Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2002
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2002

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

Search the Archive

Re: NIntegrate

  • To: mathgroup at smc.vnet.net
  • Subject: [mg36527] Re: NIntegrate
  • From: "Rob_jack" <rob_jackNSP at libero.it>
  • Date: Wed, 11 Sep 2002 03:27:47 -0400 (EDT)
  • References: <alkhi3$ve$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

"Rob_jack" <rob_jackNSP at libero.it> wrote

There is an error in the previous message.

This is the just msg:

**********************************************
In[1]=f[y_,z_]: = (y^4/( (1+(8.44*10^-4)^2 * (1+z)^2 y^2) (Exp[y]+1) )

In[2]=F[z_]: = NIntegrate[f[y,z], {y, 0, Infinity}
**********************************************

Mathematica 4.0 says:

****************************************************************************

NIntegrate: : inum : Integrand 1.07577/( 1+7.12336*10^-7 (1. + z)^2 ) is not
numerical at  {y}={1.}.
****************************************************************************


What is it??

Moreover, I have used the following procedure:
******************************************************
If z<<1,

f[y,z]=Sum[(-1)^n*a^n*(1+z)^2n *(y^(2n+4)/(Exp[y]+1),{n, 0, N} ]

here a=(8.44*10^-4)^2

therefore:

F[z]:=Sum[(-1)^n*a^n*(1+z)^2n *(y^(2n+4)/(Exp[y]+1) * Gamma[2 n+5]*Zeta[2
n+5],{n, 0, N} ]


this series (N<+oo) only approximates the function for small z, and me it
interests the behavior of F for z in the range 800-1300.

Thx in advance.

Rob_jack













  • Prev by Date: X(NumLock)=Mod2 (was Re: Profiler for Mathematica)
  • Next by Date: Re: NIntegrate
  • Previous by thread: NIntegrate
  • Next by thread: Re: NIntegrate