Re: NIntegrate about singular point

*To*: mathgroup at smc.vnet.net*Subject*: [mg126108] Re: NIntegrate about singular point*From*: A Retey <awnl at gmx-topmail.de>*Date*: Wed, 18 Apr 2012 03:53:32 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <jmjf8p$89v$1@smc.vnet.net>

Hi, > I am dealing with an integral as following > > h = 1 > m = 1 > n = 1 > t = 1 > k = 4.0269 > kk = 4.0284 > Plot[x^(m + n - 1)/(x*Sinh[x*h] - > kk*Cosh[x*h])*((x + > kk)*((-1)^(m + n)*Exp[x*(2.*(-0.2) + h)] - ((-1)^m + (-1)^n)* > Exp[-x*h]) + (x - kk)*Exp[-x*(2.*(-0.2) + h)]), {x, 0, 40}] > NIntegrate[ > x^(m + n - 1)/(x*Sinh[x*h] - > kk*Cosh[x*h])*((x + > kk)*((-1)^(m + n)*Exp[x*(2.*(-0.2) + h)] - ((-1)^m + (-1)^n)* > Exp[-x*h]) + (x - kk)*Exp[-x*(2.*(-0.2) + h)]), {x, 0, > Infinity}] > > It has a singular point when the denominator is zero. > At begining, I try to solve it in another system, yet with Nan result. > > It is the first time for me to use Mathematica. > Can it solve the integral with singular point? I think that NIntegrate handles many singularities automatically if the integral actually converges. I haven't checked, but I suspect that in this case the integral is probably simply not converging. It's easy enough though to get the cauchy principle value, if that is what you're after: expr = x^(m + n - 1)/(x*Sinh[x*h] - kk*Cosh[x*h])*((x + kk)*((-1)^(m + n)*Exp[x*(2.*(-0.2) + h)] - ((-1)^m + (-1)^n)* Exp[-x*h]) + (x - kk)*Exp[-x*(2.*(-0.2) + h)]); singularity = x /. FindRoot[Denominator[expr] == 0, {x, 4}] NIntegrate[expr, {x, 0, Infinity}, Exclusions -> {singularity}, Method -> "PrincipalValue"] you might want to have a look at tutorial/NIntegrateOverview in the documentation center to learn more details about the various options and methods of NIntegrate. hth, albert