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MathGroup Archive 2007

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Re: simplifications

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74527] Re: simplifications
  • From: roland franzius <roland.franzius at uos.de>
  • Date: Sat, 24 Mar 2007 05:19:34 -0500 (EST)
  • Organization: Universitaet Hannover
  • References: <eu1rpt$lf6$1@smc.vnet.net>

dimitris wrote:
> Hello.
> 
> Below I use Mathematica to get some definite integrals along with
> simplifications
> of the results.
> 
> ln[9]:=
> f1 = FullSimplify[Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x, 0, Pi}]]
> Out[9]=
> (1/8)*Pi*(2*Pi + 8*I*ArcSinh[1] + Log[((2 - Sqrt[2])^(9*I)*(2 +
> Sqrt[2])^(7*I))/(10 - 7*Sqrt[2])^(3*I)] +
>    Log[(6 + 4*Sqrt[2])^(-6*I)] - I*Log[10 + 7*Sqrt[2]])
> 
> In[10]:=
> f2 = FullSimplify[Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x, -2*Pi,
> 2*Pi}]]
> Out[10]=
> -Pi^2
> 
> In[11]:=
> f3 = FullSimplify[Tr[(Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x,
> #1[[1]], #1[[2]]}] & ) /@ Partition[Range[0, Pi, Pi/2], 2, 1]]]
> Out[11]=
> Pi^2/4
> 
> In[12]:=
> f4 = FullSimplify[Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x, 0, Pi/2,
> Pi}]]
> Out[12]=
> Pi^2/4
> 
> In[13]:=
> f5 = FullSimplify[Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x, 0, 5*Pi}]]
> Out[13]=
> (5*Pi^2)/4
> 
> In[14]:=
> f6 = FullSimplify[Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x, -Pi, Pi}]]
> Out[14]=
> (1/4)*Pi*(2*Pi + 8*I*ArcSinh[1] + Log[((2 - Sqrt[2])^(9*I)*(2 +
> Sqrt[2])^(7*I))/(10 - 7*Sqrt[2])^(3*I)] +
>    Log[(6 + 4*Sqrt[2])^(-6*I)] - I*Log[10 + 7*Sqrt[2]])
> 
> No matter what I try I couldn't simplify f1 and f6 directly to Pi^2/4
> and Pi^2/2.


The following evaluation shows that you are running into the complexity 
of ArcTan riemann surface problems

res = Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x, -Pi, Pi}];

(res // TrigToExp // FullSimplify // TrigToExp) //. {Pi/4 Log[b_] + Pi/4
         Log[c_] + d___ :> Pi/4  Log[ b c] + d} // FullSimplify


(1/2 + I/2)*Pi^2


You should consult the ArcTan page at mathworld. Formal Log and ArcTan 
expressions from integrals dont deserve much confidence. In this case I 
use a Log formula which is defnitely wrong in the complex domain.

-- 

Roland Franzius


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