       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:=
> 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 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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