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MathGroup Archive 2007

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simplifications

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74517] simplifications
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Fri, 23 Mar 2007 19:17:32 -0500 (EST)

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 only thing I could show (good news of course but this is not my
point!)  was

In[20]:=
Block[{Message}, FullSimplify[
   (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]]) == Pi^2/4,
   ComplexityFunction -> (Count[{#1}, _Log | _ArcSinh, Infinity] & )]]
Out[20]=
True

In[23]:=
Block[{Message}, FullSimplify[
   (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]]) == Pi^2/2,
   ComplexityFunction -> (Count[{#1}, _Log | _ArcSinh, Infinity] & )]]
Out[43]=
True

Any ideas???

PS

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



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