Liouville's theorem (differential algebra)

Basic theorem

Suppose ${\displaystyle F}$ and ${\displaystyle G}$ are differential fields with ${\displaystyle \mathrm{Con}(F)=\mathrm{Con}(G),}$ and that ${\displaystyle G}$ is an elementary differential extension of ${\displaystyle F.}$ Suppose ${\displaystyle f\in F}$ and ${\displaystyle g\in G}$ satisfy ${\displaystyle Dg=f}$ (in words, suppose that ${\displaystyle G}$ contains an antiderivative of ${\displaystyle f}$). Then there exist ${\displaystyle c_1,\dots ,c_n\in \mathrm{Con}(F)}$ and ${\displaystyle f_1,\dots ,f_n,s\in F}$ such that $${\displaystyle f=c_1\frac{D{f}_1}{f_1}+\cdots +c_n\frac{D{f}_n}{f_n}+Ds.}$$

In other words, the only functions that have "elementary antiderivatives" (that is, antiderivatives living in, at worst, an elementary differential extension of ${\displaystyle F}$) are those with this form. Thus, on an intuitive level, the theorem states that the only elementary antiderivatives are the "simple" functions plus a finite number of logarithms of "simple" functions.

A proof of Liouville's theorem can be found in section 12.4 of Geddes, et al. See Lützen's scientific bibliography for a sketch of Liouville's original proof (Chapter IX. Integration in Finite Terms), its modern exposition and algebraic treatment (ibid. §61).

Examples

As an example, the field ${\displaystyle F:=\mathbb{C}(x)}$ of rational functions in a single variable has a derivation given by the standard derivative with respect to that variable. The constants of this field are just the complex numbers ${\displaystyle \mathbb{C};}$ that is, ${\displaystyle \mathrm{Con}(\mathbb{C}(x))=\mathbb{C},}$

The function ${\displaystyle f:=\frac{1}{x},}$ which exists in ${\displaystyle \mathbb{C}(x),}$ does not have an antiderivative in ${\displaystyle \mathbb{C}(x).}$ Its antiderivatives ${\displaystyle \ln x+C}$ do, however, exist in the logarithmic extension ${\displaystyle \mathbb{C}(x,\ln x).}$

Likewise, the function ${\displaystyle \frac{1}{x^2+1}}$ does not have an antiderivative in ${\displaystyle \mathbb{C}(x).}$ Its antiderivatives ${\displaystyle {\tan }^{-1}(x)+C}$ do not seem to satisfy the requirements of the theorem, since they are not (apparently) sums of rational functions and logarithms of rational functions. However, a calculation with Euler's formula ${\displaystyle e^{i\theta }=\cos \theta +i\sin \theta }$ shows that in fact the antiderivatives can be written in the required manner (as logarithms of rational functions). $${\displaystyle \begin{array}{cc}{e}^{2i\theta }& =\frac{e^{i\theta }}{e^{-i\theta }}=\frac{\cos \theta +i\sin \theta }{\cos \theta -i\sin \theta }=\frac{1+i\tan \theta }{1-i\tan \theta }\\ \theta & =\frac{1}{2i}\ln \left(\frac{1+i\tan \theta }{1-i\tan \theta }\right)\\ {\tan }^{-1}x& =\frac{1}{2i}\ln \left(\frac{1+ix}{1-ix}\right)\end{array}}$$

Relationship with differential Galois theory

Liouville's theorem is sometimes presented as a theorem in differential Galois theory, but this is not strictly true. The theorem can be proved without any use of Galois theory. Furthermore, the Galois group of a simple antiderivative is either trivial (if no field extension is required to express it), or is simply the additive group of the constants (corresponding to the constant of integration). Thus, an antiderivative's differential Galois group does not encode enough information to determine if it can be expressed using elementary functions, the major condition of Liouville's theorem.