       Re: ArcCos[x] with x > 1

• To: mathgroup at smc.vnet.net
• Subject: [mg49304] Re: ArcCos[x] with x > 1
• From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
• Date: Tue, 13 Jul 2004 04:32:35 -0400 (EDT)
• 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

```"Carl K. Woll" <carlw at u.washington.edu> wrote:
> "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.
>
> snipped
>
> >
> > In fact, Mathematica 5.0 gets the indefinite integral wrong:
> >
> >   Simplify[Integrate[r ArcCos[1/(2r)] Sqrt[4r^2 + 1], r],
> >     1/Sqrt > 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) -  (% /. 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?

Just as Paul said: Compare the integrand with the derivative of the
supposed indefinite integral. Out shows that the difference is not zero.

In:=
expr = r*Sqrt[1 + 4*r^2]*ArcCos[1/(2*r)];
Simplify[D[Integrate[expr, r], r] - expr]

Out=
(* snip of a messy expression in terms of r *)

In:=
FullSimplify[%, 1/2 < r < 1/Sqrt]

Out=
r/3 - Sqrt[-1 + 16*r^4]/(12*r)

which is clearly not identically zero.

But Paul's case is for version 5, as is what I showed above. So at least
in the current version, some indefinite integrals are wrong.

David Cantrell

> 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. 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.
>
> 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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