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Integrals of Rational Functions by Partial Fractions

Date: 12/20/2007 at 21:29:06
From: Ramiro
Subject: Integrals of rational functions

It is theoretically possible to write any rational expression f(x)/g
(x) as a sum of rational expressions whose denominators involve 
powers of polynomials of degree not greater than two.  Specifically, 
if f(x) and g(x) are polynomials and the degree of f(x) is less than 
the degree of g(x), then it can be proved that 

f(x)/g(x) = F_1 + F_2 + ... + F_r

such that each term F_k of the sum has one of the forms

     A                         Ax + B
-----------       or      -----------------   
(ax + b)^n                (ax^2 + bx + c)^n

for real number A and B and a nonnegative integer n, where ax^2 + bx 
+ c is irreducible in a sense that this quadratic polynomial has no 
real zeros (that is, b^2 - 4ac < 0).  In this case, ax^2 + bx + c 
cannot be expressed as a product of two first-degree polynomials 
with real coefficients.

The sum F_1 + F_2 + ... + F_r is the partial fraction decomposition 
of f(x)/g(x), and each F_k is a partial fraction.

Here is an example:

Evaluate 
         4x^2 + 13x - 9
Integral ---------------  dx
         x^3 + 2x^2 - 3x

Solution:

x^3 + 2x^2 - 3x = x(x+3)(x-1)

4x^2 + 13x - 9       A       B         C
---------------  =  --- + ------- + -------
x^3 + 2x^2 - 3x      x     x + 3     x - 1

solving for A, B, and C and then integrate.

Another example:

Evaluate 
         3x^3 - 18x^2 + 29 x - 4
Integral ------------------------ dx
            (x + 1)(x - 2)^3


Solution:

3x^3 - 18x^2 + 29 x - 4     A     B       C         D
------------------------ = --- + --- + ------- + -------
   (x + 1)(x - 2)^3        x+1   x-2   (x-2)^2   (x-2)^3

multiply both sides by (x+1)(x-2)^3 and solve for A, B, C, and D 
consequentially.

I know how to do this, but I just don't see why it works.



Date: 12/21/2007 at 10:31:54
From: Doctor Peterson
Subject: Re: Integrals of rational functions

Hi, Ramiro.

"Why" has many different meanings, so you may have to clarify what 
kind of answer you want--a proof, or a method, or a feeling for the 
reasonableness of the process.

You can find a couple levels of answer here:

  Proof of the Partial Fractions Theorem for Quadratic Factors
    http://mathforum.org/library/drmath/view/51687.html 

The first question there focuses on quadratic factors, but the second 
answer deals with all cases.  That, however, is rather formal and may 
not satisfy you.  A more basic answer, just waving my hands a little 
rather than actually proving anything, goes like this:

Partial fraction decomposition is a way of reversing the process of 
adding fractions.  If we were to do the same thing with numbers, we 
might try to break up a fraction with a composite denominator into a 
sum of fractions whose denominators are primes or powers of primes:

   1     1      a     b     1    -4
  -- = ----- = --- + --- = --- + ---
  18   2*3^2    2    3^2    2     9

To add, we find an LCD by taking appropriate powers of each factor, 
and then adjust to use that LCD.  To reverse this, we split up 
factors of the denominator and find appropriate numerators.

It's a little more complicated with rational expressions, because the 
goal is not just a number but an equivalent expression--one that is
equal to the original for all x.  That will be the key to the whole 
concept.

We first look for a set of simple denominators whose LCD will be the 
given one.  The easiest way to do that would look like what we did 
with 1/18:

  3x^3 - 18x^2 + 29 x - 4     ?       ?
  ------------------------ = --- + -------
     (x + 1)(x - 2)^3        x+1   (x-2)^3

The numerators, in order to make these proper fractions, can have any 
degree less than that of the denominator, so to cover all 
possibilities we would have to allow this:

  3x^3 - 18x^2 + 29 x - 4     A    Bx^2 + Cx + D
  ------------------------ = --- + -------------
     (x + 1)(x - 2)^3        x+1      (x-2)^3

This could be done; we'd multiply both sides by the denominator and 
set coefficients of each power of x equal (so that they are actually 
the same polynomial, equal for all x), and solve the resulting four 
equations for the four unknowns. 

