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R: Re: Difficult antiderivative

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  • Subject: [mg128842] R: Re: Difficult antiderivative
  • From: "Brambilla Roberto Luigi (RSE)" <Roberto.Brambilla at rse-web.it>
  • Date: Fri, 30 Nov 2012 05:54:45 -0500 (EST)
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Many thanks Alexei! You comments is very illuminating!

I've also found that the more general integral

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

"exist" (by  means of elementary functions or usual special functions
Used by Mathematica) for  m=1,3,5... but nor for m=0,2,4...

Another  antiderivative (to which the previous can be reduced)is

Integrate[ (x^m)ArcSin[k/x]/Sqrt[1-x^2],x]

In this case Mathematica solves it for
 k =|= 1  only if m is odd but for k=1 Mathematica solves 
also for m even.

I'm asking if there exist any general criterion 
(at least for simple combinations of elementary functions, as in my examples) that tell us about the existence of antiderivative
in the field of a set of chosen elementary functions.
Can I add to this set other less elementary functions (like Pailev=E9 trascendentans) in order to catch the missing antiderivative?

Cheers, Rob




-----Messaggio originale-----
Da: Alexei Boulbitch [mailto:Alexei.Boulbitch at iee.lu] 
Inviato: gioved=EC 29 novembre 2012 12.06
A: mathgroup at smc.vnet.net
Oggetto: Re: Difficult antiderivative


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.

Any help very appreciated (and considered).

Cheers, Rob

Hi, Rob,

Yes, it is a general case that indefinite integrals cannot be expressed in 
terms of some finite combination of analytical and special functions. That 
is a more mathematically correct expression of the thing you obviously have
 in mind when writing "does not exist".

Even more, most of indefinite integrals have this property, and only smaller part of them can be expressed in terms of analytical and special functions.  It is also common that the indefinite integral "does not exist" (using your expression), while the definite one does.

Have fun, Alexei


Alexei BOULBITCH, Dr., habil.
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