       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:= Integrate[PolyLog[2, Exp[I*(x - y)]], {y, 0, 2*Pi}]

Out= 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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