Liouville's theorem (differential algebra)

Remarks

The proof goes by induction on the length of the tower $F \subseteq F_1 \subseteq \cdots \subseteq E$. At each step one writes the antiderivative in terms of the last generator $t_i$ and compares coefficients. This works because derivatives of rational functions in a logarithm or an exponential have a controlled form. An algebraic step is removed by taking traces, which is where the characteristic $0$ is used.

Example: over $F = \C(x)$ with $\partial = d/dx$, the element $\alpha = 1/x$ has the antiderivative $\log x$, and indeed $\alpha = c_1 u_1'/u_1$ with $c_1 = 1$, $u_1 = x$, $v = 0$. The theorem is the tool that proves that $e^{-x^2}$, $e^x/x$ and $1/\log x$ have no antiderivative built from elementary functions.

The hypothesis on the constants cannot be dropped. Over $F = \R(x)$, the element $\alpha = 1/(1 + x^2)$ has the antiderivative $\arctan x = \frac{1}{2i} \log \frac{x - i}{x + i}$, which lies in an elementary extension that contains the new constant $i$. But $\alpha$ has no expression of the form above with real constants and $u_i, v \in \R(x)$.

Going deeper

For a student

The theorem turns the question "is this integral elementary?" into an algebraic question about the field $F$. For $F = \C(x, e^{g})$ with $g$ a nonconstant polynomial, it reduces to a first-order linear differential equation over $\C(x)$. This gives a classical criterion for rational $f$: the integral $\int f e^{g} \, dx$ is elementary exactly when $R' + g' R = f$ has a solution $R \in \C(x)$. The Risch algorithm makes this reduction systematic and decides elementary integrability in towers of logarithms, exponentials and algebraic functions. The theorem is a basic result of differential algebra, alongside the theory of Picard–Vessiot extensions.