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Re: Integrate 'learns'?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg106336] Re: [mg106269] Integrate 'learns'?
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Fri, 8 Jan 2010 04:19:09 -0500 (EST)
  • Reply-to: hanlonr at cox.net

I get the "final answer" directly

$Version

7.0 for Mac OS X x86 (64-bit) (February 19, 2009)

Integrate[x/(3 Sin[x]), {x, Pi/4, Pi/2}]

(5*Catalan)/6 + 
   (2*I*PolyLog[2, (-1)^(1/4)])/3 - 
   (23*I*Pi^2)/288 + (1/6)*Pi*
     ArcTanh[(-1)^(1/4)]

Perhaps in doing some amount of internal simplification it applies a time constraint. On your second pass it has some cached intermediate result that enables the simplication to progress further within the time constraint.


Bob Hanlon

---- Tony Harker <a.harker at ucl.ac.uk> wrote: 

=============

  If I open a clean notebook in Version 7.0 for Microsoft Windows (32-bit)
and enter 
  Integrate[x/(3 Sin[x]),{x,\[Pi]/4,\[Pi]/2}]
  the result (after a warning message) is
  (8*Catalan - I*Pi^2 + Pi*(-Log[1 - (-1)^(1/4)] + Log[1 + (-1)^(1/4)]) -
(4*I)*(PolyLog[2, -(-1)^(1/4)] - PolyLog[2, (-1)^(1/4)]))/12
  and if I then repeat the command I get no error and
  (5*Catalan)/6 - ((23*I)/288)*Pi^2 + (Pi*ArcTanh[(-1)^(1/4)])/6 +
((2*I)/3)*PolyLog[2, (-1)^(1/4)]
  which seems to be Mathematica's final answer.

  I am happy that the results are equivalent, but puzzled about what has
been saved, and where, to generate this difference. Did Mathematica ask the
audience or phone a friend? Can anyone enlighten me?

  Tony Harker




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