Solvability in differential Galois theory

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?

• 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. – the L Mar 20 '11 at 14:21