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R: Re: Difficult antiderivative

  • To: mathgroup at smc.vnet.net
  • Subject: [mg128868] R: Re: Difficult antiderivative
  • From: "Brambilla Roberto Luigi (RSE)" <Roberto.Brambilla at rse-web.it>
  • Date: Sat, 1 Dec 2012 04:34:50 -0500 (EST)
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Dear Roland,

Many thanks for your suggestion, i.e. given the indefinite integral (antiderivative) with r<a :

1)  Integrate[ ArcCosh[a/x]/Sqrt[r^2-x^2],x]

(that Mathematica8 can't solve) change the variable x->a/Cosh[u].
Doing the substitution the integral becomes
  
2)  Integrate[ u Tanh[u]/Sqrt[q^2 Cosh[u]^2-1],x]

where q=r/a (r<a). Unfortunately also this integral is unsolvable by
Mathematica8 (unless q=1). 

Alexei Boulbitch wrote (29 november) [mg128833]
....
most of indefinite integrals have this property ("does not exist") , and only smaller part of them can be expressed in terms of analytical and special functions.  
....
For some integrals you can find the solution, for others you cannot, and no general rule exists that would help you to distinguish one group from the other. 

Are (1) and (2) cases of this unhappy class?

Cheers to all, Rob

-----Messaggio originale-----
Da: Roland Franzius [mailto:roland.franzius at uos.de] 
Inviato: venerd=EC 30 novembre 2012 11.58
A: mathgroup at smc.vnet.net
Oggetto: Re: Difficult antiderivative

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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