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Elementary Function with Non-elementary Derivative

  1. Mar 19, 2013 #1
    Does such a function exist? My gut tells me that such a function should not exist, but is there a proof that all elementary functions have elementary derivatives?
  2. jcsd
  3. Mar 19, 2013 #2


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    I believe that you could make a finite list of all elementary functions and then show that they all have derivatives.
  4. Mar 19, 2013 #3


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    I think this is a question about language not math, because "elementary functions" are whatever functions people decide to call "elementary". AFAIK the phrase "elementary function" doesn't have any mathematical siignificance (unlike "analytic function", for example).

    A common reason for inventing a new "elementary" function is to give a name to the indefinite integral of some function that has a "practical" use (in physics or engineering) - which explaiins why most (if not all) of them have elementary derivatives.
  5. Mar 19, 2013 #4


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    Elementary functions were introduced by Liouville. They are well defined.

    It is currently the largest function space on which the Risch algorithm is known to be decidable.
  6. Mar 20, 2013 #5
    Roughly I agree, but with some nuances.
    First, I prefer to say : A common reason for inventing a new "special" function is to give a name to the indefinite integral ... etc.
    Second, if a "special" function becomes of common use and is known by a very large number of people, it can be said "elementary" function in the common language.
    Third, giving a name to the integral of a already known function is not the only way to invent a new special function. Many were defined as a solution of a differential equation, or as a solution of an analytic equation (for example the W Lambert function), or thanks to several other kind of definitions. For example, see pp. 20-25 in the paper "Safari in the Contry of Special Functions" :
  7. Mar 20, 2013 #6
    Impressive set of documents! I have a question about the sophomore dream: how would one show that:

    $$ \int_0^1 x^{-x} \, dx = \sum_{n=1}^\infty \frac{1}{n^n} $$
  8. Mar 21, 2013 #7

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