not sure if it is correct).
Deivide the curve $\Gamma$ into 2 parts, and handle the integration respectively. For a $\delta$ that small enough, and $B_\delta (x_0)$ the closed ball centred at $x_0$ with redius $\delta$, we write
$$ \int_\Gamma \left\{\frac{\partial}{\partial n_y} E(x+hn_0-y) - \frac{\partial}{\partial n_y} E(x+hn_0-y)\right\} \mu(y) \mathrm{d} \sigma_y\\ = \int_{\Gamma-B_{\delta}(x_0)} + \int_{\Gamma\cap B_{\delta}(x_0)} \left\{\frac{\partial}{\partial n_y} E(x+hn_0-y) - \frac{\partial}{\partial n_y} E(x+hn_0-y)\right\} \mu(y) \mathrm{d} \sigma_y. $$
For the first part, $\int_{\Gamma-B_{\delta}(x_0)}$ , since $\nabla_y E(x_0-y)$ is (uniformly) continuous as long as $y$ is away from $x_0$, we know
$$ \int_{\Gamma-B_{\delta}(x_0)} \to 0 $$
for $h$ small enough due to the continuity.
For the second part, for $h>0$, thus $x_0+hn_0-y \neq 0$ when $y$ is in $\Gamma\cap B_\delta(x_0)$, the partial derivatives of $E(x_0+hn_0-y)$ are defined, then we have
$$ \frac{\partial}{\partial n_y} E(x+hn_0-y) = \frac{y-x_0-hn_0}{|x_0+hn_0-y|}\cdot n_y = \frac{(y-x_0)\cdot n_y-hn_0 \cdot n_y}{|x_0+hn_0-y|}. $$
It's the same for the second term in the intergrand. Thus we rewrite the integration with
$$ \color{blue}{\int_{\Gamma\cap B_\delta(x_0)}\mu(y)\left\{\frac{1}{|x_0-y+hn_0|}-\frac{1}{|x_0-y-hn_0|}\right\}(y-x_0)\cdot n_y \mathrm{d}\sigma_y} \\ - \color{orange}{\int_{\Gamma\cap B_\delta(x_0)}\mu(y)h\left\{\frac{1}{|x_0-y+hn_0|}+\frac{1}{|x_0-y-hn_0|}\right\}n_y \cdot n_0 \mathrm{d}\sigma_y}. $$
The blue part should go to $0$ as $h\to 0$, and the orange part could be estimated with
$$ \int_{\Gamma\cap B_\delta(x_0)}\mu(x_0) h \frac{2}{h} \mathrm{d}\sigma_y\leq 2 C \delta \mu(x_0)\to 0 $$
by letting $\delta\to 0$, where $C$ is a constant that is determined by the lenghth of $\Gamma$ that contained in $B_\delta(x_0)$. [ However, the estimations I made here is more of intuition than of being rigorous ]
Thus we conclude that
$$ \lim_{h\to 0}u(x_0+hn_0-y)-u(x_0-hn_0-y) = 0, $$
and, in addtion to being continuous in $\mathbb{R}^2-\Gamma$, $u$ is also continuous on $\Gamma$ in some sense.