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Its fairly well known that not every elementary function has an elementary anti-derivative. The common examples of this are $\exp(-x^2)$ and $\sin(x)/x$. The general workaround to this problem is to create new functions like $\mathrm{Erf}(X)$ and $\mathrm{Si}(x)$ to go alongside the more elementary functions. I was wondering if there is a way to add "enough" non-elementary functions so that every function has an anti-derivative that can be expressed via other functions. The typical solution to this question is to introduce Riemann integration as an operator on functions, and then declare that the set of functions is closed under this operator. That definitely works, but I'm looking for something that is more enumerable. Like a simple countable set of functions that I can build everything else out of. Has anyone come up with a set of functions like this, or proved that making one is impossible?

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    $\begingroup$ en.wikipedia.org/wiki/Hypergeometric_function $\endgroup$ – Jack D'Aurizio Jul 22 '15 at 21:52
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    $\begingroup$ If your set of functions in analytic, then yes, just use Laurent series, truncated to finite order. $\endgroup$ – Alex R. Jul 22 '15 at 22:28
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    $\begingroup$ @AlexR.: I don't see how Laurent series give a countable set of functions that meet Joel Turnblade's requirements. $\endgroup$ – Rob Arthan Jul 22 '15 at 23:38
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    $\begingroup$ @RobArthan: Isn't $\{x^k\}_{k=0}^\infty$ countable? I'm assuming OP wants antiderivatives to be generated as (possibly infinite) linear combinations of such a set. $\endgroup$ – Alex R. Jul 22 '15 at 23:41
  • $\begingroup$ I was assuming the OP was looking for a countable set of functions that would generate in the algebraic (finitistic) sense an an algebra of functions closed under antiderivation. Over to the OP to clarify. $\endgroup$ – Rob Arthan Jul 22 '15 at 23:55

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