But this would not yield something easy to integrate.  In order to 
accomplish that, we'd rather have the second fraction have only a 
constant in the numerator (that is, we'd like to drop B and C).  But 
if you try this, you find that when you try solving for the unknowns 
you would have four equations but only two variables--not enough to 
expect a solution.  We need those four variables, but we want them in 
a nicer form.

So what if we wrote the big numerator in terms of powers of (x-2) 
rather than of x:

  3x^3 - 18x^2 + 29 x - 4     A    B(x-2)^2 + C(x-2) + D
  ------------------------ = --- + ---------------------
     (x + 1)(x - 2)^3        x+1         (x-2)^3

Now we still have something we can solve, and that covers all possible 
numerators; but when we simplify that last fraction we get something 
easy to integrate:

  3x^3 - 18x^2 + 29 x - 4     A     B       C         D
  ------------------------ = --- + --- + ------- + -------
     (x + 1)(x - 2)^3        x+1   x-2   (x-2)^2   (x-2)^3

And that's the standard form we're looking for.

So, why does it work?  It allows for all possible numerators and 
provides enough variables to solve for, while yielding a useful form.

Let me know if you'd like to discuss this further, because it's 
something that isn't explained enough--too often we accept it without 
questioning.  In writing this, I discovered some details I'd probably 
never thought about!


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 12/22/2007 at 17:36:37
From: Ramiro
Subject: Integrals of rational functions

First of all, thank you, Doctor Peterson, for replying.  I think I 
see it now ... but am still not sure.  Anyway, I looked at the article
from your archives that you provided and I understand the first part
of the answer by Dr. Fenton.  As for the second part of his answer, he
quoted

"The first is the Division Algorithm, which says that given two 
polynomials f(x) and g(x), you can divide f(x) by g(x) to get a 
quotient q(x) and a remainder r(x) whose degree is strictly less 
than the degree of the divisor g(x): that is,

  f(x) = g(x)q(x) + r(x), and 0 <= deg(r(x)) < deg(g(x) .

Next, by repeated applications of this algorithm, you obtain the 
Euclidean Algorithm for finding the greatest common divisor d(x) of 
two polynomials f(x) and g(x). The key idea we need is that d(x) can
be written

   d(x) = f(x)r(x) + g(x)s(x) .

If you're not familiar with these ideas, you'll need to read up on 
them. See

   Euclidean and Division Algorithms
     http://mathforum.org/dr.math/problems/terry11.26.97.html    

You might also want to review these ideas for integers first: see

   The Official Euclidean Algorithm
     http://mathforum.org/dr.math/problems/julie.11.16.00.html    

and

   Explaining the Euclidean Algorithm
     http://mathforum.org/dr.math/problems/megan10.27.98.html  "

I understand this part:  "f(x) = g(x)q(x) + r(x), and 0 <= deg(r(x)) 
< deg(g(x)", but I don't see this part: "Next, by repeated 
applications of this algorithm, you obtain the Euclidean Algorithm 
for finding the greatest common divisor d(x) of two polynomials f(x) 
and g(x). The key idea we need is that d(x) can be written

   d(x) = f(x)r(x) + g(x)s(x)"

I looked up

   Euclidean and Division Algorithms
     http://mathforum.org/dr.math/problems/terry11.26.97.html    

and I seemed to be a little lost in the middle:
"First the Division Algorithm.  It is based on the theorem that 
given two polynomials F(x) and D(x) with D(x) nonzero, there exist 
polynomials Q(x) and R(x) such that F(x) = Q(x)*D(x) + R(x), and the 
degree of R is less than the degree of D.  Q is called the quotient, 
and R the remainder, when F is divided by D.

