Is the set of elementary functions which do not have elementary integrals bigger than set of elementary functions which have elementary integrals?

Question

It increasingly seems to me that the functions that have elementary integrals are quite rare in comparison to the ones that don't have them. Even raising an elementary function to a different power may result in it not having an elementary integral .

Ex. $\sqrt{\arctan (x)}$

Also many seemingly simple functions do not have elementary integrals.

Ex. $\frac {\sin (x)}{x}$ or $ \sin \left( \frac{1}{x} \right) $

So my question is that can we write a formal proof to prove/disprove that the set of elementary functions which do not have elementary integrals is bigger than set of elementary functions which have elementary integrals?

Answer 1

This might not answer your question precisely, but you might be interested by discussions around Liouville's theorem in differential algebra. Here is a link to the Wikipedia page of this theorem.

In a few words, the goal is to formalise the situation by saying that your "known" functions lie in some field $K$. For example, $K$ could be $\mathbb R(X)$ the field of rationnal fractions over $\mathbb R$. Then, adding some new functions like the logarithm is the same as looking at field extensions which have a certain property. I let you read that by yourself hoping that this might help you.

However I don't think this will answer your question fully, i.e. giving a way to take into account a comprehensive list of usual functions and then characterising perfectly those which integral is still some usual function.

Answer 2

They have the same cardinality

It is not clear whether all real/complex constants are allowed, or only integers. Let $K$ be the set of allowed constants.

Let $E_i$ and $E_{ni}$ be the sets of elementary functions that respectively do and don't have an elementary integral and let $E_a = E_i\ \cup\ E_{ni}$.

Lower bound on $E_i$ and $E_{ni}$

Since for every constant $c \in K$, there is a function $f(x) = c$, we have: $$|E_i| \ge |K|$$ Also, for every constant $c \in K$, there is a function $f(x) = \sin(1/x)+c$, so: $$|E_{ni}| \ge |K|$$

Upper bound on $E_a$ if $K = \mathbb{N}$

In this case every elementary function can be written using a text formula, for example sin(2*x)^(x/3). If there were more than countable infinite elementary functions, this wouldn't be possible, so:

$$|E_{a}| \le |K|$$

Upper bound on $E_a$ if $K = \mathbb{R}$

Let's say a form of an elementary function is an elementary function with the constants written as $c_i$. So for example $c_1\cdot x^{c_2}$. Just as in the previous section, we can write every form as a text formula, like for example c1*x^c2. This gives an upper bound of $|\mathbb{N}|$ on the number of forms.

Now to go from a form to an elementary function, we have to fill in the constants. Let's say some form has $n$ constants, then there are $|\mathbb{R}|^n$ ways to choose the constants. It is well known that $|\mathbb{R}|^n = |\mathbb{R}|$.

To get an upper bound on $E_a$, we can multiply the number of forms by the number of ways to choose constants. So $|\mathbb{N}| \cdot |\mathbb{R}| = |\mathbb{R}|$ and we get:

$$|E_{a}| \le |K|$$

Upper and lower bound combined

If $K = \mathbb{N}$ or $K = \mathbb{R}$ we proved these formulas to be true: $$ \begin{align} |E_i| &\ge |K| \\ |E_{ni}| &\ge |K| \\ |E_a| &\le |K| \\ |E_a| &= |E_i| + |E_{ni}| \end{align} $$ It's not hard to see that it follows that: $$|E_{a}| = |E_i| = |E_{ni}| = |K|$$