Mathematica 9 is now available
Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2012

[Date Index] [Thread Index] [Author Index]

Search the Archive

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



  • Prev by Date: Re: nonlinearmodelfit problem
  • Next by Date: Re: What characters are allowed in mathematica variable names? i.e. how
  • Previous by thread: Re: NIntegrate about singular point
  • Next by thread: Re: NIntegrate about singular point