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MathGroup Archive 2004

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Re: ArcCos[x] with x > 1

  • To: mathgroup at smc.vnet.net
  • Subject: [mg49260] Re: ArcCos[x] with x > 1
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Sat, 10 Jul 2004 02:48:31 -0400 (EDT)
  • Organization: The University of Western Australia
  • References: <cclev9$kb3$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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]. 

In version 5.0 Mathematica gives

 (1/48) (2 - 4 Sqrt[3] + 3 Sqrt[3] Pi - 4 Log[2 + Sqrt[3]])

which is also incorrect. Numerically, 

  NIntegrate[r ArcCos[1/(2r)] Sqrt[*r^2 + 1], {r, 1/2, 1/Sqrt[2]}]
  0.114539

while the above result is 0.12767. The correct exact answer is

 (1/48) (-2/3 Pi - 2 Sqrt[3] + 3 Sqrt[3] Pi - 4 Log[2 + Sqrt[3]])

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.
 
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


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