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?
Best Answer
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).