Consider the hypersingular integral on circle I(c, s, f ) = Z c+2π c = f (x)/ sin2 x−s 2 dx, f (x) ∈ C∞c, c + 2π, c ∈ R, s ∈ (c, c + 2π) defined as the Hadamard finite-part integral. Classical rectangle rule In(c, s, f ) = h Xn−1 i=0 fC( ˆxi)/ sin2 ˆxi−s 2 with fC(ˆxi) = f ( ˆxi), h = 2π/n, ˆxi = xi + 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 ˜In(c, s, f ) = h Xn−1 i=0 fC(ˆxi)/ sin2 ˆxi−s 2 − 4 f (s)π2/ h sin2 ξπ 2 , ξ ∈ −1, 1 and ˆIn(c, s, f ) = h Xn−1 i=0 fC(ˆxi)/ sin2 ˆxi−s 2 − 4 f (s)π2/ h sin2 ξπ 2 − 4 f ′(s)π tan (ξ + 1)π/ 2 We get the numerical quadrature formulas ˜In(c, s, f ) have the spectral accuracy as ˆIn(c, s, f ) − I(c, s, f ) = O(hμ), μ ≥ 0 with the special function π2 /sin2 ξπ 2 and tan (ξ+1)π/ 2 equals to zero. Numerical examples are provided to valid our theorem.
Jin Li (Wed,) studied this question.