Are Jacobi elliptic functions liouvillian? I mean: could a Jacobi elliptic function be expressed in terms of a "finite number of arithmetic operations (+ – × ÷), exponentials, constants, solutions of
algebraic equations (a generalization of nth roots), and indefinite integrals of such elements"?
 A: Here's an answer found in http://dx.doi.org/10.1088/1751-8113/46/45/455203 (the subject is Differential Galois theory):

for instance, one obtains some kind of
  solvability (virtual) for the Galois group whenever one obtains Liouvillian
  solutions, and in this case one says that the differential equation is integrable.
  This means for example that when one obtains Airy functions, the
  differential equation is not integrable, while when one obtains Jacobi elliptic
  functions, the differential equation is integrable

A: I think I have a proof that the non-constant elliptic functions are not Liouvillian (be careful that the elliptic functions are the inverse of the elliptic integrals).

*

*The Liouvillian functions $L$ is a set defined recursively :

$f(z)  = 1$ and $g(z)  =z$ are in $L$.
If $f(z),g(z)$ are in $L$ then

*

*$a f(z)+b g(z)$, $f(z)g(z)$ are in $L$


*if $f(z)$ is non-constant then $\exp f(z)$, $\log f(z), 1/f(z)$ are in $L$


*$\int f(z)dz$ is in $L$


*$h(z)$ is in $L$ if $\sum_{n=0}^d a_n h(z)^n $ is in $L$



*Then Let $M$ be the constants plus the set of almost everywhere analytic functions such that for every sector $$S_{a,\theta,b} = \{z \in \mathbb{C}, |z-a| >0,  \text{arg}(e^{i \theta} (z-a)) \in (-b,b)\}$$ there is a sub-sector $S_{a',\theta',b'}\subseteq S_{a,\theta,b}$ such that : $f(z)$ is analytic and injective on $S_{a',\theta',b'}$.
From this you can show by induction that if $f(z) \in L$ then $f(z) \in M$,
and since the elliptic functions (the doubly periodic meromorphic functions that are non-constant) are not in $M$, they are not in $L$ neither.

*

*If $f(z) \in M$ and non-constant then $1/f(z),\exp f(z),\log f(z)$  are in $M$.


*If $f(z) \not\in M$ then $\sum_{n=0}^d c_n f(z)^n \not\in M$


*If $f(z),g(z) \in M$ then $ af(z)+bg(z), f(z)g(z)$ are in $M$


*A little work shows that if $f(z) \in M$ then  $\int f(z)dz \in M$
