Exercise 29, § 1

Let $f$ and $g$ be two rational functions of $z$ (elements of $\mathbf{C}(z)$). Show that, for the primitive $\int f(z) e^{g(z)} \, dz$ to be an elementary function (exerc. 28) it is necessary and sufficient that there exists a rational function $r \in \mathbf{C}(z)$ such that $f = r' + r g'$. (Put $t = e^z$ and consider the differential field $F = \mathbf{C}(z, t)$, which is an elementary extension of $\mathbf{C}(z)$ since $t'/t = g'$. Show first that $t$ is transcendental over $\mathbf{C}(z)$; arguing by contradiction, consider a Galois extension $K$ of $\mathbf{C}(z)$ containing $t$, and the transforms of the equation $t'/t = g'$ by the Galois group of $K$; one will obtain an equation $g' = u'/u$ with $u \in \mathbf{C}(z)$, and it is impossible for $g'$ to have other than simple poles in $\mathbf{C}$. Applying exerc. 28 one has $f t = \sum_j \gamma_j \frac{u_j'}{u_j} + v'$, with $\gamma_j \in \mathbf{C}$, and the $u_j$ and $v$ in $F$; one can reduce to the case where the $\gamma_j$ are linearly independent over $\mathbf{Q}$. Then, applying exerc. 27 to the extension $\mathbf{C}(z)(t)$, show that one has $f t = v' + h$, with $h \in \mathbf{C}(z)$ and $v' \in \mathbf{C}(z)[t]$. Conclude that $v$ is necessarily of degree 1 in $t$, and consequently $f$ is equal to the coefficient of $t$ in $v'$)

Deduce from this that the primitives $\int e^{-z^2} \, dz$ and $\int e^z \, dz/z$ are not elementary functions (a theorem of Liouville), by examining in the relation $f = r' - r g'$ the behaviour of the two sides at a neighbourhood of a pole of $r$.*