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MathGroup Archive 2007

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Re: simplifications

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74540] Re: simplifications
  • From: Peter Pein <petsie at dordos.net>
  • Date: Sun, 25 Mar 2007 01:25:11 -0500 (EST)
  • References: <eu1rpt$lf6$1@smc.vnet.net>

dimitris schrieb:
> 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[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.
> 
...

> 
> Can somebody guess why we f1 and f6 do not simplify directly as the
> other fi?
> 
> 

Hi Dimitris,

maybe the minimuf of the denominator causes this trouble.

My first attempt was
In[1]:=
Apply[
 Timing[
  ClearCache[];
  MemoryConstrained[
   FullSimplify[Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x, ##1}]],
  512*1024^2 (* give Mathematica 512MB to waste *)]] & ,
 {{0, Pi/2, Pi}, {-Pi, -Pi/2, Pi/2, Pi}}, {1}]
Out[1]=
{{ 46.531*Second, Pi^2/4},
 {301.516*Second, $Aborted}}

so this is not an option.

The use of ComplexExpand gives (better) results:

In[2]:=
Timing[
 ClearCache[];
 FullSimplify[ComplexExpand[
  Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x, #1, Pi}]]]]& /@ {0, -Pi}
Out[2]=
{{43.797*Second, Pi^2/4},
 {39.562*Second, Pi^2/2}}

But the best way to do this seems to be a substitution:

In[3]:=
Timing[
 ClearCache[];
 Integrate[
  TrigExpand[x*(Sin[x]/(1 + Cos[x]^2))*Dt[x] /.
   x -> 2*ArcTan[t] /. Dt[t] -> 1],
  {t, Limit[Tan[x/2], x -> #1], Infinity}]]& /@ {0, -Pi}
Out[3]=
{{9.297*Second, Pi^2/4},
 {9.219*Second, Pi^2/2}}


Peter


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