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