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R: R: Re: Difficult antiderivative

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  • Subject: [mg128949] R: R: Re: Difficult antiderivative
  • From: "Brambilla Roberto Luigi (RSE)" <Roberto.Brambilla at>
  • Date: Wed, 5 Dec 2012 03:13:38 -0500 (EST)
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Dear Roland,

given the integral

(1)  Integrate[ ArcCosh[a/x]/Sqrt[r^2-x^2],x]   (0<r<a)

Applying your substitution x->r Cos[u] I have found (p=a/r<1)

(2)  Integrate[ ArcCosh[ p Sec[u] ],u]

With your previous substitution x-> a/Cosh[u] I have found

(3)  -p(NIntegrate[u Tanh[u]/Sqrt[Cosh[u]^2 -p^2],u]

And verified that (1),(2) and (3) give the same results when numerically eva
luated on a finite interval (b,r)  with (b<r<a). Unfortunately none of them
 seems to be solvable in the realm (rather large!) of the Mathematica8 funct
May be one has to use higher order trascendent functions? Or, wisely, stop h
Many thanks again for your attention, Rob

-----Messaggio originale-----
Da: Franzius, R. [mailto:roland.franzius at] 
Inviato: marted=EC 4 dicembre 2012 11.13
A: Brambilla Roberto Luigi (RSE)
Oggetto: Re: R: Re: Difficult antiderivative

Am 30.11.2012 14:45, schrieb Brambilla Roberto Luigi (RSE):
> Dear Roland,
> Many thanks for your suggestion, i.e. given the indefinite integral (antid
erivative) 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 o
nly smaller part of them can be expressed in terms of analytical and special
> ....
> For some integrals you can find the solution, for others you cannot, and n
o general rule exists that would help you to distinguish one group from the
> Are (1) and (2) cases of this unhappy class?

There was probably a typo somewhere.

I didn't store the procedure and cam't reproduce the complete polylog 

The most simple algebraic form may be

Sqrt(r^2-x^2)  /. x->r Cos[u]
to get  a Sin[u] which is cancled by the derivative

So you are left with

Integrate[ ArcCosh[ p Csc[u] ], u  ]

This function is a bit strange, the Argument of ArcCosh has to be >1 

  0 < Cos[u] < Min[1,p]

Apply TrigToExp and with some manipulations on the Log you get  a term

Integrate[Log[Cos[u]],u ]

an insolvable rest which my look like

Integrate[ Log[1 + Sqrt[ 1- q^2 Cos[u]^2] ], u ]


Roland Franzius

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