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MathGroup Archive 1997

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Re: barfing on an integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg6187] Re: [mg6160] barfing on an integral
  • From: seanross at worldnet.att.net
  • Date: Thu, 27 Feb 1997 02:53:05 -0500
  • Sender: owner-wri-mathgroup at wolfram.com

peter wrote:
> 
> Hello All
> 
> I'm trying to check my solution to approximations of integrals by the
> method of steepest decent.  The integral i'm approximating is:
> 
>    I(x) = int e(x*t - e^t) dt     from t=0 to t=infty
> 
> my approximation is:
> 
>    J(x) = sqrt(2 Pi/x) * e^{1.5 x log x}
> 
> The trouble is that MMA is barfing on I(x).  If I try to do it
> analytically with Integrate[] it gives me a whole bunch of E^Infty
> terms.  Most of which go to zero, but I'm having trouble getting MMA to
> actually evaluate the limit.  N[%] doesn't seem to work, and MMA doesn't
> seem to realize that e^infty / e^e^infty is truly zero.
> 
> Then I tried using NIntegrate on I(x), and it gave me a whole bunch of
> underflow errors, and said that the integrand is probably oscillatory
> (which it most certainly is not).
> 
> I'm in a very awkward position.  My approximation is only valid for large
> values of x (actually, I think anything over 10 will do).  Yet, MMA barfs
> for large values of x.
> 
> Is there some way I can check my approximation??
> 
> Much thanks.
> 
> Peter
> 
> --
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You may have to use manual transformation rules for various 
combinations. I too have been unable to get the symbolic routines in 
mathematica to do some of the simplest things.  Sometimes the problem 
can be solved with complexexpand since mathematica keeps open the 
possibility that every symbol is complex but this doesn't seem to be the 
problem here.  There is still no substitute for pencil and paper.  Just 
as a thought, have you defined your integral as a limit?
ie.  Limit[Integral[f[x],{x,0,a}],a->Infinity]?  Alternately, have you 
tried to reformulate your problem in terms of inverse variables and then 
use zero as the integral limit?  For example, break up the integral into 
the ranges {0 to 1} and {1 to infinity}, then take the infinity part 
and transform x->1/x and integrate from {0 to 1}, then transform the 
answer back.  Perhaps mathematica would have an easier time evaluating 
zeros than infinities.


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