29.3 Integration Using Partial Fractions 901 ◆ EXAMPLE 29.19 Find −2x−1 x 2 (x 2 +1) dx. SOLUTION The integrand can be broken into a sum as follows, −2x − 1 x 2 (x 2 + 1) = A x + B x 2 + Cx + D x 2 + 1 A repeated factor contributes multiple fractions, as shown. 3 We can solve for A, B, C, and D. One method is to clear denominators and evaluate both sides at four different x-values. This leaves us with four equations and four unknowns. We can show that −2x − 1 x 2 (x 2 + 1) = −2 x + −1 x 2 + 2x + 1 x 2 + 1 . −2x − 1 x 2 (x 2 + 1) dx =−2 1 x dx − 1 x 2 dx + 2x + 1 x 2 + 1 dx =−2ln|x|− x −1 −1 + 2x x 2 + 1 dx + 1 x 2 + 1 dx =−2ln|x|+ 1 x + ln(x 2 + 1) + arctan x + C ◆ General Procedure for Partial Fraction Decomposition of P(x) Q(x) If P(x) Q(x) is improper, do long division to express it as the sum of a polynomial and a proper rational function. Factor Q(x) into the product of linear and irreducible quadratic factors. (x − a) is a factor of Q(x) if Q(a) = 0 Rewrite the proper rational function as a sum of simpler rational functions as follows. For every nonrepeated linear factor (x − a) of Q there is a term A x−a For a repeated linear factor (x − a) n of Q there are n partial fractions. All numerators are constant; denominators consist of (x − a) raised to successively higher powers. E.g., for (x − a) 3 a factor of Q there are fractions A x−a + B (x−a) 2 + C (x−a) 3 . For a nonrepeated irreducible quadratic factor of Q, (x 2 + bx + c), there’s a term Ax+B x 2 +bx+c . If the quadratic factor of Q is repeated, there are multiple partial fractions, each with a linear expression in the numerator. E.g., (x 2 + 4) 2 as a factor of Q would result in the partial fractions Ax+B x 2 +4 + Cx+D (x 2 +4) 2 . Finding numerical values for the constants is an algebra problem. Integrating the partial fractions coming from linear factors of Q is straightforward. Some quadratics and repeated quadratics are straightforward while others require both completing the square and trigono- metric substitution. 3 The denominators are x and x 2 , not x and x. 902 CHAPTER 29 Computing Integrals PROBLEMS FOR SECTION 29.3 In Problems 1 through 6, write out the partial fraction decomposition of each rational function. You need not determine the coefficients; just set them up. 1. (a) x 2 +3 x(x−1)(x+5) (b) x x 3 +x 2. (a) 3 x 3 −4x (b) 4 x 3 +2x 3. (a) x 3 x 3 −4x (b) 3x+1 x 4 +2x 2 +1 4. (a) x+5 x 2 +3x−4 (b) x+5 x 2 −4 5. 3 x 3 +4x 6. x 2 +1 (x 2 +x+1)(x−1) In Problems 7 through 17, evaluate the integrals. 7. 3x+9 x 2 −6x+5 dx 8. 2 x(x−1) 2 dx 9. 1 0 x 2 2x+3 dx 10. 2 x 4 −1 dx 11. 3x 2 +3 (x 2 −1)(x−2) dx 12. 1 x(x−3) dx 13. e 2x (e x +2)(e x −1) 2 dx Hint: Use substitution. Do not attempt partial fractions until the integral is in the form P(x) Q(x) where P(x)and Q(x) are polynomials. 14. x −2 arctan xdx 15. x 2 arctan xdx 16. x 3 √ 1−x 2 dx 17. ln(x 2 − 1)dx 18. 2x 3 −2x 2 +4x+8 (x−2) 2 (x 2 +3) dx 19. x 3 x 2 +x−6 dx 29.4 Improper Integrals 903 29.4 IMPROPER INTEGRALS Mathematicians make a habit of testing and pushing the boundaries of their theorems and definitions. In this section we’ll push the boundaries of the definition of the definite integral. The definition of the definite integral b a f(x)dx requires that the interval of integration, [a, b], is finite and the integrand, f(x),isdefined and bounded everywhere on [a, b]. In this section we extend our notion of integral to include integrals for which one or both of these conditions break down. Such an integral is called an improper integral. The