!0 f(x)dx + lim!0 f(x)dx. The latter is sometimes also referred to as improper integrals of the. k=1 k p converges for p>1 and diverges otherwise.


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1 Chpter 7 Improper integrls 7. Introduction The gol of this chpter is to meningfully extend our theory of integrls to improper integrls. There re two types of soclled improper integrls: the first involves integrting function over n infinite domin nd the second involves integrnds tht re undefined t points within the domin of integrtion. In order to integrte over n infinite domin, we R b consider its of the form b! f(x)dx. If the integrnd is not defined t c ( <c< b) then we split the integrl nd consider the its R b f(x)dx = R c! f(x)dx + R b! f(x)dx. The ltter is sometimes lso referred to s improper integrls of the c+ second kind. Such situtions occur, for exmple, for rtionl functions f(x) =p(x)/q(x) whenever q(x) hs zeroes in the domin of integrtion. The notions of convergence nd divergence s discussed in Chpter & in the context of sequences nd series will be very importnt to determine these its. Improper integrls (of both types) rise frequently in pplictions nd in probbility. By relting improper integrls to infinite series we derive the lst convergence test: the Integrl Comprison test. As n ppliction we finlly prove tht the pseries P k= k p converges for p> nd diverges otherwise. 7.2 Integrtion over n infinite domin We will see tht there is close connection between certin infinite series nd improper integrls, which involve integrls over n infinite domin. We hve lredy encountered exmples of improper integrls in Section 3.8 nd in the context of rdioctive decy in Section 8.4. Recll the following definition: Definition (Improper integrl (first kind)). An improper integrl of the first kind is n integrl performed over n infinite domin, e.g. Z f(x) dx. 89
2 9 Chpter 7. Improper integrls The vlue of such n integrl is understood to be it, s given in the following definition: Z f(x) dx = b! f(x) dx. We evlute n improper integrl by first computing definite integrl over finite domin pple x pple b, nd then tking it s the endpoint b moves off to lrger nd lrger vlues. The definite integrl cn be interpreted s n re under the grph of the function. The essentil question being ddressed here is whether tht re remins bounded when we include the infinite til of the function (i.e. s the endpoint b moves to lrger vlues.) For some functions (whose vlues get smll enough, fst enough) the nswer is yes. Definition 2 (Convergence). If the it b! f(x) dx exists, we sy tht the improper integrl converges. Otherwise we sy tht the improper integrl diverges. With these definitions in mind, we cn compute number of clssic integrls Exmple: Decying exponentil Recll tht the improper integrl of decying exponentil converges we hve seen this erlier, in Section 3.8.5, nd gin in pplictions in Sections 7.3 nd Here we recpitulte this importnt result in the context of improper integrls. Suppose tht r> nd let Then b! b r e rt Z e rt dt b! e rt dt. = r (e rb e )= b! r ( b! e rb {z } ) = r, where we hve used the fct tht b! e rb =for r>. Thus the it exists (is finite) nd hence the integrl converges. More precisely, it converges to the vlue /r Exmple: The improper integrl of /x diverges We now consider clssic nd counterintuitive result, nd one of the most importnt results in this chpter. Consider the function y = f(x) = x.
3 7.2. Integrtion over n infinite domin 9 Exmining the grph of this function for positive x, e.g. for the intervl (, ), we know tht vlues decrese to zero s x increses 26. The function is not only bounded, but lso flls to rbitrrily smll vlues s x increses (see Figure 7.). Nevertheless, this is insufficient y y= /x y=/x 2 x Figure 7.. In Sections nd 7.2.3, we consider two functions whose vlues decrese long the x xis, f(x) =/x nd f(x) =/x 2. We show tht one, but not the other encloses finite (bounded) re over the intervl (, ). To do so, we compute n improper integrl for ech one. The hevy rrow is ment to remind us tht we re considering res over n unbounded domin. to gurntee tht the enclosed re remins finite! We mde similr observtion in the context of series in Section.3.. We show this in the following clcultion. Z Z b b dx = dx = x b! x ln(x) = (ln(b) ln()) b! b! = ln(b) = b! The fct tht we get n infinite vlue for this integrl follows from the observtion tht ln(b) increses without bound s b increses, tht is the it does not exist (is not finite). Thus, the re under the curve f(x) =/x over the intervl pple x ppleis infinite. We sy tht the improper integrl of /x diverges (or does not converge). We will use this result gin in Section Exmple: The improper integrl of /x 2 converges Now consider the relted function y = f(x) = x 2 nd the corresponding integrl Z x 2 dx. 26 We do not chose the intervl (, ) becuse this function is undefined t x =. Here we wnt to emphsize the behviour t infinity, not the blow up tht occurs close to x =.
