Re: Indefinite integration problem on ArcTanh(f(Cos))

• To: mathgroup at smc.vnet.net
• Subject: [mg70251] Re: Indefinite integration problem on ArcTanh(f(Cos))
• From: "David W. Cantrell" <DWCantrell at sigmaxi.net>
• Date: Tue, 10 Oct 2006 06:11:57 -0400 (EDT)
• References: <efdks2\$12a\$1@smc.vnet.net> <efgbpj\$qg0\$1@smc.vnet.net>

```"David W.Cantrell" <DWCantrell at sigmaxi.net> wrote:
> "ANDREW PALFREYMAN" <andrew_ppp at hotmail.com> wrote:
> > Hi.
> > I'm integrating Sin[t] / (1 + k Cos[2t]) , over 0 < k < 1, using
> > Mathematica 4.1. I'm plotting spot values k=0.1, 0.8, 0.95 over  0 < t
> > < 10.
> >
> > This represents the acceleration of a charged particle in an
> > alternating electric field, where k represents a particular modulation
> > of its effective inertia. Although the velocity plots (the integral dt)
> > look fine (symmetrical about zero), the plots for position (integrating
> > velocity dt) are clearly wrong; I get a massive expression (about 20
> > lines!) which will not differentiate successfully back to the original
> > velocity expression, and they are decidedly asymmetrical about zero.
> > The position plot ought to look like a sinewave as k->0, but there's no
> > resemblance.
>
> I don't know what answer version 4.1 gives, but the current version's
> answer seems to be correct. Did you think of trying The Wolfram
> Integrator
>
> <http://integrals.wolfram.com/index.jsp> ?
>
> It's free and gives the same result for the indefinite integral as the
> current version of Mathematica.
>
> Anyway, if 0 < k < 1, perhaps the nicest way to write the antiderivative
> is
>
> - ArcTan[Sqrt[2k/(1-k)] Cos[t]] / Sqrt[2k(1-k)].

Belatedly, I realize that I did not address your concern. I apologize. I
now see that you wanted a _second_ antiderivative.

Using version 5.2, if we get an antiderivative for Sin[t]/(1 + k Cos[2t])
and then get an antiderivative of that result, we do indeed obtain a messy
expression. (I used 0 for both constants of integration.) The expression
has an imaginary part and the real part of the expression is not periodic,
due to jump discontinuities. But if we take just the real part of the
expression and get rid of the jumps, then we obtain, as I suppose you
wanted, a periodic real function. The result is shown below my signature.
The first line, (Pi^2/Sqrt[2*k*(1 - k)])*Round[t/(2*Pi)]*Sign[Cos[t]], was
devised by me to get rid of the spurious jump discontinuities. I suggest
that you try plotting my result using various values of k between 0 and 1.

I suspect that the result could be expressed in a nicer way, but offhand I
don't know what that might be.

Regards,
David W. Cantrell

----------------------------------------------

(Pi^2/Sqrt[2*k*(1 - k)])*Round[t/(2*Pi)]*Sign[Cos[t]] +

Re[(-(1/(2*Sqrt[2]*Sqrt[(-(-1 + k))*k])))*
(2*t*ArcTan[Sqrt[2]*Sqrt[-(k/(-1 + k))]*Cos[t]] -
4*ArcSin[Sqrt[Sqrt[2] + (I*Sqrt[(-(-1 + k))*k])/
k]/2^(3/4)]*ArcTan[
((-I + Sqrt[2]*Sqrt[-(k/(-1 + k))])*Tan[t/2])/
Sqrt[(1 + k)/(-1 + k)]] +
4*ArcSin[Sqrt[Sqrt[2] - (I*Sqrt[(-(-1 + k))*k])/
k]/2^(3/4)]*ArcTan[
((I + Sqrt[2]*Sqrt[-(k/(-1 + k))])*Tan[t/2])/
Sqrt[(1 + k)/(-1 + k)]] -
I*t*Log[(1/2)*(2 + I*Sqrt[2]*E^(I*t)*
(Sqrt[-1 + 1/k] - Sqrt[1 + 1/k]))] -
2*I*ArcSin[Sqrt[Sqrt[2] + (I*Sqrt[(-(-1 + k))*k])/
k]/2^(3/4)]*Log[(1/2)*(2 + I*Sqrt[2]*
E^(I*t)*(Sqrt[-1 + 1/k] - Sqrt[1 + 1/k]))] +
I*t*Log[(1/2)*(2 - I*Sqrt[2]*E^(I*t)*
(Sqrt[-1 + 1/k] + Sqrt[1 + 1/k]))] +
2*I*ArcSin[Sqrt[Sqrt[2] - (I*Sqrt[(-(-1 + k))*k])/
k]/2^(3/4)]*Log[(1/2)*(2 - I*Sqrt[2]*
E^(I*t)*(Sqrt[-1 + 1/k] + Sqrt[1 + 1/k]))] +
I*t*Log[(1/2)*(2 + E^(I*t)*Sqrt[-2 + 2/k]*
(-I + Sqrt[(1 + k)/(-1 + k)]))] -
2*I*ArcSin[Sqrt[Sqrt[2] - (I*Sqrt[(-(-1 + k))*k])/
k]/2^(3/4)]*Log[(1/2)*
(2 + E^(I*t)*Sqrt[-2 + 2/k]*(-I +
Sqrt[(1 + k)/(-1 + k)]))] -
I*t*Log[(1/2)*(2 + E^(I*t)*Sqrt[-2 + 2/k]*
(I + Sqrt[(1 + k)/(-1 + k)]))] +
2*I*ArcSin[Sqrt[Sqrt[2] + (I*Sqrt[(-(-1 + k))*k])/
k]/2^(3/4)]*Log[(1/2)*
(2 + E^(I*t)*Sqrt[-2 + 2/k]*
(I + Sqrt[(1 + k)/(-1 + k)]))] +
PolyLog[2, -((E^(I*t)*(-I + Sqrt[(1 + k)/
(-1 + k)]))/(Sqrt[2]*
Sqrt[-(k/(-1 + k))]))] - PolyLog[2,
(E^(I*t)*(-I + Sqrt[(1 + k)/(-1 + k)]))/
(Sqrt[2]*Sqrt[-(k/(-1 + k))])] -
PolyLog[2, -((E^(I*t)*(I + Sqrt[(1 + k)/
(-1 + k)]))/(Sqrt[2]*
Sqrt[-(k/(-1 + k))]))] + PolyLog[2,
(E^(I*t)*(I + Sqrt[(1 + k)/(-1 + k)]))/
(Sqrt[2]*Sqrt[-(k/(-1 + k))])])]

```

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