Proving that $\lim_{j \to i} Z_{ij} = [\ln(\frac{\Delta s_i}{2})-1]\Delta s_i$

If I have the following integral equation $ $ \phi(\vec{x})=\frac{1}{\pi}\int [\phi\frac{\partial (\ln r)}{\partial n} -\ln(r) \frac{\partial \phi}{\partial n}] ds$ $

An approximate solution of $ \phi$ is obtained numerically by dividing the boundary into a finite number of segments ,N.

So we can write $ $ \phi(\vec x_j)=\sum_{i=1}^{N} [\phi(\vec x_i)\frac{\partial \ln(r_{ij})}{\partial n} -\ln(r_{ij})\frac{\partial \phi}{\partial n}(\vec x_i)]\Delta s_i $ $ Where $ \Delta s_j$ represents the boundary segment length and $ r_{ij}$ is the distance between the $ i^{th}$ and the $ j^{th}$ segment

So It’s easy to write $ $ \frac{\partial \ln(r_{ij})}{\partial n}=\frac{1}{r_{ij}}[-\frac{(x_i -x_j)}{r_{ij}}(\frac{\Delta y}{\Delta x})_j +\frac{(y_i -y_j)}{r_{ij}}(\frac{\Delta x}{\Delta y})_j ]\Delta s_j$ $ And $ $ Z_{ij}=[\ln(r_{ij})]\Delta s_j$ $

Prove that $ $ \lim_{j \to i} \frac{\partial \ln(r_{ij})}{\partial n}=[\frac{(-x_{ss} y_s + x_s y_{ss})_i}{2}]\Delta s_i$ $

My try $ $ \lim_{j \to i} \frac{\partial \ln(r_{ij})}{\partial n}=\lim_{h\to 0}\frac{-(x_i -x_{i+h})(y’_{i+h})+(y_i – y_{i+h})(x’_{i+h})}{(x_i -x_{i+h})^2 +(y_i -y_{i+h})^2}$ $

Using $ $ x_{i+h}=x_i +h x’_i +(h^2/2) x”_i$ $ and $ $ x’_{i+h}=x’_i +hx”_i$ $ Hence we get the required result

My question How to prove that $ $ \lim_{j \to i} Z_{ij} = [\ln(\frac{\Delta s_i}{2})-1]\Delta s_i$ $

Thanks in advance .