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MathGroup Archive 2004

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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.


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