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Re: Difficult antiderivative
*To*: mathgroup at smc.vnet.net
*Subject*: [mg128853] Re: Difficult antiderivative
*From*: Roland Franzius <roland.franzius at uos.de>
*Date*: Fri, 30 Nov 2012 05:58:25 -0500 (EST)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*Delivered-to*: l-mathgroup@wolfram.com
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*References*: <k94hra$bk4$1@smc.vnet.net>
Am 28.11.2012 09:26, schrieb Brambilla Roberto Luigi (RSE):
> Using Mathematica (v5.1 to v.8) it is possible to obtain (in few
> seconds) the following antiderivative
>
>
>
> (1) Integrate[ x ArcCosh[a/x]/Sqrt[r^2-x^2],x]
>
>
>
> (a rather lengthy expression).
>
>
>
> It happens that I need
>
>
>
> (2) Integrate[ ArcCosh[a/x]/Sqrt[r^2-x^2],x]
> or equivalently
>
>
>
> (3) Integrate[ x^2 ArcCosh[a/x]/Sqrt[r^2-x^2],x]
> I have tried a lot of tricks (by part integration etc..) but I was not
> able to find it.
>
> Q.: may be that the antiderivative does not exist?
>
> The numerical integral (b<r<a)
>
>
>
> NIntegrate[ ArcCosh[a/x]/Sqrt[r^2-x^2],{x,b,r}]
>
> Don't give any problem.
x/Sqrt[a-x^2 ] is a simple derivative of Sqrt[a-x^2].
Substitutuing x-> a/Cosh[u], with the right assumptions about r>x>0, I
get an expression with Logs and Polylogs. But itd to tedious to check it
by resubstitution and FullSimplifying the derivative.
A crucial point is to replace
Sqrt[r^2- a^2/Cosh[u]^2 ] -> Sqrt[q^2 Cosh[u]^2-1]
in order to get a managable expression.
--
Roland Franzius
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