5.7 Improper Integrals

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1 458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the moon: R+H R Gm 2 d = Gm R+H R = Gm R Gm R + H Notice tht s the height H grows very lrge, the second term in this nswer becomes very smll nd the totl work pproches Gm R. We cn write: Gm lim H R Gm = Gm R + H R Here we re tking limit of n epression tht rose s the vlue of definite integrl, so we cn lso write: Gm Gm R H R Gm R+H Gm R + H H R 2 d We could write this lst integrl, t lest informlly, s: R Gm 2 We cll this new type of integrl n improper integrl becuse the intervl of integrtion is infinite, violting n ssumption we mde when d originlly developing the definite integrl b f () d using Riemnn sums tht the length of the intervl of integrtion,, b, ws finite. Emple. Represent the re of the infinite region between f () = 2 nd the -is for (see mrgin) s n improper integrl. Solution. We cn represent the re of region (which hs infinite length) s: 2 d We don t yet know whether this re is finite or infinite. Prctice. Represent the volume swept out when the infinite region between f () = nd the -is for 4 is revolved bout the -is (see mrgin) using n improper integrl. Generl Strtegy for Improper Integrls In the lifting--pylod emple bove, we defined our first improper integrl s the limit of proper integrl over finite intervl s the length of the intervl becme lrger nd lrger.

2 5.7 improper integrls 459 Our generl pproch to evlute improper integrls over infinitely long intervls s well s nother type of improper integrl introduced lter in this section will mimic this strtegy: Shrink the intervl of integrtion so you hve (proper) definite integrl you cn evlute, then let the intervl grow to pproch the desired intervl of integrtion. The vlue of the improper integrl will be the limiting vlue of the (proper) definite integrls s the intervls grow to the intervl you wnt, provided tht this limit eists. Infinitely Long Intervls of Integrtion To evlute n improper integrl on n infinitely long intervl: replce the infinitely long intervl with finite intervl evlute the integrl on the finite intervl let the finite intervl grow longer nd longer, pproching the originl infinitely long intervl Emple 2. Evlute d (see mrgin). 2 Solution. The intervl, ) is infinitely long, but we cn evlute the integrl on finite intervls such s, 2,,,, nd, more generlly,, where is some mssive positive number: 2 2 d = 2 2 d = 2 d = nd, more generlly, = = = 2 2 = 2 = d 2 = = 9 = = d = so: The vlue of the improper integrl is. d 2 = We sy tht the improper integrl d in the Emple 2 is 2 convergent nd tht it converges to. Furthermore, from Emple, we know tht this improper integrl represents the re of n infinitely long region. We now hve n emple which you my find highly counterintuitive of region with infinite length but finite re. Not ll improper integrls converge, however.

3 46 pplictions of definite integrls Emple. Evlute ech improper integrl. (See mrgin for grphicl interprettions of these integrls s res of unbounded regions.) () + 2 d (b) d (c) cos() d Solution. () Replcing the upper limit of the improper integrl with mssive positive number : d d 2 = π 2 rctn() rctn() so the improper integrl is convergent nd converges to π 2. (b) Replcing the upper limit of the improper integrl with mssive positive number : d d ln() ln() = Becuse this limit diverges, we sy the improper integrl is divergent or tht it diverges. (c) Once gin replcing with in the upper limit of the integrl: lim cos() d sin() sin() As grows without bound, the vlues of sin() oscillte between nd, never pproching single vlue, so the limit does not eist; we sy tht this improper integrl diverges. Prctice 2. Evlute: () d (b) sin() d Definition: For ny integrble function f () defined for ll nd ny integrble function g() defined for ll b: b f () d g() d = lim N b N f () d g() d If the limit in question eists nd is finite, we sy tht the corresponding improper integrl converges or is convergent nd define the vlue of the improper integrl to be the vlue of the limit. If the limit in question does not eist, we sy tht the corresponding improper integrl diverges or is divergent.

