So suppose you have an open, simply-connected, and bounded subset $D$ of $\mathbb{R}^2$ with the boundary $\partial D$. I am interested in the integral of the normal vector along the boundary, i.e., $$\int_{\partial D} \nu(y) ds(y).$$ Suppose the boundary $\partial D$ has the parameterization $x(t)$. I constrain myself to the instances when $x(t)$ is piece-wise differentiable, i.e., the boundary is allowed to have corners.
Am correct in thinking that my integral makes sense even when the boundary has corners because the integral is just an improper integral at that condition? Or is there more to it (like the corners can't extend off to infinity and so on)? I am struggling with the language that this particular question/answer should use.
If you have a parametrization $x:[a,b]\to \partial D$ that is piecewise differentiable with bounded derivative, no improper integrals get involved. Once you partition the interval $[a,b]$ into subintervals $[a_i,b_i]$ on which $x$ is differentiable, you can write $$\int_{\partial D} \nu(y)\,ds(y) = \sum_i \int_{a_i}^{b_i} \nu(x(t))|x'(t)|\,dt$$ If $x'$ becomes unbounded near the corners, then you indeed have an improper integral there. But, at least in theory, any curve of finite length can be parametrized by arclength, which makes $|x'(t)|\equiv 1$. (Whether it's practical to find such a parametrization is another story.)