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Re: A tough Integral
*To*: mathgroup at smc.vnet.net
*Subject*: [mg28681] Re: A tough Integral
*From*: sotirisgk1 at aol.com (SotirisGK1)
*Date*: Tue, 8 May 2001 02:51:16 -0400 (EDT)
*References*: <9cb93f$ci2@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Tough no more.
Simply have 2 physicists work on it. I am the second one.
Integrate it numerically to x=10 which is easily produced to 70 digits or more
in less than 5 min (for any cpu over 100 Mhz).(you may have to break it from 0
to 4 then from 4 to 10) Then you are left with the 10 to infinity part.
Follow the logic above by Richard Easther in expanding the 1/(x^2+Cos[x]) in
geometric series up to the 31st say term ala
1/x^2*Sum[(-1*Cos[x]/x^2)^n,{n,0,30}]
The remainder is of the order of 1/x^64 or so which from x larger than 10 when
integrated produces contribution a lot less than 10^-61 .
Now use TrigReduce on this sum to have the effect that all the cosine powers
are eliminated and replaced by cosines of multiples of x (eg Cos[30x] etc).
At this level all that is left to do is evaluate the integral of expressions of
the form
Cos[k*x]/x^n with k,n positive integers . Mathematica does that analytically
but produces the result in the form of HypergeometricPFQ functions and other
constants or logarithms that can be evaluated at arbitrary precision if you use
the Block[{$MaxExtraPrecision = n}, expr]
command where the expression will be the N[of integral from 10 to infinity
expressed in Hypergeometric functions and other stuff,60].
This may take a good few hours on mathematica to integrate the
trigonometrically reduced above sum of 31 terms since each term will generate a
good number of other cosines leading to some 100-300 or more individual
integrals of the form Cos[kx]/x^n each of which will be expressed in terms of
hypergeometric functions eventually.
You can in the end be looking at having a large sum of various constants of
Euler and Pi and Logs and Hypergeometric functions at various arguments all of
which are doable at 60 digits.
Now good luck in doing all this. It may take you 2-3 hours depending on speed
and understanding of what is going on but i bet its totally doable at a
practical level. If that was a homework for a few days time its certainly
solved within the first day of assignment.
Ps: Thanks for making us think and learn mathematrica a little bit more than we
did before.
Now back to my challenge of the 1/n^(n*x) sum i posted the other day that
nobody has responded yet about. :-)
SotirisGK1 at aol.com (aka Singularity_Chaser)
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