integral ∞ 1 1 x 2 dx is improper because the interval of integration is unbounded; 1 0 x −1/2 dx is improper because the integrand is unbounded at x = 0. What motivates us to try to make sense of improper integrals? A perfectly good answer is “to see if we can.” But in fact, physicists and statisticians work with such integrals on a regular basis. For example, probabilities are computed by calculating the area under a distribution p(x), called a probability density function. The probability of a randomly chosen data point lying to the right of the marker “a” on the distribution is given by ∞ a p(x) dx, where p(x) is a nonnegative function. p(x) x a Figure 29.3 The probability of lying somewhere must be 1, (100 %), so if p(x) is a probability density function, then ∞ −∞ p(x) dx = 1. Physicists use improper integrals in their models with great regularity, moving particles in “from infinity” in particle physics, and computing integrals such as 1 0 1+x 1−x dx when studying aerodynamics. The integral 1 0 1+x 1−x dx is improper because the integrand is unbounded, “blowing up” at x = 1. We’ll make sense of improper integrals by taking up the types of improprieties, infinite interval and unbounded integrand, one at a time. Our method of dealing with an impropriety at an endpoint of integration is to excise it, constructing an ordinary integral with an endpoint that we allow to approach the trouble spot. If the appropriate limit exists and is finite, we define the improper integral to be that limit. If the impropriety does not occur at an endpoint, then we split up the integral so that it does. We’ll begin with a few examples. Infinite Interval of Integration ◆ EXAMPLE 29.20 Compute ∞ 1 1 x 2 dx, if possible. SOLUTION ∞ 1 1 x 2 dx corresponds graphically to the area under 1 x 2 and is bounded on the left by x = 1. We approach this problem by looking at the proper integral b 1 1 x 2 dx, and letting b increase without bound. As b increases the accumulated area increases; the question is whether or 904 CHAPTER 29 Computing Integrals not the accumulated area increases without bound. b 1 1 x 2 dx = −1 x b 1 =− 1 b + 1 lim b→∞ b 1 1 x 2 dx = lim b→∞ − 1 b + 1 = 1 So the area of the shaded region approaches 1 as b increases without bound. We say ∞ 1 1 x 2 dx is convergent, and converges to 1. x y b1 y = 1 x 2 Figure 29.4 ◆ ◆ EXAMPLE 29.21 Compute ∞ 1 1 x dx, if possible. SOLUTION Again, we integrate from 1 to b (we compute a proper integral) and take the limit as b →∞. b 1 1 x dx = ln x b 1 = ln b − ln 1 = ln b lim b→∞ b 1 1 x dx = lim b→∞ ln b =∞ Wesay ∞ 1 1 x dx diverges. x y b1 y = 1 x Figure 29.5 ◆ REMARK It is not at all clear from looking at the graphs of 1 x 2 and 1 x that ∞ 1 1 x 2 dx should converge to 1 and ∞ 1 1 x dx should be unbounded. In fact, it might surprise you. After all, as x →∞both 1 x and 1 x 2 approach 0. The crucial difference is that 1 x 2 approaches 0 much faster. What should not surprise you is that 0 < ∞ 1 1 x 3 dx ≤ 1, because 0 < 1 x 3 ≤ 1 x 2 for x ≥ 1. More generally, if f is continuous and f(x)≥0 for all x ≥ a, then for b>a b a f(x)dx increases with b. Therefore, lim b→∞ b a f(x)dxis either finite or increases without bound. In other words, if lim b→∞ b a f(x)dx is bounded, then ∞ a f(x)dx converges, provided f is nonnegative on [a, ∞). 