4 5.7 improper integrls 46 Functions Undefined t n Endpoint of the Intervl of Integrtion Consider the grph of on the intervl (, (see mrgin) nd compre this region to the grph from Emple 2. It ppers we cn generte the new region by reflecting the old region cross y = nd dding rectngle (of re ) t the bottom, so we might resonbly ssume tht the integrl d is finite number. This integrl is over finite intervl,,, but we hve new problem: the integrnd is undefined t =, one of the endpoints of the intervl of integrtion. This violtes nother ssumption we mde when developing the definition of definite integrl s limit of Riemnn sums. If the function you wnt to integrte is unbounded t one of the endpoints of n intervl of finite length, s in this sitution, you cn shrink the intervl of integrtion so tht the function is bounded t both endpoints of the new, smller intervl, then evlute the integrl over the smller intervl, nd finlly let the smller intervl grow to pproch the originl intervl. Emple 4. Evlute d. Solution. The function is not defined t =, the lower endpoint of integrtion, but the function is bounded on intervls such s.6,,.9, nd, more generlly, on the intervl c, for ny c > :.6.9 nd, in generl: d = 2 d = 2 c.6 = = 2.2 =.8.9 = = 2.6 =.4 d = 2 = 2 2 c = 2 2 c c so, tking the limit s c decreses towrd : lim d 2 2 c = 2 c + c c + which is wht you should hve epected bsed on the grph. A region of re plus rectngle of re should hve n re of + = 2. Definition: For ny function f () defined nd continuous on (, b nd ny function g() defined nd continuous on, b): b b b f () d c + g() d c b c c f () d g() d

5 462 pplictions of definite integrls See Problems for prctice with integrls of this type. If the limit eists, we sy the integrl converges nd define the vlue of the integrl to be the vlue of the limit. If the limit does not eist, we sy tht the integrl diverges. Prctice. Show tht () d = 6 nd (b) d diverges. If n integrnd is unbounded t one or more points inside the intervl of integrtion, you cn split the originl improper integrl into two or more improper integrls over subintervls where the integrnd is unbounded t only one endpoint of ech subintervl. Testing for Convergence: The P-Test nd the Comprison Test Sometimes we cre only whether or not n improper integrl converges. We now consider two methods for testing the convergence of n improper integrl. Neither method gives you the ctul vlue of the integrl, but ech enbles you to determine whether or not certin improper integrls converge. The Comprison Test for Integrls enbles you to determine the convergence (or divergence) of certin integrls by compring them with other (esier) integrls. The P-Test involves specil cses often used with the Comprison Test for Integrls. P-Test for integrls: For ny >, the improper integrl converges if p > nd diverges if p. = p d Proof. It is esiest to consider three cses rther thn two: p =, p > nd p <. If p = then: p d = d d ln ( ) ln() ln() so the improper integrl diverges. For the other two cses, p =, so: lim p p+ p d p + p p p If p >, then p < so lim p p = nd: p+ p d p + p+ p + = p p which is finite number, so the improper integrl converges. If p <, then p > p so lim p = nd: p d so the improper integrl diverges. p+ p + p+ = p +

6 5.7 improper integrls 46 Emple 5. Determine the convergence or divergence of ech integrl. () 5 2 d (b) d (c) 8 d Solution. () The integrl mtches the form required by the P-Test with p = 2 >, so the improper integrl converges. The P-Test does not tell us the vlue of the integrl. (b) The integrl mtches the form required by the P-Test with p = 2 <, so the improper integrl diverges. (c) This is not n improper integrl, so the P-Test does not pply, but: 8 8 d = 8 d = 2 2 = = = 9 2 so the vlue of the integrl is 4.5. The following Comprison Test enbles us to determine the convergence or divergence of n improper integrl of positive function by compring this function with functions whose improper integrls we lredy know converge or diverge. Comprison Test for Integrls of Positive Functions: Suppose f () nd g() re defined nd integrble for ll with f () g(). Then: g() d converges f () d converges. f () d diverges g() d diverges. The proof involves strightforwrd ppliction of the definition of n improper integrl nd vrious fcts bout limits, but the grph in the mrgin provides geometriclly intuitive wy of understnding why these results must hold. If g() d converges, then the re under the grph of g() is finite, so the (smller) re under the grph of f () must lso be finite, nd f () d must converge s well. If f () d diverges, then the re under the grph of f () is infinite, so the (bigger) re under the grph of g() must lso be infinite, nd g() d must lso diverge. Just s importnt s understnding wht this Comprison Test does tell us is relizing wht the Comprison Test does not tell us. If g() d diverges, or if f () d converges, the Comprison Test tells us bsolutely nothing bout the convergence or divergence of the other integrl. Geometriclly, if g() d diverges, then the re under the grph of g() is infinite, but the (smller) re under the

