Proving that special functions do not have closed-form expression

To be a bit more precise, the part of mathematics concerned with closed-form integrals and closed form solutions to ODE is called "Differential Algebra." This is not for the faint-hearted, and was begun by RITT and his student KOLCHIN at Columbia. One of the current leaders in this study is called "The Bozh" by my friend Dmitry, anyway Michael Boshernitzan of Rice. I have a book by Kaplansky called, and I think this is clever, "Differential Algebra."

A differential field is simply one which has a derivation, $D(uv) = u D(v) + v D(u).$ One can then decide whether a new item, an indefinite integral especially, is in the field. Thus the meaning of the phrase "closed form" is entirely up to the individual mathematician: a function is elementary if it is in the differential field being considered. Put another way, a function is elementary if you say it is. In the same way, if a child brings home a stray puppy, the parents tell the child not to give it a name as it will be gone soon. Your elementary functions are just those to which you have assigned a name.

There is a side note here: modern computer algebra systems use Ritt's algorithm, and many types of improvements based on including Grobner bases into the mix, for deciding whether some integral has a closed form, as far as the CAS is concerned. Note that one may do this in steps, first decide whether it is in the traditional basket of elementary functions we learn in high-school, next add in a few less traditional, see if that changes things.

There is a type of needlework called CROSS STITCH, for which a sampler is a piece small enough to be framed and hung on a wall. A book with text examples at BOOK. We could commission such a thing and send it to J.M. It would read

$$\begin{array}{c}\text{Elementary is in the}\cr\text{Eye of the Beholder}\end{array}$$

or maybe

$$\begin{array}{c}\text{Closed Form is in the}\cr\text{Eye of the Beholder}\end{array}$$

The "field" of mathematics that deals with such a question is, coincidentally, field theory. A function has a closed-form representation if and only if it belongs to a certain tower of function fields.

In practice, given a function $f: \mathbb{F}\to\mathbb{F}$, we want to determine if there exists an $n\in\mathbb{N}$ and a sequence of fields $\{K_i\}_{i=0}^{n}$ such that:

• $K_{i+1}$ is a simple field extension $K_i$ (described below)
• $K_0\equiv \mathbb{F[x]}$
• $f\in K_n$

The field extension must be of the form $K_{i+1}=K_i[g]$ where either:

• $g$ is algebraic over $K_i$
• $g=\exp(h)$ for $h\in K_i$
• $g=\ln(h)$ for $h\in K_i$

An elementary introduction can be found here