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It is well known that the function $f(x) = e^{-x^2}$ has no elementary anti-derivative.

The proof I know goes as follows: Let $F = \mathbb{C}(X)$. Let $F \subseteq E$ be the Picard-Vessiot extension for a suitable homogeneous differential equation for which $f$ is a solution. Then one may calculate $G(E/F)$ and show it is connected and not abelian.

On the other hand, a calculation shows that if $K$ is a differential field extension of $F$ generated by elementary functions then the connected component of $G(K/F)$ is abelian, so it is impossible for an anti-derivative of $f$ to be contained in such a field $K$.

However, in classical Galois theory we can do much better, there, we know that a polynomial equation is solvable by radicals if and only if the corresponding Galois group is solvable.

So to my question - is an analog of this is available in differential Galois theory? Is there a general method to determine by properties of $G(F/E)$ if $F$ is contained in a field of elementary functions?

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The analogue to "solvable by radicals" in differential Galois theory is "solvable by quadratures". The theorem says that a PV-extension is Liouvillian (adjoining primitives and exponentials) iff the connected component of the differential Galois group is solvable. See "A first look at differential algebra" by Hubbard and Lundell, for an expository account.

I slightly misread the question at first, thinking you were looking for the analogue of solvbility by radicals in differential algebra. When it comes to determining if the primitive of a function is elementary or not, the characterization is given by Liouville's theorem. Now, for the general case of differential equations solvable in terms of elementary functions, there is a generalization of Liouville's theorem, that you can find in the article "Elementary and Liouvillian Solutions of Linear Differential Equations", by M.F. Singer and J. Davenport (link to Singer's papers here).

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  • $\begingroup$ Thank you for the link! Maybe I am missing something, but if I understand this correctly, Liouvillian extensions of $\mathbb{C}(X)$ can contain functions that are not elementary. Is this correct? If so, then this is not enough to answer the question which differential equations are solvable by elementary functions. $\endgroup$
    – the L
    Commented Mar 20, 2011 at 14:21
  • $\begingroup$ Thank you, actually the wikipedia article has the answer to my specific question: "an antiderivative's differential Galois group does not encode enough information to determine if it can be expressed using elementary functions", thus providing a negative answer to my question. $\endgroup$
    – the L
    Commented Mar 20, 2011 at 15:06
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It's probably not quite what you're looking for, but a necessary and sufficient condition for existence of elementary antiderivatives is given in this nice write-up on Dave Rusin's web, here. He includes a number of references which could prove useful to people interested in such questions.

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  • $\begingroup$ Dear Todd Trimble, Surely you would give the link to this page pf MathAtlas on Differential Algebra math.niu.edu/~rusin/known-math/index/12HXX.html. $\endgroup$
    – agt
    Commented Mar 20, 2011 at 14:11
  • $\begingroup$ Well, yes. I simply copied and pasted the URL of the page I was actually after, which unfortunately sends one to the previous page. Don't ask me why. Anyway, using my link, one can scroll down to "Selected Topics", and find the desired page by clicking the topic which is third from the bottom. $\endgroup$
    – Todd Trimble
    Commented Mar 20, 2011 at 14:26

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