Re: get rid of I in the result of an integral
- To: mathgroup at smc.vnet.net
- Subject: [mg116538] Re: get rid of I in the result of an integral
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Fri, 18 Feb 2011 04:37:45 -0500 (EST)
Ruth Lazkoz S=E1ez wrote: > Hi, > > When I do Integrate[1/((1 + y)^2 Sqrt[-x^2 + y^2]), {y, x, Infinity}, > Assumptions -> x^2 < 1 && 1 > x > 0 && Element[x, Reals]] > > I get > > (-1 + x^2 - I Sqrt[1 - x^2] ArcSec[x])/(-1 + x^2)^2 > > and I would like the result expresed in terms of ArcSech[x] so that I > does not appear. > > I know I can tell Mathematica to do the replacement afterwards, but I > want to show the result to a colleague who is not so familiar with > mathematica, and if it were possible to get the result in one go just by= > adding some extra assumption or so, it would be more convincing. > > Help will be appreciated. Best, > > Ruth If you TrigExpand and then (Full)Simplify, that seem to work. InputForm[ii == Integrate[1/((1 + y)^2*Sqrt[-x^2 + y^2]), {y,x,Infinity}, Assumptions->1>x>0]] Out[13]//InputForm== (-1 + x^2 - I*Sqrt[1 - x^2]*ArcSec[x])/(-1 + x^2)^2 InputForm[FullSimplify[TrigToExp[ii], Assumptions->0<x<1]] Out[16]//InputForm== (-1 + x^2 + Sqrt[1 - x^2]*Log[(1 + Sqrt[1 - x^2])/x])/(-1 + x^2)^2 Daniel Lichtblau Wolfram Research