       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:=
f1 = FullSimplify[Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x, 0, Pi}]]
Out=
(1/8)*Pi*(2*Pi + 8*I*ArcSinh + Log[((2 - Sqrt)^(9*I)*(2 +
Sqrt)^(7*I))/(10 - 7*Sqrt)^(3*I)] +
Log[(6 + 4*Sqrt)^(-6*I)] - I*Log[10 + 7*Sqrt])

In:=
f2 = FullSimplify[Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x, -2*Pi,
2*Pi}]]
Out=
-Pi^2

In:=
f3 = FullSimplify[Tr[(Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x,
#1[], #1[]}] & ) /@ Partition[Range[0, Pi, Pi/2], 2, 1]]]
Out=
Pi^2/4

In:=
f4 = FullSimplify[Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x, 0, Pi/2,
Pi}]]
Out=
Pi^2/4

In:=
f5 = FullSimplify[Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x, 0, 5*Pi}]]
Out=
(5*Pi^2)/4

In:=
f6 = FullSimplify[Integrate[x*(Sin[x]/(1 + Cos[x]^2)), {x, -Pi, Pi}]]
Out=
(1/4)*Pi*(2*Pi + 8*I*ArcSinh + Log[((2 - Sqrt)^(9*I)*(2 +
Sqrt)^(7*I))/(10 - 7*Sqrt)^(3*I)] +
Log[(6 + 4*Sqrt)^(-6*I)] - I*Log[10 + 7*Sqrt])

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:=
Block[{Message}, FullSimplify[
(1/8)*Pi*(2*Pi + 8*I*ArcSinh + Log[((2 - Sqrt)^(9*I)*(2 +
Sqrt)^(7*I))/(10 - 7*Sqrt)^(3*I)] +
Log[(6 + 4*Sqrt)^(-6*I)] - I*Log[10 + 7*Sqrt]) == Pi^2/4,
ComplexityFunction -> (Count[{#1}, _Log | _ArcSinh, Infinity] & )]]
Out=
True

In:=
Block[{Message}, FullSimplify[
(1/4)*Pi*(2*Pi + 8*I*ArcSinh + Log[((2 - Sqrt)^(9*I)*(2 +
Sqrt)^(7*I))/(10 - 7*Sqrt)^(3*I)] +
Log[(6 + 4*Sqrt)^(-6*I)] - I*Log[10 + 7*Sqrt]) == Pi^2/2,
ComplexityFunction -> (Count[{#1}, _Log | _ArcSinh, Infinity] & )]]
Out=
True

Any ideas???

PS

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

```

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