Re: integrate
- To: mathgroup at smc.vnet.net
- Subject: [mg72547] Re: integrate
- From: "dimitris" <dimmechan at yahoo.com>
- Date: Sat, 6 Jan 2007 03:50:33 -0500 (EST)
- References: <emgg3i$2ca$1@smc.vnet.net><en04jk$21i$1@smc.vnet.net>
For two weeks I din't have access to the internet.
Thanks for your response.
Regards
Dimitris
Ï/Ç David W. Cantrell Ýãñáøå:
> "dimitris" <dimmechan at yahoo.com> wrote:
> For your third question:
> > Integrate[Sqrt[x^2 + 1]/Sqrt[1 + x^6], {x, 0, 10}]
> > (-(-1)^(1/6))*EllipticF[I*ArcSinh[10*(-1)^(1/3)], (-1)^(2/3)]
> >
> > N[%]
> > -0.10016641038463325 - 1.6857503548125956*I
> >
> > NIntegrate[Sqrt[x^2 + 1]/Sqrt[1 + x^6], {x, 0, 10}]
> > 2.056349237110889
> >
> > --------> Any ideas to "help" Mathematica to give a correct answer?
>
> I have an answer, but getting there was not pleasant. I hope others
> will have better suggestions.
>
> Part 1:
> I wanted to know why the symbolic answer was wrong, so I looked at.
> In[7]:= Integrate[Sqrt[(1 + x^2)/(1 + x^6)], x]
> Out[7]=
> (-(-1)^(1/6))*Sqrt[1 - (-1)^(1/3)*x^2]*Sqrt[1 + (-1)^(2/3)*x^2]*
> Sqrt[1/(1 - x^2 + x^4)]*EllipticF[I*ArcSinh[(-1)^(1/3)*x], (-1)^(2/3)]
>
> The product of the algebraic factors involving x should be just 1, but
> I was not able to get Mathematica to give that:
>
> In[8]:=
> FullSimplify[Sqrt[1 - (-1)^(1/3)*x^2]*Sqrt[1 + (-1)^(2/3)*x^2]*
> Sqrt[1/(1 - x^2 + x^4)], x > 0]
> Out[8]=
> Sqrt[1 - (-1)^(1/3)*x^2]*Sqrt[(1 + (-1)^(2/3)*x^2)/(1 - x^2 + x^4)]
>
> Maybe I didn't try hard enough. Can someone suggest how I should get
> the above to simplify to 1, perhaps assuming x > 0? Anyway, I then
> simplified Out[7] myself to
>
> In[9]:= -(-1)^(1/6)*EllipticF[I*ArcSinh[(-1)^(1/3)*x], (-1)^(2/3)]
>
> and then plotted the real part of that from x = 0 to 10. A jump
> discontinuity is apparent, at roughly x = 2.45. If we ignore that
> discontinuity and try to evaluate the integral from x = 0 to 10 by the
> Fundamental Theorem, we get the same wrong answer that Mathematica
> got. So now I think I know what Mathematica did wrong. But I'm a bit
> surprised. Even when Mathematica returns antiderivatives having such
> discontinuities, it normally takes such discontinuities correctly into
> account when doing _definite_ integrals. But it didn't do so here, so
> there is indeed a bug to be exterminated.
>
> Part 2:
> I thought I would help Mathematica by taking the discontinuity, caused
> by a branch cut, into account by hand. But unfortunately it wasn't
> obvious exactly where the cut was. So I went to the Wolfram Functions
> site and looked under EllipticF. It was not helpful. The only relevant
> statement was "Branch cut locations: complicated." I then gave up on
> determining the exact location. But if I could have done that, it
> would have been nicer than what follows. (Anyone know exactly where
> the discontinuity is?)
>
> Part 3:
> I sought to transform the original integral into another one which
> Mathematica might handle better. My first thought was to use the
> substitution u = x^2, which led to
>
> In[16]:= 1/2*Integrate[Sqrt[(1 + u)/(u*(1 + u^3))], u]
> Out[16]=
> (-1)^(1/6)*Sqrt[1 - (-1)^(1/3)/u]*Sqrt[1 + (-1)^(2/3)/u]*u^(3/2)*
> Sqrt[1/(u - u^2 + u^3)]*EllipticF[I*ArcSinh[(-1)^(1/3)/Sqrt[u]],(-1)^(2/3)]
>
> Simplifying by hand and putting the result back in terms of x gave
>
> In[17]:= (-1)^(1/6)*EllipticF[I*ArcSinh[(-1)^(1/3)/x], (-1)^(2/3)].
>
> Although different from In[9], this result still has a discontinuity.
> But thankfully, it's in a different place, roughly x = 0.4. That
> allows us to combine In[9] and In[17], using the former to integrate
> from x0 to 1 and the latter to integrate from 1 to x1. Thus, assuming
> x0 < 1 < x1, we can state that
>
> Integrate[Sqrt[(1 + x^2)/(1 + x^6)], {x, x0, x1}]
>
> is
>
> In[21]:=
> int = (-1)^(1/6)*(-2*EllipticF[I*ArcSinh[(-1)^(1/3)], (-1)^(2/3)] +
> EllipticF[I*ArcSinh[(-1)^(1/3)*x0], (-1)^(2/3)] +
> EllipticF[I*ArcSinh[(-1)^(1/3)/x1], (-1)^(2/3)]);
>
> In[22]:= int /. {x0 -> 0, x1 -> 10}
> Out[22]=
> (-1)^(1/6)*(EllipticF[I*ArcSinh[(1/10)*(-1)^(1/3)], (-1)^(2/3)] -
> 2*EllipticF[I*ArcSinh[(-1)^(1/3)], (-1)^(2/3)])
>
> which is the precise symbolic result desired.
>
> In[23]:= N[%]
> Out[23]= 2.0563492371150103 + 3.3306690738754696*^-16*I
>
> David W. Cantrell