# Prove $f(x)=\int\frac{e^x}{x}\mathrm dx$ is not an elementary function

How do I prove that the exponential integral $$f(x)=\int \frac{e^x}{x}\mathrm dx$$ is not an elementary function?

Also, what are the general methods and tricks to prove that an integral or solution to an equation is not an elementary function?

• The general method is the Differential Galois Theory or the Galois theory of differential equations. There is an extremely nice exposition of the theory in a book by Andy Majid published by the AMS---at the end, IIRC, there is a complete argument that $\int e^{x^2}$ is not elementary, as an example. – Mariano Suárez-Álvarez Oct 28 '11 at 11:21
• Risch's services would be of great assistance. – J. M. is a poor mathematician Oct 28 '11 at 11:21
• en.wikipedia.org/wiki/Risch_algorithm – Peđa Terzić Oct 28 '11 at 11:23

## 1 Answer

All we need for this is a theorem of Liouville (1835): Suppose that $f$ and $g$ are rational functions with $f\neq 0$ and $g$ non-constant. Then $$\int f(x) e^{g(x)} dx$$ is an elementary function if and only if there exists a rational function $r$ such that $f=r'+g'r.$

Here we have $g(x)=x$ and $f(x) = 1/x.$ Assume there exists a rational function $r$ such that $1/x = r' + r \text{ }$ (1). Denote the multiplicity of the pole at $0$ by $m$ (where $m\geq 1$ so that both sides of (1) agree when $x\to 0$), so that $\displaystyle r(x) = \frac{p(x)}{x^m Q(x)}$ where $p, Q$ are polynomials with no common factors and $Q$ is not divisible by $x.$

Substituting this form into (1) and multiplying both sides by $x^m$ yields $$x^{m-1} = \frac{p(x)+ p'(x)}{Q(x)} - \frac{Q'(x) p(x) }{Q^2(x)} - \frac{m}{x Q(x)} .$$ Taking limits of both sides as $x\to 0$ illustrates the lunacy in our assumption that such a rational function exists, and hence $\displaystyle \int \frac{e^x}{x} dx$ is non-elementary.

• The fun is then of course hidden in the proof of Liouville's theorem. We could also say that «we only need the theorem that $\int e^x/x$ cannot be expressed in elementary terms»! :) – Mariano Suárez-Álvarez Oct 28 '11 at 13:56