7 464 pplictions of definite integrls grph of f () could be either finite or infinite, so we cn t conclude nything bout the convergence or divergence of f () d. Likewise, if f () d converges, then the re under the grph of f () is finite, but the (bigger) re under the grph of g() could be either finite or infinite, so we cn t conclude nything bout the convergence or divergence of g() d. Emple 6. Determine whether ech of these integrls is convergent or divergent by compring it with n pproprite integrl tht you lredy know converges or diverges. () d (b) Solution. () We know tht 5 > nd so: + sin() 9 2 d (c) d > < + 5 < < < 7 The numertor of the originl integrnd is constnt nd the dominnt term in the denomintor of tht integrnd is, so it should mke sense to compre the originl integrnd with. The numertor of the originl integrnd fluctutes between 2 nd 4 while the dominnt (nd only) term in its denomintor is 2, so it should mke sense to compre the originl integrnd with 2. We lso know, by the P-Test with p = >, tht d con- verges, so 7 d = 7 d lso converges. By the Com- 7 d must converge s + 5 prison Test, the smller integrl well. (b) We know tht sin(), so: 2 + sin() 4 < + sin() = 4 2 By the P-Test with p = 2 >, 2 d converges, so 4 2 d = 4 d lso converges. By the Comprison Test, the smller 2 + sin() integrl 2 d must converge s well. (c) We know tht u is n incresing function, so: 5 < 5 < 5 > By the P-Test with p = 2 <, integrl 6 6 d 5 d must lso diverge. diverges, so the bigger

8 5.7 improper integrls Problems In Problems 26, evlute ech improper integrl, or show why it diverges d e π 2 π d 4. 5 ln() d 2 d e ( 2) d. ( + 2) 2 d 2. d 4. 4 d d 8. 2 sin() d 2. tn() d 22. tn() d d 26. π 5 ln() 2 d 2 e d + 2 d ( 2) 2 d + 2 d ( + 2) d d 2 d 2 8 d sin() d 2 d d 2 d e π 7 4 d d + d 7 + ln() d + cos() 2 d d d d 2 d 2 d d d 45. Emple (b) showed tht d grew rbitrrily lrge s grew rbitrrily lrge, so no finite mount of pint would cover the region bounded by the -is nd the grph of f () = for > : Show tht the volume of the solid obtined when the region grphed bove is revolved bout the -is: In Problems 27 44, determine whether ech improper integrl converges or diverges, but do not evlute the integrl d d d d d 4 7 d is finite, so the -dimensionl trumpet-shped region cn be filled with finite mount of pint. Does this present contrdiction?

9 466 pplictions of definite integrls 46. Determine whether or not the volume of the solid obtined by revolving the region between the - is nd the grph of f () = sin() for for (see below) bout the -is is finite. 5. Use the figure below left to help determine which A A is lrger: d or k. k= 47. Compute the volume of the solid obtined when the region in the first qudrnt between the positive -is nd the grph of f () = 2 + (see below) is revolved bout the -is. 52. Use the figure bove right to help determine A A which is lrger: d or k. k=2 5. Use the figure below left to help determine which is lrger: A A 2 d or k= k Compute the volume of the solid obtined when the region in the first qudrnt between the positive -is nd the grph of f () = e is revolved bout the -is. 49. Compute the volume of the solid obtined when the region in the first qudrnt between the positive -is nd the grph of f () = 2 + (see below) is revolved bout the y-is. 5. Compute the volume of the solid obtined when the region in the first qudrnt between the positive -is nd the grph of f () = e is revolved bout the y-is. 54. Use the figure bove right to help determine which is lrger: A 2 d or A k=2 k 2. The Lplce trnsform of function f (t) is defined using n improper integrl involving prmeter s: F(s) = e st f (t) dt Lplce trnsforms re often used to solve differentil equtions. 55. Compute the Lplce trnsform of the constnt function f (t) =. 56. Compute the Lplce trnsform of f (t) = e 4t. 57. Define function g(t) by: { if t < 2 g(t) = if t 2 Compute the Lplce trnsform of g(t). 58. Define function h(t) by: { if t < h(t) = if t Compute the Lplce trnsform of h(t).

10 5.7 improper integrls 467 b 59. Devise Q-Test to determine whether q d converges or diverges for ny number b >. 6. Use the result of the previous problem to test the e π convergence of d nd d. 5.7 Prctice Answers ( ) 2. π d = π d 2. () d d = 2 2 (b) Replcing with in the upper limit of the integrl: sin() d sin() d cos() + cos() cos() This limit does not eist (the vlues of cos() oscillte between nd 2 nd never pproch ny fied number) so the improper integrl diverges.. () The integrl is improper t its upper limit, where =, so: d = c lim ( ) 2 d c 2 c c c 2 c = + 2 = 6 (b) The integrl is improper t its lower limit, where =, so: d d ln ( ) c + c ln() ln(c) = c + c c + so the integrl diverges.

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