Integrating Rational Functions
Dave L. Renfro
AP-Calculus list group (non-Usenet) that I thought
would be useful to archive in sci.math.
Kaleb Allinson wrote:
http://mathforum.org/kb/message.jspa?messageID=4641191
> Could someone help with g(x) = (x*abs(x)) / (x^2 + 1)
> Find integral from 0 to 1 of g(x)
>
> I've figured that from 0 to 1 g(x) = (x^2) / (x^2 + 1)
> I've tried u-substitution and integration by parts,
> but nothing seems to be working.
I've decided to begin a new thread because
what I wrote should be of more general interest.
Something useful to keep in mind is that *any*
antiderivative of a rational function (with
real number coefficients) can be expressed
as a linear combination of a rational function,
finitely many lograrithms of polynomials of
degrees 1 or 2, and finitely many inverse
tangents of polynomials of degree 1, all
of which will have real number coefficients.
TECHNICAL NOTE: The coefficients will be
sums, differences, products, quotients,
and/or square roots of the original
coefficients and/or of the zeros of the
denominator of the rational function.
In particular, if the denominator of
the original rational function can't be
expressed in a certain form (in terms of
radicals, for example), then the coefficients
of the antiderivative might also have this
defect. On the other hand, if the zeros of
the denominator are constructible (expressible
in radical form, expressible in real-radical
form, are algebraic, etc.), then so too will
be the coefficients involved in (at least
one of) the antiderivative(s).
This result is something that many texts
don't do a good job of pointing out, in my
opinion. It's a nice way of summarizing
several techniques to get a fairly strong
general result that is also important in
several areas of mathematics, such as in
symbolic integration theory and in complex
variables. Incidentally, any rational function
of sin(x) and cos(x) can be turned into
a rational function of u by the substitution
u = tan(x/2), so any antiderivative of a
rational function of sin(x) and cos(x) can
also be expressed in a certain precise way.
I'm not sure if every step below is part
of the BC syllabus, but every step should
have been covered in your college calculus
classes. [In this post I'm addressing a
general point that all calculus teachers
should know, whether or not they actually
cover all aspects of it when teaching a
high school calculus course. As for the
problem the original poster asked, it
requires very little of what's below.
After long division, it's about as simple
of an inverse tangent form that you
can have (no completing the square,
no u-substitution and further algebraic
manipulations), being (-1) / (x^2 + 1).]
STEP 1: Using long division, express the
rational function as polynomial plus a
proper rational function. ["Proper" means
the degree of the denominator is strictly
larger than the degree of the numerator.]
STEP 2: Obtain the partial fraction expansion
of the proper rational function.
STEP 3: Integrate the polynomial and integrate
each term in the partial fraction expansion:
Terms of the form A/(ax+b), with a not zero,
give you (A/a)*ln|ax+b|.
Terms of the form A / (ax+b)^n, with a not
zero and n = 2, 3, 4, ..., are handled by
the substitution u = ax+b, and you'll get
a rational function for the result.
Terms of the form (Ax + B) / (ax^2 + bx + c),
where ax^2 + bx + c is an irreducible (over
the real numbers) quadratic, are handled in
a multi-step process that should be outlined
somewhere in the text. [Complete the square
in the denominator, then make an appropriate
u-substitution to get it into the form
(A'u + B') / (u^2 + d^2), then handle
A'u / (u^2 + d^2) and B' / (u^2 + d^2)
separately -- the former by the substitution
w = u^2 + d^2 and the latter is an inverse
tangent form.]
Terms of the form (Ax + B) / (ax^2 + bx + c)^n,
where ax^2 + bx + c is an irreducible (over
the real numbers) quadratic and n = 2, 3, ...,
are handled as above. For the term
A'u / (u^2 + d^2)^n, the same substitution
w = u^2 + d^2 works. For the term
B' / (u^2 + d^2)^n, there's a reduction
formula that you'll find in integral tables.
Its derivation involves integration by
parts, which I sometimes carried out in
class and other times assigned as an extra
credit homework problem.
Incidentally, one of the historical reasons
why the Fundamental Theorem of Algebra (FTA)
was important is precisely because we need
it in order to carry out the above (in theory).
In particular, what we need is the fact that
every polynomial with real coefficients can be
factored into linear and irreducible quadratic
factors over the real numbers, which follows
easily from the fact that every polynomial has
at least one complex root. This fact was known
to follow easily from FTA by mathematicians
in the 1600-1700's, before FTA was proved in the
early 1800's (see <http://tinyurl.com/zgnx3>
for a nice historical survey of FTA), although
the connection between this fact and FTA is
often not made in precalculus texts.
----------------------------------------------
Here's a proof that FTA implies every polynomial
with real coefficients can be factored into
linear and irreducible quadratic factors over
the real numbers.
We first observe that every polynomial with
degree n has n complex roots counting multiplicity.
This follows quickly from the easy-to-prove fact
that "r is a root of polynomial P(x) implies (x-r)
is a factor of P(x)". [Proof of fact: Use the
division theorem to write P(x) = Q(x)*(x-r) + R(x)
in quotient/remainder form for the divisor (x-r).
Now plug x=r into both sides.]
Next, note that any polynomial with real coefficients
has complex roots that show up in complex conjugate
pairs. [Proof: Show that the complex conjugate of
a sum is the sum of the complex conjugates, and
similarly for a product. Now suppose z is a solution
to the polynomial P. Then P(z) = 0. Take the complex
conjugate of both sides. Using the distributivity
of complex conjugation over sums and products,
you'll get P(z*) = 0, where z* is the complex
conjugate of z.]
Now suppose we have a polynomial P(x) with real
coefficients. Each real root r generates the
factor (x-r) and each non-real complex conjugate
pair of roots, a +/- bi where a and b are real,
generates the irreducible quadratic factor
(x-a)^2 + b^2. Note that this quadratic has
real coefficients. [I got this quadratic factor
by writing x = a+bi, subtracting a, squaring, then
getting all the terms on one side of the equation.]
----------------------------------------------
An interesting problem to pose after all this is:
PROBLEM: What is an antiderivative of 1 / (x^4 + 1)?
NOTE: If 4 is replaced by 1, we get a logarithm.
If 4 is replaced by 2, we get an inverse tangent.
If 4 is replaced by 3, we get something that is
about in the middle of difficulty for typical
homework problems on partial fractions.
Besides being much more demanding than you'd
expect from its relatively simple form, this
problem is also historically significant. None
other than Leibniz himself was unable to carry
out this integration, and he used this (in 1702)
as an example for the failure of the Fundamental
Theorem of Algebra. For some discussion of these
issues (in particular, for Leibniz's failure and
for the fact that these issues were motivated
by the integration of rational functions),
see the beginning of Section 19.4 (pp. 411-412)
in Morris Kline's "Mathematical Thought From
Ancient to Modern Times", 1972 (p. 411 is at
google-books -- <http://tinyurl.com/lpdcl>).
Many of the following google-books search results
also discuss these issues:
http://books.google.com/books?as_q=Leibniz+1702+algebra
Dave L. Renfro
G. A. Edgar
Note it omits the proof that the partial fraction decomposition works.
Calcluls texts also omit this interesting tidbit.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
José Carlos Santos
> Note it omits the proof that the partial fraction decomposition works.
> Calcluls texts also omit this interesting tidbit.
Spivak at least tells his readers that a proof can be found at van der
Waerden's Algebra.
Best regards,
Jose Carlos Santos