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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?

Harry Peter
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ziutek
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    Questions related to this are part of the content of http://en.wikipedia.org/wiki/Differential_Galois_theory . Almost certainly there is no finite set of new functions you can add to the elementary functions so that all elementary functions have antiderivatives expressible in terms of elementary functions and the functions you toss in. – Travis Willse Sep 01 '14 at 14:32

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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.

IV_
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You actually only need to add one function to the set, but it's a very weird function. Define a function $f(R, x)$ such that $R$ is an encoding of any formula you like involving elementary functions, integrals, $x$ as a symbol, and $f$ itself (it is trivial to encode most parts of an elementary function in the integer part of $R$, and any real number constants in the function's formula can then be encoded as an index, and each of the real number constants' digits can then be alternated in fractional part of $R$; for instance, if you need to encode $a = a_1a_2a_3\ldots$ and $b = b_1b_2b_3\ldots$, the part of $R$ after the decimal point will look like $.a_1b_1a_2b_2a_3b_3\ldots$, and $a$ can be encoded as the index 1 and $b$ as the index 2), and $f(R, x)$ is defined as the result of evaluating the formula represented by $R$ on the input $x$.

This can also be done with two one-argument functions $f_1$ and $\textrm{spaceout}$, where $f_1$ behaves like $f$ except it takes its input with the digits alternating between the representations of $R$ and $X$, and the other, $\textrm{spaceout}(x)$, spaces the digits, so that for instance $13.12$ becomes $1030.102$; then, you can simply do the conversion as if you were using the two-argument form, and replace each $f(R, x)$ with $f_1(\textrm{spaceout}(R) + \textrm{spaceout}(x)/10)$

  • What does "the part of $R$ after the decimal point" mean if $R$ is an integer? – LSpice Mar 25 '23 at 22:01
  • ah oops, I should be clearer; the integer part encodes the actual structure of the formula, R as a whole is a real number – Morgan H Mar 26 '23 at 23:30