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MathGroup Archive 2007

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Re: Integrate

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74476] Re: Integrate
  • From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
  • Date: Thu, 22 Mar 2007 01:19:11 -0500 (EST)
  • References: <etikok$j7r$1@smc.vnet.net> <etnkdg$gp5$1@smc.vnet.net> <etqnvl$tq$1@smc.vnet.net>

"dimitris" <dimmechan at yahoo.com> wrote:
> Hello.
>
> It is interesting to get responses in this thread because for some
> time I believe
> I was alone!
>
> Easily you can get an antiderivative real in the ntegration range:
>
> (*INs*)
> f[x_]=Log[Sin[x]^2]*Tan[x]
> integrand = f[x]*dx /. x -> ArcSin[Sqrt[u]] /. dx ->
> D[ArcSin[Sqrt[u]], u]
> ff=Integrate[integrand, {u, 0, Sin[z]^2}, Assumptions -> 0 < z < Pi]
> Simplify[D[ff, z]] /. z -> x
>
> (*OUTs*)
> Log[Sin[x]^2]*Tan[x]
> Log[u]/(2*(1 - u))
> (1/12)*(-Pi^2 + 6*PolyLog[2, Cos[z]^2])

Thanks, Dimitris! That's a far simpler result than the one I had given, and
it's easily obtained using Mathematica!

David

> 0
> Log[Sin[x]^2]*Tan[x]
>
> Plot[ff/.z->x,{x,0,Pi}];
>
> Dimitris
>
> =CF/=C7 David W.Cantrell =DD=E3=F1=E1=F8=E5:
> > "dimitris" <dimmechan at yahoo.com> wrote:
> > > Hello again!
> > >
> > > Of course sometimes things work quite unexpectedly!
> > >
> > > Consider again the integral (no! I am not obsessed with it!)
> > >
> > > Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi}]
> > > Integrate::idiv: Integral of Log[Sin[x]^2]*Tan[x] does not converge
> > > on {x,0,Pi}.
> >
> > Obsessed or not, it is a curious bug, being platform dependent.
> >
> > You might be interested in an antiderivative, not directly obtainable
> > from Mathematica AFAIK, which is valid over the whole real line:
> >
> > Letting u = Abs[Cos[x]],
> >
> > Integrate[Log[Sin[t]^2]*Tan[t], {t, 0, x}]
> >
> > is
> >
> > Log[2]^2 - Pi^2/3 + 2 Log[1 + u] Log[(1 + 1/u)/2] +
> > 4 Log[Sqrt[2/(1 + u)]] Log[Sqrt[(1 - u)/2]] + 2 PolyLog[2, 1/(1 + u)]
> > + PolyLog[2, 2 - 2/(1 + u)] + PolyLog[2, 1 - 2/(1 + u)]
> >
> > I'm not sure whether the result above could be simplified further or
> > not.
> >
> > David W. Cantrell


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