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Re: Numerical integration over halfinfinite intervals
 To: mathgroup at smc.vnet.net
 Subject: [mg60282] Re: Numerical integration over halfinfinite intervals
 From: "JensPeer Kuska" <kuska at informatik.unileipzig.de>
 Date: Sat, 10 Sep 2005 06:46:34 0400 (EDT)
 Organization: Uni Leipzig
 References: <dfrgt6$fsg$1@smc.vnet.net>
 Sender: ownerwrimathgroup at wolfram.com
Hi,
SequenceLimit[list] returns the approximation
given by Wynn's epsilon \
algorithm to the limit of a sequence whose first
few terms are given by list. \
Warning: Wynn's epsilon algorithm can give finite
results for divergent \
sequences.
may help you.
Regards
Jens
"Alan" <info at optioncity.REMOVETHIS.net> schrieb im
Newsbeitrag news:dfrgt6$fsg$1 at smc.vnet.net...
I do a lot of 1D NIntegrates over halfinfinite
domains [0,Infinity).
 Sometimes you can simply put Infinity as an
upper
 bound and Mathematica will return an answer. In
that
 case, I don't have a problem.

 But, sometimes, this fails or takes too
 long, and I am forced to truncate the integral.

 Let's assume my integral converges, is not zero,
and my
 integrand is relatively smooth with a few
derivatives, at least.
 Suppose each finite truncated integral can be
successfully computed to
 the same fixed PrecisionGoal.

 Given that, my questions:

 Is it possible to extrapolate these truncated
results to a
 limit with a known precision? If so, how, and
how does that
 precision relate to the fixed PrecisionGoal
above?

 Thanks,
 alan


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