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Re: ArcCos[x] with x > 1

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  • Subject: [mg49291] Re: ArcCos[x] with x > 1
  • From: "Carl K. Woll" <carlw at>
  • Date: Mon, 12 Jul 2004 02:11:31 -0400 (EDT)
  • Organization: University of Washington
  • References: <cclev9$kb3$> <cco3io$4ig$>
  • Sender: owner-wri-mathgroup at

"Paul Abbott" <paul at> wrote in message
news:cco3io$4ig$1 at
> In article <cclev9$kb3$1 at>, jaegerm at 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].


> 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

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

> 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
> AUSTRALIA                  

Carl Woll

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