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From: Grischika on 24 Jul 2008 04:54
On 23 =C9=C0=CC, 13:26, Valeri Astanoff <astan...(a)gmail.com> wrote:
> Good day,
>
> Neither Mathematica 6 nor anyone here can integrate this:
>
> In[1]:= Integrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], {t, 0, Pi}]
> Out[1]= Integrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], {t, 0, Pi}]
>
> In[2]:= NIntegrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], {t, 0, Pi}]
> Out[2]= 6.72288
>
> I know the exact result:
>
> In[3]:= =9A(1/5^(3/4))*(Sqrt[2]*(10*EllipticE[(1/10)*(5 - Sqrt[5])] -
> =9A =9A =9A =9A 10*EllipticK[(1/10)*(5 - Sqrt[5])] + (5 + 3*Sqrt[5])*
> =9A =9A =9A =9A EllipticPi[(1/10)*(5 - 3*Sqrt[5]), (1/10)*(5 - Sqrt[5])])=
)//N
> Out[3]= 6.72288
>
> but I would like to prove it.
>
> Thanks in advance to the samaritan experts...
>
> V.Astanoff

Hello.
You can try to take indefinite integral:

eq=Integrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], t]

Here Mathematica gives you an answer:

((2/5 + (4*I)/5)*Cos[t/2]^4*((2 + I)*Sqrt[1 - 2*I]*
EllipticE[I*ArcSinh[Sqrt[1 - 2*I]*Tan[t/2]], -3/5 + (4*I)/5]*
Sec[t/2]^2*Sqrt[1 + (1 - 2*I)*Tan[t/2]^2]*
Sqrt[1 + (1 + 2*I)*Tan[t/2]^2] -
I*((6 - 2*I)*Sqrt[1 - 2*I]*EllipticF[
I*ArcSinh[Sqrt[1 - 2*I]*Tan[t/2]], -3/5 + (4*I)/5]*Sec[t/2]^2*
Sqrt[1 + (1 - 2*I)*Tan[t/2]^2]*Sqrt[1 + (1 + 2*I)*Tan[t/2]^2] -
4*Sqrt[1 - 2*I]*EllipticPi[1/5 + (2*I)/5,
I*ArcSinh[Sqrt[1 - 2*I]*Tan[t/2]], -3/5 + (4*I)/5]*Sec[t/2]^2*
Sqrt[1 + (1 - 2*I)*Tan[t/2]^2]*Sqrt[1 + (1 + 2*I)*Tan[t/2]^2] +
(2 + I)*(Tan[t/2] + 2*Tan[t/2]^3 + 5*Tan[t/2]^5))))/
Sqrt[5 - 4*Cos[t] + Cos[2*t]]

Then find Limits:
Limit[eq, t ->0]

gives 0,

Limit[eq, t -> Pi]

gives

(2/5 + I/5)*Sqrt[2/5 + (4*I)/5]*
((1 + 2*I)*Sqrt[5]*EllipticE[-3/5 - (4*I)/5] -
5*EllipticE[-3/5 + (4*I)/5] + (4 - 4*I)*Sqrt[5]*
EllipticK[-3/5 - (4*I)/5] - (10 - 10*I)*EllipticK[8/5 - (4*I)/5] -
(4 + 8*I)*EllipticPi[1/5 + (2*I)/5, -3/5 + (4*I)/5] -
4*Sqrt[5]*EllipticPi[1 - 2*I, -3/5 - (4*I)/5])

so, the result is
N@%

-6.72287972344033 - 9.947771772989*^-15*I

The problem only with the sign of result.


