Re: ArcCos[x] with x > 1
- To: mathgroup at smc.vnet.net
- Subject: [mg49291] Re: ArcCos[x] with x > 1
- From: "Carl K. Woll" <carlw at u.washington.edu>
- Date: Mon, 12 Jul 2004 02:11:31 -0400 (EDT)
- Organization: University of Washington
- References: <cclev9$kb3$1@smc.vnet.net> <cco3io$4ig$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
"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[2]. snipped > > 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 1/2<r<1/Sqrt[2]. 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