Re: Polylogarithm Integration - Bis
- To: mathgroup at smc.vnet.net
- Subject: [mg46160] Re: Polylogarithm Integration - Bis
- From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
- Date: Mon, 9 Feb 2004 05:54:07 -0500 (EST)
- References: <c02l20$71b$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Bill Rowe <readnewsciv at earthlink.net> wrote: > On 2/6/04 at 4:15 AM, D at D.gov (D) wrote: > > >Some more example for the integral: > > >Integrate[PolyLog[2, Exp[I*(x - y)]], {y, 0, 2*Pi}] > > >with x=5/2, I get 0 as expected. But with x=2.5 I get > > >19.634954084936204 + I*1.7763568394002505`*^-15 > > >which is not close to zero! > > When you use x = 5/2 you are asking Mathematica to use exact numbers. So, > Integrate returns the correct exact result of 0. But when you use 2.5 you > are asking Mathematica to use machine precision numbers. Integrate does > not seem to be coded to avoid problems with loss of precision that occurs > with machine precision numbers. You miss the point. Such a huge discrepancy cannot be explained in this simple manner. See below. > If you want to use machine precision, use > NIntegrate instead of Integrate. > > When I use NIntegrate with x = 2.5 on this problem I get > > -1.551478671979467*^-6 - 7.615056976939538*^-7*I > > and a warning that numerical integration is converging too slowly. > Clearly, this result is close to zero. And coupled with the warning > strongly indicates there are issues with numerical precision for this > particular integral. Unfortunately you snipped the part where D had said: "With x=-2.5 I get 44.964016795241896 + I*1.7763568394002505`*^-15 etc. I have only one conclusion: there is a bug with Integrate or PolyLog or both." So now consider the following, obtained using version 5.0: In[1]:= Integrate[PolyLog[2, Exp[I*(x - y)]], {y, 0, 2*Pi}] Out[1]= If[x >= 2*Pi || x <= 0, Pi^3 - Pi*Log[-E^((-I)*x)]^2 + 2*I*Pi^2*Log[1 - E^((-I)*x)] - 2*I*Pi^2*Log[1 - E^(I*x)], Integrate[PolyLog[2, E^(I*(x - y))], {y, 0, 2*Pi}, Assumptions -> !(x >= 2*Pi || x <= 0)]] In[2]:= % /. x -> -5/2 Out[2]= Pi^3 - 2*I*Pi^2*Log[1 - E^(-((5*I)/2))] + 2*I*Pi^2*Log[1 - E^((5*I)/2)] - Pi*Log[-E^((5*I)/2)]^2 In[3]:= FullSimplify[%] Out[3]= (1/4)*(5 - 4*Pi)^2*Pi The above is essentially the same result as D got, but using exact numbers. In[4]:= N[%] Out[4]= 44.964016795241896 But the answer should presumably be 0 instead: In[5]:= Integrate[PolyLog[2, Exp[I*(-5/2 - y)]], {y, 0, 2*Pi}] Out[5]= 0 So I must agree with D that there is indeed a bug somewhere here. David Cantrell