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

```

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