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MathGroup Archive 2006

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Re: sum of integrals over patial intervals != integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71822] Re: sum of integrals over patial intervals != integral
  • From: Peter Pein <petsie at dordos.net>
  • Date: Thu, 30 Nov 2006 06:05:23 -0500 (EST)
  • References: <ekh5hc$rn5$1@smc.vnet.net> <ekjet2$ccj$1@smc.vnet.net>

dimitris schrieb:
> Also
> 
> In[1]:=
> $Version
> Out[1]=
> "4.0 for Microsoft Windows (April 21, 1999)"
> 
> In[5]:=
> f[x_] := Log[Sin[x]^2]*Tan[x];
> 
> In[20]:=
> Off[$MaxExtraPrecision::meprec]
> 
> In[21]:=
> Integrate[f[x], {x, 0, Pi}]
> N[%, 40]
> Out[21]=
> Pi^2/3 + 1/2*(-3*Log[2]^2 - Log[4]^2 + Log[2 - Sqrt[2]]^2 +
> Log[16]*Log[2 + Sqrt[2]] - Log[2 + Sqrt[2]]^2 -
>     4*PolyLog[2, -(1/Sqrt[2])] - 4*PolyLog[2, 1/Sqrt[2]] + 4*PolyLog[2,
> 1 - 1/Sqrt[2]] - 2*PolyLog[2, 1/4*(2 - Sqrt[2])] -
>     4*PolyLog[2, 2/(2 + Sqrt[2])] - 2*PolyLog[2, 1/4*(2 + Sqrt[2])])
> Out[22]=
> -9.8933845188332`0.3443*^-92
> 
> Regards
> Dimitris
> 
> Î?/Î? Bob Hanlon έγÏ?αÏ?ε:
>> Works in my version:
>>
>> $Version
>>
>> 5.2 for Mac OS X (June 20, 2005)
>>
>> f[x_]:=Log[Sin[x]^2]Tan[x];
>>
>> Integrate[f[x],{x,0,Pi}]
>>
>> 0
>>
>>
>> Bob Hanlon
>>
>> ---- Peter Pein <petsie at dordos.net> wrote:
>>> Dear group,
>>>
>>> I wanted Mathematica to show, that for f[x_]:=Log[Sin[x]^2]Tan[x],
>>> Integrate[f[x],{x,0,Pi}]==0, because f[x]+f[Pi-x]==0.
>>>
>>> Mathematica says Integrate[f[x],{x,0,Pi}] does not converge, but
>>> Integrate[f[x],{x,0,Pi/2}] and Integrate[f[x],{x,Pi/2,Pi}] evaluate to
>>> -Pi^2/12 resp. P^2/12 and the sum is zero. The more general integral
>>> Integrate[f[x],{x,0,z},Assumptions->Pi/2<z<=Pi] evaluates explicitly (?).
>>>
>>> What did I do wrong?
>>> http://people.freenet.de/Peter_Berlin/Mathe/komisch.nb
>>>
>>> TIA,
>>> Peter
>>>
> 

Dear Dimitris, dear Bob,

Thank you very much for your answers!

I would like to know, whether the development version finds the simple
substitution u==Sin[x]^2:

f[x_] = Log[Sin[x]^2]*Tan[x];
(* u==Sin[x]^2 *)
new = f[x]*Dt[x] /. x -> ArcSin[{-1, 1}*Sqrt[u]] /. Dt[u] -> 1
--> {Log[u]/(2*(1 - u)), Log[u]/(2*(1 - u)}

F[z_] = Apart[Integrate[new[[1]], {u, 0, Sin[z]^2},
   Assumptions -> 0 < z < Pi]]
--> -Pi^2/12 + 1/2*PolyLog[2, Cos[z]^2]

Simplify[F'[x] == f[x]]
--> True


Thanks again,
Peter


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