4 92 Chpter 7. Improper integrls Then b! b x 2 dx = ( x ) = b! b! b =. Thus, the it exists, nd is. In contrst to the exmple in Section 7.2.2, this integrl converges. We observe tht the behviours of the improper integrls of the functions /x nd /x 2 re very different. The former diverges, while the ltter converges. The only difference between these functions is the power of x. As shown in Figure 7., tht power ffects how rpidly the grph flls off to zero s x increses. The function /x 2 decreses much fster thn /x. Consequently /x 2 hs sufficiently s infinite til, such tht the re under its grph does not become infinite  not n esy concept to digest! This observtions leds us to wonder wht power p is needed to mke the improper integrl of function /x p converge. We nswer this question below When does the integrl of /x p converge? Here we consider n rbitrry power, p, tht cn be ny rel number. We sk when the corresponding improper integrl converges or diverges. Let Z x p dx. For p =we hve lredy estblished tht this integrl diverges (see Section 7.2.2), nd for p =2we hve seen tht it is convergent (see Section 7.2.3). By similr clcultion, we find tht b! x p ( p) b = b! p b p. Thus, this integrl converges provided tht the term b p does not blow up s b increses. For this to be true, we require tht the exponent ( p) should be negtive, i.e. p< or p>. In this cse, we hve p. To summrize our result, Z dx converges if p>, nd diverges if p pple. xp Exmples: (i) The integrl Z p x dx, diverges. We see this from the following rgument: p x = x 2, so p = 2 generl result, this integrl diverges. <. Thus, by the
5 7.3. Appliction: Present vlue of continuous income strem 93 (ii) The integrl Z x. dx, converges. Here p =. >, so the result implies convergence of the integrl. 7.3 Appliction: Present vlue of continuous income strem Here we discuss the vlue of n nnuity, which is kind of svings ccount tht gurntees continuous strem of income. You would like to py P dollrs to purchse n nnuity tht will py you n income f(t) every yer from now on, for t>. In some cses, we might wnt constnt income every yer, in which cse f(t) would be constnt. More generlly, we cn consider the cse tht t ech future yer t, we sk for income f(t) tht could vry from yer to yer. If the bnk interest rte is r, how much should you py now? Solution If we invest P dollrs (the principl i.e., the mount deposited) in the bnk with interest r then the mount A(t) in the ccount t time t (in yers), will grow s follows: A(t) =P + r n nt, where r is the nnul interest rte (e.g. 5%) nd n is the number of times per yer tht interest is compound (e.g. n = 2mens interest compounded twice per yer, n = 2 mens monthly compounded interest, etc.). Define h = r n. Then t time t, we hve tht A(t) =P ( + h) h rt h = P ( + h) h i rt Pe rt for lrge n or smll h. Here we hve used the fct tht when h is smll (i.e. frequent intervls of compounding) the expression in squre brckets bove cn be pproximted by e, the bse of the nturl logrithms. Recll tht h i e = ( + h) h. h! This result ws obtined in first semester clculus course by selecting the bse of exponentils such tht the derivtive of e x is just e x itself. Thus, we hve found tht the mount in the bnk t time t will grow s A(t) =Pe rt, continully compounded interest. (7.) Hving estblished the exponentil growth of n investment, we return to the question of how to set up n nnuity for continuous strem of income in the future. Rewriting
6 94 Chpter 7. Improper integrls Eqn. (7.), the principle mount tht we should invest in order to hve A(t) to spend t time t is P = A(t)e rt. Suppose we wnt to hve f(t) spending money for ech yer t. We refer to the present vlue of yer t s the quntity P = f(t)e rt, i.e. we must py P now, in the present, to get f(t) in future yer t. Summing over ll the yers, we find tht the present vlue of the continuous income strem is P = LX t= f(t)e rt {z} t Z L f(t)e rt dt, where L is the expected number of yers left in the lifespn of the individul to whom this nnuity will be pid, nd where we hve pproximted sum of pyments by n integrl (of continuous income strem). One problem is tht we do not know in dvnce how long the lifespn L will be. As crude pproximtion, we could ssume tht this income strem continues forever, i.e. tht L. In such n pproximtion, we hve to compute the integrl: P = Z f(t)e rt dt. (7.2) The integrl in Eqn. (7.2) is n improper integrl (i.e. integrl over n unbounded domin), s we hve lredy encountered in Section We shll hve more to sy bout the properties of such integrls, nd bout their technicl definition, existence, nd properties in Chpter 7. We refer to the quntity P = Z s the present vlue of continuous income strem f(t). Exmple: Setting up n nnuity f(t)e rt dt, (7.3) Suppose we wnt n nnuity tht provides us with n nnul pyment of, from the bnk, i.e. in this cse f(t) = $, is function tht hs constnt vlue for every yer. Then from Eqn (7.3), P = Z e rt dt = By previous clcultion in Section 3.8.5, we find tht P = r, Z e rt dt. e.g. if interest rte is 5% (nd ssumed constnt over future yers), then P =.5 = $2,. Therefore, we need to py $2, tody to get, nnully for every future yer.
