Normal derivative of the autonomous single layer potential

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Let $G$ be the fundamental solution of the Laplace equation. Let $\Omega$ be an open and bounded subset of $\mathbb{R}^n$ regular enough. It is known that:

\begin{equation} \int_{\partial \Omega}\frac{\partial G(x-y)}{\partial \nu(y)}d\sigma(y)= \begin{cases} -1 \mbox{ if } x\in \Omega,\\ -1/2 \mbox{ if } x\in \partial\Omega,\\ 0 \mbox{ if } x\in \mathbb{R}^n\setminus\bar\Omega.\\ \end{cases} \end{equation} Where $\nu(y)$ is the unit outer normal to $\partial \Omega$ in $y$. In other words this is the jump formula for the double layer potential when the density is constant.

How about the value of the normal derivative of the single layer potential? Let $x_0\in\partial\Omega$. I want to compute: \begin{equation} \int_{\partial \Omega}\frac{\partial G(x-y)}{\partial \nu(x_0)}d\sigma(y) \end{equation}

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OK, here is the formula (slightly different notation than yours): Let $$ \mathcal{Sl}f(x)=\int_{\partial\Omega}f(y)E(x,y)\,dS(y). $$ Then $$ \frac{\partial}{\partial \nu}\mathcal{Sl}f_{\pm}(x)=\frac{1}{2}\bigl(\mp f+N^{\#}f\bigr), $$ where $$ N^{\#}f(x)=2\int_{\partial\Omega}f(y)\frac{\partial E}{\partial \nu_x}(x,y)\,dS(y). $$