       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
>
> --
> Birthdays are good for you:  A federal funded project has recently determined
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>      I BOYCOTT ANY COMPANY THAT USES MASS ADVERTISING ON THE INTERNET

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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