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