The proof of this theorem goes like this:

If F(x) = 0, then set Q(x) = 0 and R(x) = 0, and we are done."

If F(x) = Q(x)*D(x) + R(x), and F(x), I don't see the reason for 
setting Q(x) = R(x) = 0--what are we trying to prove by using these 
statements, and why "we are done"?

Sorry.  I think I have to stop here as this message is getting long.  
However, I will get back to you.  Thank you for the time!



Date: 12/22/2007 at 23:24:19
From: Doctor Peterson
Subject: Re: Integrals of rational functions

Hi, Ramiro.

The part you quote at the end here from Dr. Rob is the beginning of
his proof of the Division Algorithm, which you said at the top you
understood.  This is far more "obvious" than the Euclidean Algorithm
(in the form stated by Dr. Fenton, but a formal proof can be hard to
follow--that's why I offered a less formal discussion to help you see 
the main ideas.  I'm wondering if you're thinking he's already gone 
past the DA and is proving the EA; we don't yet know that F(x) =
Q(x)*D(x) + R(x), as you imply, but are proving it is always true.

Let's take some examples (which is often a good way to understand a
proof, or the theorem itself, before you get into the details).

He's giving three cases: either the dividend F is zero (which has no
degree); or it has degree less than that of the divisor D (this is his
"f<d"); or it has degree greater than or equal to D ("f>=d").

First case: if F(x) = 0 (for all x), then the quotient and remainder
are both zero:

  F(x) = Q(x)*D(x) + R(x)
  0    =  0  *D(x) + 0

This is true regardless of what the divisor, D(x), is.  That's what he
means by saying "we're done"; this is the easy case, where there's no
real work to do.  It's like saying, "It's easy to divide zero by
anything; the quotient is always zero, so there's nothing left to do."

Second case: if F(x) has degree less than D(x).  Then we can use F
itself as the remainder, since its degree is small enough; then the
quotient is 0.  I'll take the divisor as x^2+2x+3 and the dividend as x+1:

  F(x) = Q(x)*D(x)      + R(x)
  x+1  =  0  *(x^2+2x+3) + (x+1)

Third case: if the degree of F is at least as great as that of D, then
we do long division as usual (in fact, the term division algorithm
means exactly this: the method we use to divide polynomials).  This
time I'll take F(x) = x^2+2x+3 and D(x) = x+1, which has smaller degree):
        _________x_+_1_
  x + 1 ) x^2 + 2x + 3
          x^2 +  x
          --------
                 x + 3
                 x + 1
                 -----
                     2

He describes the first step of this in formal terms: we divide the
leading terms, in this case x^2 / x = x, and use that as the first
term of the quotient, T0(x) = (A/B)*x^(f-d).  The rest of the proof
shows that this process will end up by producing the desired quotient
and remainder.  He then gives an example, dividing F(x) = x^4 - 3*x^2
+ 2*x by D(x) = x^3 + x^2 - 2, which he shows in the same form as the
proof, rather than in the usual form:

                     _______________________x_-_1_
  x^3 + x^2 + 0x - 2 ) x^4 + 0x^3 - 3x^2 + 2x + 0
                       x^4 +  x^3 + 0x^2 - 2x
                       ----------------------
                             -x^3 - 3x^2 + 4x + 0
                             -x^3 -  x^2 + 0x + 2
                             --------------------
                                   -2x^2 + 4x - 2

After that, he goes on to prove the Euclidean Algorithm, which is
probably what is of more interest to you.