From: Bob F on 24 Jul 2008 05:04
On Jul 23, 4:26 am, Valeri Astanoff <astan...(a)gmail.com> wrote:
> Good day,
>
> Neither Mathematica 6 nor anyone here can integrate this:
>
> In[1]:= Integrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], {t, 0, Pi}]
> Out[1]= Integrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], {t, 0, Pi}]
>
> In[2]:= NIntegrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], {t, 0, Pi}]
> Out[2]= 6.72288
>
> I know the exact result:
>
> In[3]:= (1/5^(3/4))*(Sqrt[2]*(10*EllipticE[(1/10)*(5 - Sqrt[5])] -
> 10*EllipticK[(1/10)*(5 - Sqrt[5])] + (5 + 3*Sqrt[5])*
> EllipticPi[(1/10)*(5 - 3*Sqrt[5]), (1/10)*(5 - Sqrt[5])])=
)//N
> Out[3]= 6.72288
>
> but I would like to prove it.
>
> Thanks in advance to the samaritan experts...
>
> V.Astanoff

You can use the TrigExpand[] function for the Cos[2t] to get the
equivalent Cos[t]^2 - Sin[t]^2, and if you do this Mathematica 6.0.3
on a Mac comes up with

1/5 Sqrt[2 + 4 I] (-5 I EllipticE[-(3/5) - (4 I)/5] +
(2 + I) Sqrt[5] EllipticE[-(3/5) + (4 I)/5] -
(12 - 4 I) EllipticK[-(3/5) - (4 I)/5] +
(6 - 2 I) Sqrt[5] EllipticK[8/5 - (4 I)/5] +
4 I Sqrt[5] EllipticPi[1/5 + (2 I)/5, -(3/5) + (4 I)/5]
+
(8 + 4 I) EllipticPi[1 - 2 I, -(3/5) - (4 I)/5])

The equivalence to your expression is left up to you...but this does
evaluate numerically to the same as what you had.

-Bob

From: Jens-Peer Kuska on 24 Jul 2008 05:05
Hi,

expr = 5 - 4*Cos[t] + Cos[2*t] // TrigExpand
Integrate[Sqrt[expr], {t, 0, Pi}]

??

Regards
Jens

Valeri Astanoff wrote:
> Good day,
>
> Neither Mathematica 6 nor anyone here can integrate this:
>
> In[1]:= Integrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], {t, 0, Pi}]
> Out[1]= Integrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], {t, 0, Pi}]
>
> In[2]:= NIntegrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], {t, 0, Pi}]
> Out[2]= 6.72288
>
> I know the exact result:
>
> In[3]:= (1/5^(3/4))*(Sqrt[2]*(10*EllipticE[(1/10)*(5 - Sqrt[5])] -
> 10*EllipticK[(1/10)*(5 - Sqrt[5])] + (5 + 3*Sqrt[5])*
> EllipticPi[(1/10)*(5 - 3*Sqrt[5]), (1/10)*(5 - Sqrt[5])]))//N
> Out[3]= 6.72288
>
> but I would like to prove it.
>
> Thanks in advance to the samaritan experts...
>
>
> V.Astanoff
>

From: Valeri Astanoff on 25 Jul 2008 06:14
On 24 juil, 10:53, "David W.Cantrell" <DWCantr...(a)sigmaxi.net> wrote:
[...]
Good day,

Thank you, David (along with Alois, Kevin & Grischka)
for your advice and suggestions.

Though I still wonder why

(1/5)*Sqrt[2+4*I]*(-5*I*EllipticE[-(3/5)-(4*I)/5]+
(2+I)*Sqrt[5]*EllipticE[-(3/5)+(4*I)/5]-
(12-4*I)*EllipticK[-(3/5)-(4*I)/5]+
(6-2*I)*Sqrt[5]*EllipticK[8/5-(4*I)/5]+
4*I*Sqrt[5]*EllipticPi[1/5+(2*I)/5,-(3/5)+(4*I)/5]+
(8+4*I)*EllipticPi[1-2*I,-(3/5)-(4*I)/5])

doesn't fullsimplifies to

(1/5^(3/4))*(Sqrt[2]*(10*EllipticE[(1/10)*(5-Sqrt[5])]-
10*EllipticK[(1/10)*(5-Sqrt[5])]+(5+3*Sqrt[5])*
EllipticPi[(1/10)*(5-3*Sqrt[5]),(1/10)*(5-Sqrt[5])]))

(I have to say I'm not very familiar with the Elliptics!)

Thanks anyway

Valeri


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