Re: ArcCos[x] with x > 1

*To*: mathgroup at smc.vnet.net*Subject*: [mg49267] Re: ArcCos[x] with x > 1*From*: John Doty <jpd at whispertel.LoseTheH.net>*Date*: Sat, 10 Jul 2004 02:48:37 -0400 (EDT)*References*: <cclev9$kb3$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

jaegerm at ibmt.fhg.de wrote: > Hi, > > 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]. How does Mathematica calculate ArcCos[x] with x > 1 which > is outside the function's domain [-1,1]? By using Euler's Exp[I x] == Cos[x] + I Sin[x] > In Mathematica's > opinion, ArcCos[2] is a purely imaginary number, so why isn't > i*ArcCos[2] reduced to a real expession? Because (I ArcCos[2]) is a perfectly good analytic expression, and Mathematica has no built-in preference for the representation you want. > How do I get an analytical > expression for the imaginary part of ArcCos[2]? ComplexExpand[I ArcCos[2]] will convert your result to the representation you prefer. -jpd