[Date Index]
[Thread Index]
[Author Index]
Re: Polylogarithm Integration - Bis
*To*: mathgroup at smc.vnet.net
*Subject*: [mg46202] Re: Polylogarithm Integration - Bis
*From*: Bill Rowe <readnewsciv at earthlink.net>
*Date*: Tue, 10 Feb 2004 00:06:59 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
On 2/9/04 at 5:54 AM, DWCantrell at sigmaxi.org (David W. Cantrell)
wrote:
>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.
>Unfortunately you snipped the part where D had said:
>"With x=-2.5 I get
>44.964016795241896 + I*1.7763568394002505`*^-15
Yes. It didn't occur to me to check both values.
>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.
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.
--
To reply via email subtract one hundred and four
Prev by Date:
**Re: Re: Find last NZ in list**
Next by Date:
**Re: Find last NZ in list**
Previous by thread:
**Re: Polylogarithm Integration - Bis**
Next by thread:
**Re: Polylogarithm Integration - Bis**
| |