Re: Indefinite integration problem on ArcTanh(f(Cos))
- To: mathgroup at smc.vnet.net
- Subject: [mg69959] Re: Indefinite integration problem on ArcTanh(f(Cos))
- From: dimmechan at yahoo.com
- Date: Thu, 28 Sep 2006 06:15:50 -0400 (EDT)
- References: <efdks2$12a$1@smc.vnet.net>
Dear Andrew
I will concentrate on the mathematical aspects of your question.
In what follows I have converted everything in InputForm.
$VersionNumber
5.2
ac[k_, t_] = Sin[t]/(1 + k*Cos[2*t])
Sin[t]/(1 + k*Cos[2*t])
vel[k_, t_] = FullSimplify[Integrate[ac[k, t], t], 0 < k < 1]
ArcTanh[(Sqrt[2]*Cos[t])/Sqrt[(-1 + k)/k]]/(Sqrt[2]*Sqrt[(-1 + k)*k])
FullSimplify[D[vel[k, t], t], 0 < k < 1] == ac[k, t] (*check*)
True
pos[k_, t_] = FullSimplify[Integrate[vel[k, t], t], 0 < k < 1]
(1/(8*Sqrt[2]*Sqrt[(-1 + k)*k]))*(4*t*(Log[1 - I*Sqrt[2]*Sqrt[-(k/(-1 +
k))]*Cos[t]] -
Log[1 + I*Sqrt[2]*Sqrt[-(k/(-1 + k))]*Cos[t]]) + RootSum[k +
2*#1^2 + k*#1^4 & ,
(1/(k + #1^2))*(-2*k*t*Log[1 - E^(I*t)*#1] + 2*I*k*PolyLog[2,
E^(I*t)*#1] + Sqrt[2]*Sqrt[k/(-1 + k)]*t*Log[1 - E^(I*t)*#1]*
#1 - Sqrt[2]*k*Sqrt[k/(-1 + k)]*t*Log[1 - E^(I*t)*#1]*#1 -
I*Sqrt[2]*Sqrt[k/(-1 + k)]*PolyLog[2, E^(I*t)*#1]*#1 +
I*Sqrt[2]*k*Sqrt[k/(-1 + k)]*PolyLog[2, E^(I*t)*#1]*#1 -
2*t*Log[1 - E^(I*t)*#1]*#1^2 + 2*I*PolyLog[2, E^(I*t)*#1]*#1^2 -
Sqrt[2]*Sqrt[k/(-1 + k)]*t*Log[1 - E^(I*t)*#1]*#1^3 +
Sqrt[2]*k*Sqrt[k/(-1 + k)]*t*Log[1 - E^(I*t)*#1]*#1^3 +
I*Sqrt[2]*Sqrt[k/(-1 + k)]*PolyLog[2, E^(I*t)*#1]*#1^3 -
I*Sqrt[2]*k*Sqrt[k/(-1 + k)]*PolyLog[2, E^(I*t)*#1]*#1^3) & ] -
RootSum[k + 2*#1^2 + k*#1^4 & , (1/(k + #1^2))*(-2*k*t*Log[1 -
E^(I*t)*#1] + 2*I*k*PolyLog[2, E^(I*t)*#1] -
Sqrt[2]*Sqrt[k/(-1 + k)]*t*Log[1 - E^(I*t)*#1]*#1 +
Sqrt[2]*k*Sqrt[k/(-1 + k)]*t*Log[1 - E^(I*t)*#1]*#1 +
I*Sqrt[2]*Sqrt[k/(-1 + k)]*PolyLog[2, E^(I*t)*#1]*#1 -
I*Sqrt[2]*k*Sqrt[k/(-1 + k)]*PolyLog[2, E^(I*t)*#1]*#1 -
2*t*Log[1 - E^(I*t)*#1]*#1^2 + 2*I*PolyLog[2, E^(I*t)*#1]*#1^2 +
Sqrt[2]*Sqrt[k/(-1 + k)]*t*Log[1 - E^(I*t)*#1]*#1^3 -
Sqrt[2]*k*Sqrt[k/(-1 + k)]*t*Log[1 - E^(I*t)*#1]*#1^3 -
I*Sqrt[2]*Sqrt[k/(-1 + k)]*PolyLog[2, E^(I*t)*#1]*#1^3 +
I*Sqrt[2]*k*Sqrt[k/(-1 + k)]*PolyLog[2, E^(I*t)*#1]*#1^3) & ])
FullSimplify[D[pos[k,t]]< k < 1]
-((I*(Log[1 - I*Sqrt[2]*Sqrt[-(k/(-1 + k))]*Cos[t]] - Log[1 +
I*Sqrt[2]*Sqrt[-(k/(-1 + k))]*Cos[t]]))/
(2*Sqrt[2]*Sqrt[(-(-1 + k))*k]))
FullSimplify[% - vel[k, t], 0 < k < 1] (*check*)
0
Plot[ac[1/10, t], {t, 0, 10}]
Plot[vel[1/10, t], {t, 0, 10}]
Everything seems fine up to here, both physically and mathematically.
However the obtained pos[k,t] function expressed in terms of polylogs
has imaginary and real parts.
Information[PolyLog]
"PolyLog[n, z] gives the nth polylogarithm function of z. PolyLog[n, p,
z] gives the Nielsen generalized
polylogarithm function of z for parameters n and p."
Attributes[PolyLog] = {Listable, NumericFunction, Protected,
ReadProtected}
The obtained indefinite integral is correct as you see but it does not
make sense physically
(at least to me).
I do not know if and how you can persuade Mathematica to give another
indefinite integral.
Anyway for history
Plot[Re[pos[1/10, t]], {t, 0, 10}]
Plot[Im[pos[1/10, t]], {t, 0, 10}]
I hope others' replies to be more helpful for you.
Regards
Dimitris
P.S. I have worked also with another symbolic system and I got almost
similar results.
f:=sin(t)/(1+k*cos(2*t));
f := sin(t)/(1+k*cos(2*t))
plot(subs(k=1/10,f),t=0..10);
g:=int(f,t);
g :=
1/2*2^(1/2)/((-1+k)*k)^(1/2)*arctanh(k*cos(t)*2^(1/2)/((-1+k)*k)^(1/2))
plot(subs(k=1/10,g),t=0..10);
h:=int(g,t);
(I do not present the result due to their length)
hh:=simplify(h) assuming(0<k,k<1);
(I do not present the result due to their length)
plot(subs(k=1/10,Re(hh)),t=0..10);
plot(subs(k=1/10,Im(hh)),t=0..10);