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Liouville's original criterion for elementary anti-derivatives states:

If $f,g$ are rational, nonconstant functions, then the antiderivative of $f(x)\exp{g(x)}$ can be expressed in terms of elementary functions if and only if there exists a rational function $h(x)$ such that $f=h'+hg'$.

Of particular note for my purpose is the fact that $f,g$ must be rational functions. Motivated by this recent question, I'm interested in the following weakening of Liouville's problem:

If $f,g$ are rational non-constant functions, under what conditions can the antiderivative of $\sqrt{f(x)}\exp{g(x)}$ be expressed in terms of elementary functions?

This presumably falls under the heading of the modern Liouville criterion, as expounded by Wikipedia here. (See also Bill Dubuqeu's short precis on the Liouville-Ritt theory of integration in finite terms here.) Alas, I don't know enough differential algebra to apply this productively, nor do I really have the patience to gain such expertise.

Motivated by a remark made in a (withdrawn) answer to the initial version of this question, however, one may come up a partial answer by assuming an antiderivative of the form $\sqrt{h(x)}\exp{g(x)}$. Differentiating, we obtain

$$\frac{d}{dx}\sqrt{h(x)}\exp{g(x)}=\left(g'(x)\sqrt{h(x)}+\frac{h'(x)}{2\sqrt{h(x)}}\right)\exp{g(x)}$$ Setting this equal to $\sqrt{f(x)}\exp g(x)$ and solving for $f(x)$, we conclude that $\sqrt{f(x)}\exp g(x)$ has an elementary antiderivative if there exists a rational function $h(x)$ such that $$f(x)=g'(x)^2h(x)+g'(x)h'(x)+\frac{h'(x)^2}{4h(x)}.$$

What this does not show, however, is that the condition above is necessary rather than sufficient.

Semiclassical
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  • It's much easier to write $h=f^{1/2}e^gq$ such that $h'=f^{1/2}e^g$ so the ODE becomes $q'+(g'+(\log f)'/2)q=1$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Aug 26 '21 at 08:22
  • Even if $g = 0$ and $f(x) = \frac 1{P(x)}$, which would come under the topic of elliptic integrals , this would be quite a difficult problem to discuss at a preliminary level. The original Liouville theorem doesn't require anything beyond undergraduate algebra but when the discussion comes to requiring algebraic-irrational operations on functions, then one can't work in the usual setting, but must instead move to forms on a curve, and invoke Riemann surface theory. Think of it like this : adjoining $\sqrt{2}$ to $\mathbb Z$ forces you to move to $\frac{\mathbb Z[X]}{\langle X^2-2\rangle}$... – Sarvesh Ravichandran Iyer Sep 01 '21 at 05:59
  • ... and something like that occurs here as well : we are adjoining the formal symbol $\sqrt f(x)$ to our field, so we must move to the curve defined by $y^2 = f(x)$ on which the corresponding differential form and work there. But these curves could have all kinds of shapes. The important point, though, is that Liouville's theorem will apply here as well, provided you extend the notion of a derivative to this domain (and you can). So in complicated answers involving extensions, this is what I would expect to do. Nevertheless, you will require more theory than you currently have at hand. – Sarvesh Ravichandran Iyer Sep 01 '21 at 06:03
  • @TeresaLisbon That makes sense, at least in broad strokes. (I definitely don't have the expertise in algebreic geometry that seems needed here.) I wonder if this question would have been better suited for Math Overflow; if this bounty doesn't succeed at generating answers, I may repost it over there. – Semiclassical Sep 01 '21 at 15:07
  • @Semiclassical Oh, please do so! I would absolutely love it!Another answer involving a function with square-roots which I really love is over here where you should try to see what you understand and what you need to pick up. Another nice link is over here. – Sarvesh Ravichandran Iyer Sep 01 '21 at 15:46

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