       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.

In version 5.0 Mathematica gives

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

which is also incorrect. Numerically,

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

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

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

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.

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)
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Crawley WA 6009                      mailto:paul at physics.uwa.edu.au
AUSTRALIA                            http://physics.uwa.edu.au/~paul

```

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