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.
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.