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
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