A simple-layer potential is defined as
$$\Psi(M)=\iint_{S}\dfrac{\sigma(N)}{R(M,N)}dS(N)$$
where $S$ denotes a flat region in the plane $z=0$; the coordinates of $M$ and $N$ are $(\rho,\phi,z)$ and $(\rho_{0},\phi_{0},0)$,respectively. Suppose $\sigma(N)$ is sufficiently smooth on $S$.
\begin{equation} R(M,N)=\sqrt{\rho^{2}+\rho_{0}^{2}-2\rho\rho_{0}\cos(\phi-\phi_{0})+z^2} \end{equation}
$(1)$ Prove that for $M\in S$
$$\dfrac{\partial\Psi(M)}{\partial z}\Bigg|_{z\rightarrow0^{+}}=-2\pi\sigma(M)$$
This is also a well-known property simple layer
proof:
When $M \in S$, the integrand will be singular at $M$.
$$\Psi(M)=\iint_{S/B_{\epsilon}(M)}\dfrac{\sigma(N)}{R(M,N)}dS(N)+\dfrac{1}{\epsilon}\iint_{B_{\epsilon}(M)}\dfrac{\sigma(N)}{R(M,N)}dS(N)$$
When taking derivative with respect to $z$ and then putting $z\rightarrow 0$, the 1st term will be vanished. For the 2nd, $\dfrac{\partial}{\partial z}$ is equivalent to $\dfrac{\partial}{\partial \epsilon}$. So we have
$$\dfrac{1}{\epsilon}\iint_{B_{\epsilon}(M)}\dfrac{\sigma(N)}{R(M,N)}dS(N)=-\dfrac{1}{\epsilon^2}\sigma(\xi)(2\pi \epsilon^2)$$
Letting $\epsilon\rightarrow 0$, we'll finish the proof.
$(2)$ Similar with the above proof, it seems that $\dfrac{\partial^{2}\Psi(M)}{\partial z^{2}}\Bigg|_{z\rightarrow0^{+}}\rightarrow+\infty$ for $M \in S$.
*But the textbook gives an exmaple where $\dfrac{\partial^{2}\Psi(M)}{\partial z^{2}}\Bigg|_{z\rightarrow0^{+}}$ is finite for $M \in S$. Why?? I really get confused by this contradiction. This is my question. Waiting for your help. By the way, if possible, would like to recommend me some references on this question.
Thanks in advance!!*