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Re: get rid of I in the result of an integral
*To*: mathgroup at smc.vnet.net
*Subject*: [mg116537] Re: get rid of I in the result of an integral
*From*: Ruth Lazkoz SÃez <ruth.lazkoz at ehu.es>
*Date*: Fri, 18 Feb 2011 04:37:34 -0500 (EST)
Thanks for the tip, I would however like to get the result expressed
using ArcSech[x] without having to tell Mathematica explictly to do
/.ArcSec[x]->-I ArcSech[x] at the end
El 17/02/11 14:04, Bob Hanlon escribi=C3=B3:
> sol1 = Assuming[{Element[x, Reals], 0< x< 1},
> Integrate[1/((1 + y)^2 Sqrt[-x^2 + y^2]), {y, x, Infinity}]]
>
> (-1 + x^2 - I*Sqrt[1 - x^2]*ArcSec[x])/(-1 + x^2)^2
>
> sol2 = Assuming[{Element[x, Reals], 0< x< 1},
> FullSimplify[
> Integrate[1/((1 + y)^2 Sqrt[-x^2 + y^2]), {y, x, Infinity}] //
> TrigToExp]]
>
> (-1 + x^2 + Sqrt[1 - x^2]*Log[(1 + Sqrt[1 - x^2])/x])/(-1 + x^2)^2
>
> Using the form Assuming[{ _ }, _ ], makes the assumtions available to both FullSimplify and Integrate without having to repeat the assumptions.
>
> Assuming[{Element[x, Reals], 0< x< 1},
> FullSimplify[sol1 - sol2 == 0]]
>
> True
>
>
> Bob Hanlon
>
> ---- "Ruth Lazkoz S=C3=A1ez"<ruth.lazkoz at ehu.es> 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
>
>
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