Re: NIntegrate about singular point
- To: mathgroup at smc.vnet.net
- Subject: [mg126112] Re: NIntegrate about singular point
- From: danl at wolfram.com
- Date: Wed, 18 Apr 2012 03:54:55 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <jmjf8p$89v$1@smc.vnet.net>
On Tuesday, April 17, 2012 5:05:13 AM UTC-5, bowlderster wrote: > Hello, all. > 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? > > Thanks for your help That is not an integrable singularity. To see this, rationalize everything,= compute the singular point explicitly, and expand the integrand in a serie= s centered at that point. We'll get a 1/(x-pt) type of term, and this means= the Riemann integral does not exist. h = 1; m = 1; n = 1; t = 1; k = 40269/10000; kk = 40284/10000; In[326]:= expr = Rationalize[ 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)])] Out[326]= (x*((-(10071/2500) + x)/ E^((3*x)/5) + (2/E^x + E^((3*x)/5))*(10071/2500 + x)))/ (-((10071*Cosh[x])/2500) + x*Sinh[x]) In[327]:= sing = x /. First[Solve[Denominator[expr] == 0 && x >= = 0, x]] Out[327]= Root[{10071 - 2500*#1 + (10071 + 2500*#1)/E^(2*#1) & , 4.0309413727078305519742154202340325203130636422539920318199`20.\ 60201996114937}] In[328]:= N[Normal[Series[expr, {x, sing, 2}]], 20] Out[328]= \ -0.3554151801476991150413491075955450313832012232822013113183`20. + 12.9359005198882606437231966513206867737`20./ (-4.0309413727078305519842918447641817012641557622412458781615`\ 20. + x) - 0.492816984068007502148671038677316080890365779186271576225`20.* (-4.0309413727078305519842918447641817012641557622412458781615`\ 20. + x) + 0.0874544334620205088679175508145220699125102673447387653118`20.* (-4.0309413727078305519842918447641817012641557622412458781615`\ 20. + x)^2 If you are looking to approximate numerically a principal value integral, t= hat would be a different matter. Daniel Lichtblau Wolfram Research