Note that what I gave you gives more of a sense of what partial
fractions are all about, without going through any of these details.
If you want a real proof, keep working through this, but if not, you
don't really have to.  It is good preparation for later math, though,
so I encourage you to go as far as you can.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 12/23/2007 at 14:01:17
From: Ramiro
Subject: Integrals of rational functions

Hello, Doctor Peterson -

Thank you for your help.  I have a question on this integral: 
Int [1/(4 sin x - 3 cos x)] dx.

The book demonstrates the steps until it says, "Using partial
fractions, we have 1/(3u^2 + 8u - 3) = (1/10)[3/(3u-1) - 1/(u+3)]..."
but it doesn't say how they got the expression on the right.  It 
only said to use partial fractions.

This is what I tried:

3u^2 + 8u - 3 = (u+3)(3u-1)

and I expect (u+3)(3u-1) to be in the form A/(u+3) + B/(3u-1), and 
solve for A and B.  So A(3u-1) + B(u+3) = (u+3)^2 (3u-1)^2.

Let u = 1/3, 10B/3 = 0, or B = 0, and let u = -3, A = 0.  I don't 
think this is the correct way.  But where am I going wrong?  When I 
substitute A = B = 0 into the original equation, it becomes 0 = (u+3)
^2 (3u-1)^2, which is certainly incorrect!

Or should I expect 1/[(u+3)(3u-1)] = A/(x+3) + B/(3u-1)?  I still 
don't really see why partial fractions works.



Date: 12/23/2007 at 21:04:05
From: Doctor Peterson
Subject: Re: Integrals of rational functions

Hi, Ramiro.

You wrote:

>This is what I tried:
>3u^2 + 8u - 3 = (u+3)(3u-1)
>
>and I expect (u+3)(3u-1) to be in the form A/(u+3) + B/(3u-1), and 
>solve for A and B.  

That is correct, if you really meant that the left side above should
be equal to this sum (which is how you say it below), not that the
denominator alone should equal this.

>So A(3u-1) + B(u+3) = (u+3)^2 (3u-1)^2.

I don't know where you get this. It looks like you meant what you 
wrote, which is not correct.  What you want instead is

        1           A       B
  ------------- = ----- + ------
  3u^2 + 8u - 3   u + 3   3u - 1

or

         1            A       B
  --------------- = ----- + ------
  (u + 3)(3u - 1)   u + 3   3u - 1

When you multiply by the LCD, (u+3)(3u-1), you get

  1 = A(3u - 1) + B(u + 3)

  1 = 3Au - A + Bu + 3B

  1 = (3A + B)u + (-A + 3B)

and equating coefficients,

  3A +  B = 0
  -A + 3B = 1

Solve that for A and B and you have it.

>Let u = 1/3, 10B/3 = 0, or B = 0, and let u = -3, A = 0.  I don't 
>think this is the correct way.  But where am I going wrong?  When I 
>substitute A = B = 0 into the original equation, it becomes 0 = (u+3)
>^2 (3u-1)^2, which is certainly incorrect!

It looks like you have been taught a different version of the method,
where (since the equation has to be true for all u) you can choose a
couple values of u that make it easy to solve for A and B. We want to
find A and B so that the following is always true:

  1 = A(3u - 1) + B(u + 3)

We will have 3u - 1 = 0 when u = 1/3, and then we have

  1 = 0A + 10/3 B

so that B = 3/10.

We will have u + 3 = 0 when u = -3, and then

  1 = -10A + 0B

so that A = -1/10.


>Or should I expect 1/[(u+3)(3u-1)] = A/(x+3) + B/(3u-1)?  I still 
>don't really see why partial fraction works.

Yes, this is correct.  Perhaps if you tell me why you thought what you
said above was correct, I can correct some subtle misunderstanding.

I had assumed you knew the HOW and were only asking about WHY, since
you correctly described the method initially.  Does this help you to
see the "how" better?


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 12/28/2007 at 00:06:58
From: Ramiro
Subject: Integrals of rational functions

Thank you, Doctor Peterson.  You were very helpful.
Associated Topics:
College Modern Algebra

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