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29.4 Improper Integrals 911 y y = e—x2 y = e—x 1 x Figure 29.13 The rst summand is proper. Well concern ourselves with the second integral and compare with the convergent integral 1 . 2 lim 2 1 1 We will show that this limit exists and is nite. We know 2 0, so for 1, 2 increases with . Therefore, as , 1 either grows without bound or is nite. 0 2 because 2 on [1, ] 1 1 0 lim 2 lim 1 1 1 lim 1 lim so 0 2 1.Therefore 2 isconvergent.Weconcludethat 2 converges. y 1 y = e—x2 x Figure 29.14 REMARK 2 is an interesting integral. Weve concluded that it is convergent; using more advanced methods it can be shown that its value is . If we wanted to approximate 2 , we could proceed as follows. i. 2 2 2 ii. Cut the tail off of 0 2 and bound it. For instance, 6 2 6 0.0025. Even better, bound by . For instance, 912 CHAPTER 29 Computing Integrals 5 2 5 2 225 7 1012 The interested reader has many details to Þll in here. The claim is that 2 is a much better bound, a tighter Þt, than is . y y = e—x2 y = xe—x2 1 x Figure 29.15 iii. Use numerical methods to approximate the proper integral 2 after the tail, 2 , has been amputated. Notice that the graph of 2 is bell-shaped. As stands, 2 cannot be a probability density function because the area under such a function must be exactly 1. A bit of tinkering takes care of this. A standard normal distribution in statistics is described mathematically by the formula 22 . The Method of Comparison Weveusedthemethodofcomparisoninseveralinstances.Westatethecomparisontheorem below. We omit the formal proof, but the statements should seem quite reasonable; an informal argument was provided earlier in this section. C o m p a r i s o n T h e o r e m Let and be continuous functions with 0 for . If converges, then converges. If diverges, then diverges. Suppose is positive and continuous. To show converges, we must produce a larger function whose improper integral converges. To show diverges, we must produce a smaller function whose improper integral diverges. 29.4 Improper Integrals 913 Naturally, we cant produce both, so we begin by taking a guess about whether or not converges. The integral (and constant multiples of this integral) can be useful for comparison. Recall that this integral converges for 1 and diverges for 1. EXAMPLE 29.29 Is 1 1sin2 convergent? y y =1 + sin2 x x Figure 29.16 SOLUTION 0 1sin diverges, so 1 1sin diverges by comparison. REMARK Notonlycanwecompareimproperintegralswithoneanotherbutwecancompare improper integrals with innite series. See Figure 29.17. y y = 1 area = 1 area = 1 area = area = 1 1 2 3 4 5 x Figure 29.17 Theshadedcircumscribedrectangleshaveareascorrespondingtothetermsoftheharmonic series. The rectangles lie above . We can argue that the harmonic series 1 2 3 4 divergesbecause 1 1 islargerthan 1 and 1 isadivergentimproper integral. We can argue that the innite series 2 2 2 3 4 converges by com-paring it to the convergent improper integral 1 2 . 914 CHAPTER 29 Computing Integrals y y = 12 area = 1 area = 1 area = 16 1 2 3 4 5 x Figure 29.18 Similarly, after completing Exercise 29.11 we can argue that 1 1 converges for 1 and diverges for 1. We will formalize the comparison between improper integrals and innite series by the end of the next chapter. P R O B L E M S F O R S E C T I O N 2 9 . 4 In Problems 1 through 5, pinpoint all the improprieties in the integral. If necessary rewritetheintegralasasumofintegralssothateachimproprietyoccursatanendpoint and there is only one impropriety per integral. 1. (a) 0 24 2. (a) 0 2 3. (a) 24 4. (a) tan 2 5. (a) tan1 (b) 0 24 (b) 2 (b) 24 (b) tan (b) tan1 6. Show that 1 converges for 1 and diverges for 1. 7. Show that 1 1 converges for 1 and diverges for 1. 8. Show that 1 1 diverges. 9. (a) Evaluate 0 2 . (b) Evaluate 2 . 10. Show 4 2 0.0000001. Hint: Compare it to 4 2 . In Problems 11 through 36, determine whether the integral is convergent or divergent. Evaluate all convergent integrals. Be efÞcient. If lim 0, then is divergent. 29.4 Improper Integrals 915 11. 2 12. 13. cos 1 14. cos 15. 16. 1 52 17. 1 18. 1 ln 19. ln 20. 1 21. 1 22. ln 23. ln 2 24. 22 25. 1 26. 1 1 27. 5 3 28. ln 29. 3 30. arctan 31. 2 ln 32. arctan 33. tan ... - --nqh--