Thursday, 19 May 2016

calculus - Expressing indefinite integrals in terms of a predefined set of functions.



It is well known that some integrals of elementary functions cannot be expressed as elementary functions.



I was wondering if it was possible to extend the set of elementary operators by some additional set, so that all integrals of elementary functions can be expressed in terms of the new enlarged set. Of course the additional members would have to be defined as certain integrals or more generally as certain solutions to given differential equations.




It is interesting if such an extension set exists which is finite. If not finite, does it at least have some structure?


Answer



The closure under integration of the set of the elementary functions is the set of the Liouvillian functions.
[Wikipedia: Nonelementary integral]



And here different sets of functions are possible based on other differential fields. The definition of this classes of functions is given e.g. in section 1 of Davenport, J. H.: What Might "Understand a Function" Mean. In: Kauers, M.; Kerber, M., Miner, R.; Windsteiger, W.: Towards Mechanized Mathematical Assistants. Springer, Berlin/Heidelberg, 2007, page 55-65.



There is an uncountably infinite number of elementary functions non-integrable in the elementary functions. See my answer here.


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