29.4 Improper Integrals 905 Similarly, we can argue that because 0 < 1 x ≤ 1 √ x for x ≥ 1, 0 < b 1 1 x dx ≤ b 1 1 √ x dx for b ≥1. As b increases without bound b 1 1 x dx increases without bound, so b 1 1 √ x dx increases without bound as well. This type of comparison was not useful when dealing with 1 x and 1 x 2 .0< 1 x 2 ≤ 1 x for x ≥ 1, but being larger than a convergent integral is an inconclusive characteristic, as is being smaller than a divergent integral. 1 x 2 1 x 1 √x 1 √x 1 x 2 1 x 3 1 x 3 x y 1 x y 1 0 < for x ≥ 1 ≤ 1 x 0 < for x ≥ 1 ≤ Figure 29.6 Definition Consider ∞ a f(x)dx, where b a f(x)dx exists for all b ≥ a. ∞ a f(x)dx = lim b→∞ b a f(x)dx provided this limit exists and is finite. If the limit is finite, then ∞ a f(x)dx converges. Otherwise the integral diverges. a −∞ f(x)dx=lim b→−∞ a b f(x)dxprovided this limit exists and is finite, and a b f(x)dx exists for all b ≤ a. If ∞ a f(x)dx and a −∞ f(x)dx both converge, then ∞ −∞ f(x)dx = a −∞ f(x)dx + ∞ a f(x)dx. If one or both of ∞ a f(x)dx and a −∞ f(x)dx diverges, then ∞ −∞ f(x)dx diverges. ◆ EXAMPLE 29.22 Evaluate ∞ −∞ 1 1+x 2 dx if this integral is convergent. SOLUTION Split this into two improper integrals. x = 0 makes a convenient breaking point. 906 CHAPTER 29 Computing Integrals y x y = 1 1 + x 2 Figure 29.7 ∞ −∞ 1 1+x 2 dx = 0 −∞ 1 1+x 2 dx + ∞ 0 1 1+x 2 dx, provided both these integrals converge. ∞ 0 1 1 + x 2 dx = lim b→∞ b 0 1 1 + x 2 dx = lim b→∞ arctan x b 0 = lim b→∞ arctan b − arctan 0 = lim b→∞ arctan b = π 2 y x y = tan x x = – π 2 x = π 2 Figure 29.8 1 1+x 2 is an even function, so 0 −∞ 1 1+x 2 dx = π 2 , and ∞ −∞ 1 1+x 2 dx = π 2 + π 2 = π . ◆ So far the divergent improper integrals we’ve looked at diverge by growing without bound. The next example illustrates another way of diverging. ◆ EXAMPLE 29.23 Determine whether ∞ 0 sin xdxconverges or diverges. SOLUTION ∞ 0 sin xdx= lim b→∞ b 0 sin xdx = lim b→∞ − cos x b 0 = lim b→∞ − cos b + 1 29.4 Improper Integrals 907 This limit doesn’t exist. The value of − cos b +1 oscillates between 0 and 1. This improper integral diverges by oscillation. ◆ EXERCISE 29.10 Show that ∞ 1 e −x dx = 1 e . The solution is embedded in the solution to Example 29.28. Observation If lim x→∞ f(x)=0,then ∞ 0 f(x)dx diverges. If lim x→∞ f(x)=0, then work must be done in order to determine whether or not ∞ a f(x)dx diverges or converges. Unbounded and Discontinuous Integrands We begin with an example. ◆ EXAMPLE 29.24 Find 1 0 1 √ x dx. y x b 1 y = 1 √x Figure 29.9 SOLUTION This integral is improper because 1 √ x blows up at x = 0. To get around this problem we compute 1 b 1 √ x dx for b>0and take the limit as b → 0 + . lim b→0 + 1 b x − 1 2 dx = lim b→0 + 2x 1 2 1 b = lim b→0 + 2 ·1 −2 √ b = 2 ◆ ◆ EXAMPLE 29.25 Show that 1 0 1 x dx diverges. SOLUTION 1 0 1 x dx ⇒ lim b→0 + 1 b 1 x dx = lim b→0 + ln |x| 1 b = lim b→0 + 0 −ln |b|=−(−∞). y x B A 1 1 y = 1 x Figure 29.10 908 CHAPTER 29 Computing Integrals One could also argue that the integral diverges by the symmetry of the shaded regions in Figure 29.10. Area A is infinite, so area B is infinite. But 1 0 1 x dx = 1+ (the area of B). Consequently 1 0 1 x dx is divergent. ◆ Definition If f is continuous on (c, d] and discontinuous at c, then d c f(x)dx = lim b→c + d b f(x)dx provided this limit exists and is finite. If f is continuous on [c, d) and discontinuous at d, then d c f(x)dx = lim b→d − b c f(x)dx provided this limit exists and is finite. If f has a discontinuity at p, c<p<d,then d c f(x)dx = p c f(x)dx + d p f(x)dx provided both integrals are convergent. It can be shown that i. ∞ 1 1 x p dx converges for p>1and diverges for p ≤ 1, ii. 1 0 1 x p dx converges for p<1and diverges for p ≥ 1. This is left as an exercise. EXERCISE 29.11 Prove statement (i) above. EXERCISE 29.12 Prove statement (ii) above. 