We have now reduced the problem to the integration of , where degree degree , and is square-free (has only simple roots). We know by the classical algorithm that this rational function can be expressed as

where the are the **k** distinct roots of , and the
are the residues of at the roots of .
This gives ,
which has possibly complex coefficients. Can we avoid the use of complex
numbers?

The answer is yes, and depends on the following observation. We know that the are the residues of , and indeed we know an explicit formula for them: , because the roots of are simple (this is easy to prove: first note that , and that since and that is the Laurent expansion of ). Rearranging this formula for the residues we get

The key to the Rothstein-Trager method is to observe that this means
the are the roots of the following * resultant*:

where means take the resultant in terms
of **x**, leaving a polynomial in **z**.

This resultant formula looks strange, and indeed it's hard to think
how it was recognized in the first place, but it's not hard to verify.
What it is saying is that the common roots of and
are the roots of the resultant---and that's easy, because one of the
polynomials doesn't contain **z** in the first place. So this formula
is easy to understand with hindsight.

The next main observation of the Rothstein-Trager method is that it
is the * distinct* residues that give rise to uncombineable
log terms: can be
combined to . So the only way we
can avoid using complex numbers, or algebraic extensions in general,
is if there are multiple roots of the resultant.

This is true in general, and we have the following theorem, which
is theorem 11.7 in Geddes * et al*.

** Theorem** If , **b** is monic and square-free, and
, then

where the are the distinct roots of , and further the .

For a proof of this theorem see Geddes * et al.*. For now, note
that it ** solves the problem completely**. We compute a resultant,
find its distinct roots, compute some GCD's, and then we write down
the integral. * All the calculations are rational*. (Except of
course for the factorization of the resultant, but even there we can
do as well as is possible with another square-free factorization).

** Remark**. I have not stressed the fact that this method
introduces only those algebraic extensions that are necessary. For
a discussion of this, see Geddes * et al.*, but note that this
idea goes beyond the simple rational function integration performed
here and will prove useful in identifying functions which have no
elementary antiderivative in the Risch algorithm, which we will
see next week.

Thu Nov 16 13:46:20 PST 1995