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MathGroup Archive 2004

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Re: ArcCos[x] with x > 1

  • To: mathgroup at smc.vnet.net
  • Subject: [mg49306] Re: ArcCos[x] with x > 1
  • From: "Carl K. Woll" <carlw at u.washington.edu>
  • Date: Tue, 13 Jul 2004 04:32:38 -0400 (EDT)
  • Organization: University of Washington
  • References: <cclev9$kb3$1@smc.vnet.net> <cco3io$4ig$1@smc.vnet.net> <cctaff$c11$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Paul,

Sorry, I was too hasty in my post. The result returned by Mathematica 5 that
you posted is certainly incorrect.

Since it seems that Mathematica 4 always produced correct indefinite
integrals, Mathematica 5 has certainly taken a step backward in producing
indefinite integrals.

Carl Woll

"Carl K. Woll" <carlw at u.washington.edu> wrote in message
news:cctaff$c11$1 at smc.vnet.net...
> "Paul Abbott" <paul at physics.uwa.edu.au> wrote in message
> news:cco3io$4ig$1 at smc.vnet.net...
> > In article <cclev9$kb3$1 at smc.vnet.net>, jaegerm at ibmt.fhg.de wrote:
> >
> > > as the result of a definite integral (Integrate[r ArcCos[1/(2 r)] (1 +
> > > 4 r^2)^(1/2), {r, 1/2, 1/2^(1/2)}]) I received the expression
> > > i*ArcCos[2].
>
> snipped
>
> >
> > In fact, Mathematica 5.0 gets the indefinite integral wrong:
> >
> >   Simplify[Integrate[r ArcCos[1/(2r)] Sqrt[4r^2 + 1], r],
> >     1/Sqrt[2] > r > 1/2]
> >
> > returns
> >
> >   (1/12) (2r^2 + (4r^2 + 1)^(3/2) ArcCos[1/(2r)] -
> >    Log[4r^2 + Sqrt[16r^4 - 1]] - Sqrt[16r^4 - 1])
> >
> > Evaluating this result at the endpoints
> >
> >   Simplify[(% /. r -> 1/Sqrt[2]) -  (% /. r -> 1/2)]
> >
> > gives the incorrect answer above. Comparing the integrand with the
> > derivative with respect to r shows that the indefinite integral is
wrong.
> >
>
> In my experience with Integrate, the only errors that occur in its use
arise
> when Integrate is used to compute definite integrals. These errors are
> produced when the two endpoints lie on different branches of thefunctions
in
> the indefinite integral. I have never seen a case where an indefinite
> integral will produce an incorrect result, although my experience is
> strictly limited to versions of Mathematica prior to 5.0.
>
> Why do you state that the indefinite integral is incorrect? In your case,
I
> plotted the difference between the derivative of the indefinite integral
and
> the integrand and the result was zero, so that the indefinite integral is
> indeed correct, at least for the range 1/2 < r < 1/Sqrt[2]. Outside this
> range, in particular for r<1/2, the derivative of the the indefinite
> integral and the integrand are different, but this is to be expected,
since
> your simplify expression assumed that r>1/2. That is, there is no reason
to
> expect that the derivative of the indefinite integral and the integrand
> should be functionally equivalent, so that applying FullSimplify to the
> difference need not be zero. They only need to be equal on the range
> 1/2<r<1/Sqrt[2].
>
> Unfortunately, I think the use of Integrate for definite integrals may
> always have these problems with branches. After all, even if it were
> possible to rotate the branch cuts of functions on the fly, it will not
> always be possible to orient all of the branch cuts of the functions that
> arise in the indefinite integral so that they don't cross the path of
> integration.
>
> > Cheers,
> > Paul
> >
> > -- 
> > Paul Abbott                                   Phone: +61 8 9380 2734
> > School of Physics, M013                         Fax: +61 8 9380 1014
> > The University of Western Australia      (CRICOS Provider No 00126G)
> > 35 Stirling Highway
> > Crawley WA 6009                      mailto:paul at physics.uwa.edu.au
> > AUSTRALIA                            http://physics.uwa.edu.au/~paul
> >
>
> Carl Woll
>
>



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