Re: Polylogarithm Integration - Bis

*To*: mathgroup at smc.vnet.net*Subject*: [mg46246] Re: Polylogarithm Integration - Bis*From*: D <D at D.gov>*Date*: Thu, 12 Feb 2004 07:16:11 -0500 (EST)*Organization*: University of Oslo, Norway*References*: <c0figi$9s3$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Bill Rowe wrote: > There does seem to be a bug somewhere. By inspection it is clear (1/4)*(5 - 4*Pi)^2*Pi is not zero. But it isn't obvious the preceding expression should reduce to this when FullSimplify is applied. > > But I am not convinced the explanation I gave is substantially wrong, although I might have worded it more carefully. > > I said it was a problem with the coding of Integrate but to be more precise I should have said using N to convert the output of Integrate to a numerical answer isn't guaranteed to not lose numerical precision. And as a consequence, applying N to the results of Integrate can get results that are quite inaccurate. > > In any case, I still think when a numerical answer is desired it is far better to use NIntegrate rather than apply N to the results of Integrate. And I note with x = -2.5, NIntegrate gives me the same warning regarding the integral coverging to slowly. Yes, there is definitively a bug. The fact that Mathematica gives different answers with, e.g., 5/2 and 2.5 is indeed due to a choice of different algorithms (numerical vs analytical), I guess. But this is not the crucial point. The crucial point is for In[1]:= Integrate[PolyLog[2, Exp[I*(x - y)]], {y, 0, 2*Pi}] Out[1]= If[x >= 2*Pi || x <= 0, Pi3 - Pi*Log[-E^((-I)*x)]^2 + 2*I*Pi2*Log[1 - E^((-I)*x)] - 2*I*Pi2*Log[1 - E^(I*x)], Integrate[PolyLog[2, E^(I*(x - y))], {y, 0, 2*Pi}, Assumptions -> !(x >= 2*Pi || x <= 0)]] that the result is incorrect with x but correct with 5/2. In both cases Mathematica use formal algorithms, but obviously different. My guess is that there is a mistake, somewhere, with the branch cut in the definition of the polylogarithms. I tested Mathematica before buying it, because I need such integrals for my research. I will not buy it until such bugs are fixed or if someone can explain to me how to overcome the problem. I am not sure that Wolfram Research is interested in these kind of problems, because the market is too low. D.