Emergent Mind

Abstract

We consider the numerical computation of $I[f]=\intBarb_a f(x)\,dx$, the Hadamard Finite Part of the finite-range singular integral $\intb_a f(x)\,dx$, $f(x)=g(x)/(x-t){m}$ with $a<t<b$ and $m\in{1,2,\ldots},$ assuming that (i)\,$g\in C\infty(a,b)$ and (ii)\,$g(x)$ is allowed to have arbitrary integrable singularities at the endpoints $x=a$ and $x=b$. We first prove that $\intBarb_a f(x)\,dx$ is invariant under any suitable variable transformation $x=\psi(\xi)$, $\psi:[\alpha,\beta]\rightarrow[a,b]$, hence there holds $\intBar\beta_\alpha F(\xi)\,d\xi=\intBarb_a f(x)\,dx$, where $F(\xi)=f(\psi(\xi))\,\psi'(\xi)$. Based on this result, we next choose $\psi(\xi)$ such that the transformed integrand $F(\xi)$ is sufficiently periodic with period $\T=\beta-\alpha$, and prove, with the help of some recent extension/generalization of the Euler--Maclaurin expansion, that we can apply to $\intBar\beta_\alpha F(\xi)\,d\xi$ the quadrature formulas derived for periodic singular integrals developed in an earlier work of the author. We give a whole family of numerical quadrature formulas for $\intBar\beta_\alpha F(\xi)\,d\xi$ for each $m$, which we denote $\widehat{T}{(s)}_{m,n}[{\cal F}]$, where ${\cal F}(\xi)$ is the $\T$-periodic extension of $F(\xi)$.

We're not able to analyze this paper right now due to high demand.

Please check back later (sorry!).

Generate a summary of this paper on our Pro plan:

We ran into a problem analyzing this paper.

Newsletter

Get summaries of trending comp sci papers delivered straight to your inbox:

Unsubscribe anytime.