Rectangle rule to compute hypersingular integral on a Circle

Abstract

Consider the hypersingular integral on circle
$$I(c,s,f)=\int_{c}^{c+2\pi} \hskip-.34in= \ \ \ \frac{f(x)}{\sin^{2}\frac{x-s}{2}}dx,\quad f(x)\in C^{\infty}[c,c+2\pi], c\in R, s\in (c,c+2\pi)$$
defined as the Hadamard finite-part integral. Classical rectangle rule
$$I_{n}(c,s,f)=h\sum_{i=0}^{n-1} \frac{f_{C}(\hat{x}_{i})}{\sin^{2}\frac{\hat{x}_{i}-s}{2}}$$
with $f_{C}(\hat{x}_{i})=f(\hat{x}_{i}),h=2\pi/n,\hat{x}_{i}=x_{i}+h/2$ is the middle of subinterval
which can not be used to compute hypersingular integral as there are the divergence part.
In order to give the simple rectangle rule, we present the modify rectangle rule as
$$\tilde{I}_{n}(c,s,f)=h\sum_{i=0}^{n-1} \frac{f_{C}(\hat{x}_{i})}{\sin^{2}\frac{\hat{x}_{i}-s}{2}}-\frac{4f(s)\pi^2}{h\sin^2 \frac{\xi\pi}{2} },\xi\in[-1,1]$$
and
$$\hat{I}_{n}(c,s,f)=h\sum_{i=0}^{n-1} \frac{f_{C}(\hat{x}_{i})}{\sin^{2}\frac{\hat{x}_{i}-s}{2}}-\frac{4f(s)\pi^2}{h\sin^2 \frac{\xi\pi}{2} }-4f'(s)\pi\tan \frac{(\xi+1)\pi}{2}$$
We get the numerical quadrature formulas $\tilde{I}_{n}(c,s,f)$ have the spectral accuracy as
$$\hat{I}_{n}(c,s,f)-I(c,s,f)=O(h^{\mu}),\mu\ge0$$
with the special function $\frac{\pi^2}{\sin^2 \frac{\xi\pi}{2} }$ and $\tan \frac{(\xi+1)\pi}{2}$ equals to zero.
Numerical examples are provided to valid our theorem.