Re: ArcCos[x] with x > 1
- To: mathgroup at smc.vnet.net
- Subject: [mg49304] Re: ArcCos[x] with x > 1
- From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
- Date: Tue, 13 Jul 2004 04:32:35 -0400 (EDT)
- References: <cclev9$kb3$1@smc.vnet.net> <cco3io$4ig$1@smc.vnet.net> <cctaff$c11$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
"Carl K. Woll" <carlw at u.washington.edu> wrote: > "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? Just as Paul said: Compare the integrand with the derivative of the supposed indefinite integral. Out[2] shows that the difference is not zero. In[1]:= expr = r*Sqrt[1 + 4*r^2]*ArcCos[1/(2*r)]; Simplify[D[Integrate[expr, r], r] - expr] Out[1]= (* snip of a messy expression in terms of r *) In[2]:= FullSimplify[%, 1/2 < r < 1/Sqrt[2]] Out[2]= r/3 - Sqrt[-1 + 16*r^4]/(12*r) which is clearly not identically zero. > In your case, But Paul's case is for version 5, as is what I showed above. So at least in the current version, some indefinite integrals are wrong. David Cantrell > 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