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Consider the following property of a function $f$:

There exists a non-negative integer $n$ such that the $n$'th derivative of $f$ is a rational function.

Question 1: Is there a name in the literature for this property?

Question 2: Are there any good references for reading about the class of all such functions?

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    $\begingroup$ In algebraic geometry, we would usually keep track of the differential equation, rather than its solution space. Then, roughly, this is a coherent D-module with regular singularities at the poles of the rational function. $\endgroup$ Commented Apr 28, 2017 at 9:25
  • $\begingroup$ Thanks Jason, that's a good point. But I want to take the union of these solution spaces over all rational functions and all $n$. Then my feeling is that I get something bigger than a coherent D-module, and I would like a name for this large space of functions. $\endgroup$ Commented Apr 28, 2017 at 16:47
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    $\begingroup$ The rational functions have a basis consisting of the functions $x^n$ and $\frac{1}{(x - r)^n}$ (say we're working over $\mathbb{C}$) so it suffices to understand the antiderivatives of these functions. These have a basis consisting of the previous basis along with the functions $(x - r)^n \log (x - r)$. Is there anything else to say? $\endgroup$ Commented May 30, 2017 at 6:27
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    $\begingroup$ @QiaochuYuan ... What else to say? Antiderivatives of those. For example, $\int (\log x)/x\;dx$ is not elementary. On the other hand, can we see that we do not get $e^x$ ... so that we do not even get all elementary functions. $\endgroup$ Commented May 30, 2017 at 12:09
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    $\begingroup$ @Gerald: neither $\frac{\log x}{x}$ nor $\frac{x}{\log x}$ have the property. When you take the antiderivative of $x^n \log x$ you get a constant times $x^{n+1} \log x$ plus a polynomial. So no other functions are necessary. You can see this on the level of formal power series (ignoring $\frac{1}{x}$ for simplicity): taking repeated antiderivatives of rational formal power series (=polynomial-exponential coefficients) adds denominators of the form $\frac{1}{(n+1)(n+2) \dots}$ which you can reduce to linear combinations of denominators of the form $\frac{1}{n+k}$ using partial fractions. $\endgroup$ Commented May 30, 2017 at 19:25

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