Many functions are not integrable in elementary terms.
notable functions:
$e^{-x^2}$
Consequence: the error function, $\erf = \int_{-\infty}^x ...$
cannot be expressed in elementary terms.
Logarithmic integral (which is $e^u/u$ in other coordinates)
Elliptic integrals
General theory: differential fields
\section{Outline}
Proof:
- Liouville's theorem: elementary integrals have a \emph{special form}
- for specific questions, look at \emph{specific properties}
- for $e^{-x^2}$, reduces to showing that a degree~$1$ \emph{ODE}
has no rational solution, which one solves analogously to how one
shows \emph{irrationality} of $\sqrt{2}$.
State theorem
Anti-derivatives are of a special form:
if $F'=f$, then
$$F=h+\sum c_j \log g_j$$
i.e., $f$ can be integrated in elementary terms iff
$f = h'+\sum c_j \frac{g'_j}{g_j}$
This holds for any elementary field containing $f$.
Lemma:
If $fe^g$ is integrable in elementary terms, then
it has anti-derivative $Re^g$, where $R$ is a rational function
satisfying $R'+g'R=f$.
Proof: This takes some doing.
Case: $e^{-x^2}$ is not integrable in elementary terms.
Proof: By lemma, take $f=1, g=-x^2$, so sufficies to show that
$R'-2xR=1$ has no rational solutions.
Look at poles.
$1$ is constant and has no poles, hence the poles of $R'$ and $2xR$
must cancel.
Differentiation decreases the order of the pole at infinity by $1$,
and increase the order of all finite poles.
First, $R$ cannot have any poles, as $R'$ has poles of one degree
higher, hence $R$ must be a polynomial.
Second, there are no polynomial solutions, by looking at the pole at
infinity:
if $R$ is a polynomial, then $\deg R' = \deg R - 1$,
and $\deg 2xR = \deg R + 1$, so they can't add up to $1$.
\subsection{Intuition}
As a simpler case, there are no rational functions such that $R'=R$
(i.e., $R'-R=0$).
If degree of poles is too abstract, think of ``order of growth at
infinity / a finite point''.
This proof is very much like Pythagoras' proof of the irrationality of
$\sqrt{2}$.
The underlying idea is that rational numbers and rational functions
are quite similar.
the ring of integers of a field of fractions:
analogy between $\bZ \subset \bQ$
and polynomials inside rational functions.
(primes $p$ and $(z-\alpha)$)
($\bZ$ and $\bC[z]$ are both PIDs; do you only need Dedekind domain,
esp. integrally closed in fraction field)
Pythagoras:
$x^2-2=0$ has no rational solutions.
Proof:
- any solution must be an integer (by clearing denominators)
- it has no integer solutions (say, by reduction mod $4$)
More generally, if $m$ is monic, $m(x)$ has only algebraic integer solutions,
hence rational solutions are integers.
Here:
- $m(D)f=C$ has only polynomial solutions (among rational functions)
- $Df-2xf=1$ has no polynomial solutions (say, because of degree)
Exercise: Do for $e^u/u$.
\section{Proof of Lemma}
First, $g_j$ can be taken to be polynomials,
not just rational, as $\log p/q = \log p - \log q$.
Consider $\bC(X,Y)$ ($X=x,Y=e^g$)
Can assume irreducible in $Y$.
...it's pretty long and painful.
\section{References}
Rosenlicht's article "Integration in finite terms" in volume 79 of the
American Mathematical Monthly (in 1972)
JSTOR link!
http://www.math.lsa.umich.edu/~bdconrad/papers/finalint.pdf
from
http://www.math.lsa.umich.edu/~bdconrad/