4 How to Approach an Improper Integral: An Informal Summary 1. Determine all improprieties: Infinite interval, discontinuous integrand, unbounded integrand. 2. If necessary, split up the integral into the sum of integrals so that (a) each integral has only one impropriety, (b) the impropriety is exposed; it occurs at the endpoint of an integral. 3. Compute each improper integral as the limit of a proper integral. Replace the endpoint at which the impropriety occurs and compute the appropriate one-sided limit. 4. The original integral converges only if each of the summands converges independently. 4 Both exercises are also included as problems at the end of the chapter. 29.4 Improper Integrals 909 ◆ EXAMPLE 29.26 Compute ∞ −∞ 1 x 3 dx, if possible. SOLUTION This integral is improper at x = 0 and at the endpoints of integration. Split the interval up: (−∞, −1], [−1, 0], [0, 1], [1, ∞). –1 XXX 10 ∞ −∞ 1 x 3 dx = −1 −∞ 1 x 3 dx + 0 −1 1 x 3 dx + 1 0 1 x 3 dx + ∞ 1 1 x 3 dx We need to compute four different limits. We know ∞ 1 1 x 3 dx will converge, either having completed Exercise 29.11 or because 0 < 1 x 3 ≤ 1 x 2 for x ≥ 1 and ∞ 1 1 x 2 dx converges. So ∞ 1 1 x 3 is increasing and bounded. g x y = 1 x 3 Figure 29.11 (By symmetry, we know −1 −∞ 1 x 3 dx is convergent, converging to − ∞ 1 1 x 3 dx.) If you’ve completed Exercise 29.12, you know 1 0 1 x 3 dx diverges. We show this below. 1 0 1 x 3 dx = lim b→0 + 1 b 1 x 3 dx = lim b→0 + x −2 −2 1 b = lim b→0 + − 1 2 + 2 2b 2 =∞ ⇒ The integral 1 0 1 x 3 dx diverges. If any one of the summands diverges the original integral is divergent, so we’re done. ◆ 910 CHAPTER 29 Computing Integrals ◆ EXAMPLE 29.27 Compute 3 0 1 x−2 dx, if possible. SOLUTION The integrand blows up at x = 2, and 2 ∈ [0, 3]. Therefore, we must split the interval into [0, 2] and [2, 3]. 3 0 1 x − 2 dx = 2 0 1 x − 2 dx + 3 2 1 x − 2 dx You can compute either of these summands independently. Both diverge, so the original integral diverges as well. Alternatively, you can notice that 3 2 1 x − 2 dx = 1 0 1 u du and this latter integral diverges. (This was shown in Example 29.25.) y x = 2 y = 1 u y xu y = 1 x –2 3 Figure 29.12 CAUTION If you aren’t careful you might not notice that 3 0 1 x−2 dx is an improper integral. Failure to notice this makes your work fatally flawed, as you would erroneously write 3 0 1 x − 2 dx = ln |x − 2| 3 0 = ln |1|−ln |−2|=−ln 2. ◆ ◆ EXAMPLE 29.28 Determine whether ∞ 0 e −x 2 dx converges or diverges. SOLUTION We have an interesting dilemma here because we can’t find an antiderivative for e −x 2 . Therefore, we’ll argue by comparison. e −x 2 approaches zero very quickly, so we expect the integral to converge. Therefore, we’ll show that it is less than some convergent integral. For x ≥ 1, 0 <e x ≤e x 2 so 1 e x ≥ 1 e x 2 ∞ 0 e −x 2 dx = 1 0 e −x 2 dx + ∞ 1 e −x 2 dx. . INTEGRALS Mathematicians make a habit of testing and pushing the boundaries of their theorems and definitions. In this section we’ll push the boundaries of the definition of the definite integral. The definition of. the integrand is unbounded at x = 0. What motivates us to try to make sense of improper integrals? A perfectly good answer is to see if we can.” But in fact, physicists and statisticians work. function. Factor Q(x) into the product of linear and irreducible quadratic factors. (x − a) is a factor of Q(x) if Q(a) = 0 Rewrite the proper rational function as a sum of simpler rational functions