7 7.4. Integrl comprison test Integrl comprison test The integrls discussed bove cn be used to mke comprisons tht help us to identify when other improper integrls converge or diverge 27. The following importnt result estblishes how these comprisons work: Suppose we re given two functions, f(x) nd g(x), both continuous on some infinite intervl [, ). Suppose, moreover, tht t ll points on this intervl the first function is smller thn the second, i.e. pple f(x) pple g(x). Then the following conclusions cn be mde: (i) (ii) If (iii) If Z f(x) dx pple Z g(x) dx. The re under f(x) is smller thn the re under g(x). Z g(x) dx converges, then Z f(x) dx converges. If the lrger re is finite, so is the smller one. Z f(x) dx diverges, then Z g(x) dx diverges. If the smller re is infinite, so is the lrger one. These sttements hve to be crefully noted. Wht is ssumed nd wht is concluded works one wy. Tht is the order if...then is importnt. Reversing tht order leds to common error. Exmple: Determine whether the following integrl converges: Z x +x 3 dx. Solution: by noting tht for ll x> Note tht for x> pple x +x 3 pple x x 3 = x 2. pple x +x 3 pple x x 3 = x 2. Thus, Z Z x dx pple +x3 x 2 dx. Since the lrger integrl on the right is known to converge, so does the smller integrl on the left. } 27 Similr ides will be employed for the comprison of infinite series in Chpter. A recurring theme in this course is the close connection between series nd integrls, for exmple, recll the Riemnn sums in Chpter 2.
8 96 Chpter 7. Improper integrls 7.5 Integrtion of n unbounded integrnd The second kind of improper integrls refers to integrnds tht re undefined t one (or more) points of the domin of integrtion [, b]. Suppose f(x) is continuous on the open intervl (, b) but becomes infinite t the lower bound, x =. Then the integrl of f(x) over the domin [ +, b] for >hs definite vlue regrdless of how smll is chosen. Therefore, we cn consider the it f(x)dx,! + + where! + mens tht pproches from bove, i.e. > lwys holds. If this it exists nd is equl to L, then we define f(x)dx = L. If n ntiderivtive of f(x), sy F (x) is known, then the fundmentl theorem of clculus permits us to compute + f(x)dx = F (b) F ( + ). We re thus led to determine the existence (or nonexistence) of the it F ( + ).! + Exmple : Clculte the following integrl for p 6= : Solution: dx (x ) p. We interpret the integrl s the following it:! + + pple dx (x ) p =! + p (b ) p p. Thus, for p>the term p becomes rbitrrily lrge s! + nd hence the re diverges nd the integrl does not exist. Conversely, for p<the term p converges to s! + nd hence the improper integrl exists nd is p (b ). p Finlly, note tht for p =the ntiderivtive is undefined nd the integrl does not exist. Alterntively, for p =we directly see tht + dx b x =ln
9 7.5. Integrtion of n unbounded integrnd 97 5 y e e e 2 e e e 2 x 5 ln x 2 4e 2. Z e 2 Figure 7.2. Consider the improper integrl ln x 2 dx. The integrnd e 2 does not exist for x =. Nevertheless, the definite integrl exists nd equls diverges s! +. } Note tht for =the bove exmple recovers the integrnd /x p tht ws discussed in Section In prticulr, we find tht the improper integrl of the second kind Z x p dx exists for p < nd equls /( p) but does not exist for p. Conversely, in Section we observed tht the improper integrl of the first kind Z x p dx exists for p> nd equls /(p ) but does not exist for p pple. Note tht for p = neither of the integrls exists. Exmple 2: Clculte the following integrl: Z e 2 e 2 ln x 2 dx. Solution: A grph of the integrnd is shown in Figure 7.2. First, we note tht the integrnd is not defined t x =. Therefore, we split the integrl into two prts such tht
10 98 Chpter 7. Improper integrls the undefined point mrks once the upper nd once the lower bound nd we write the two integrls s it: Z e 2 e 2 ln x 2 dx =! Z e 2 ln x 2 Z e 2 dx + ln x 2! + dx, where! mens pproches from below, i.e. < lwys holds. Second, the integrnd is n even function nd the integrl runs over symmetric domin. Hence we get Z e 2 2 ln x 2! + dx. Check it! The integrl cn be solved using the substitution u = x 2 followed by n integrtion by prts. This yields 2x ln x2 2! + e 2 =4 2 e x (ln(x) )! + =4e 2 (ln( ) ).! + The fct tht the it exists nd converges to cn be seen by setting =/k nd considering the it k!: (ln( ) ) =! + k! k (ln( ) ) = (ln(k) + ). k k! k Now we cn use either de l Hôpitl s rule or simply recognize tht k grows much fster thn ln k nd hence the it converges to. Thus, we find tht the improper integrl exists nd is 4e L Hôpitl s rule This section introduces powerful method to evlute tricky its of the form x! f(x) g(x). The rule is nmed fter the French mthemticin Guillume de l Hôpitl, who published it in the 7 th century. For our purposes, the rule is often prticulrly useful to evlute the its tht rise in improper integrls of the first (unbounded domin) nd second kind (unbounded integrnd). Consider two functions, f(x) nd g(x), nd suppose tht the following four prerequisites re stisfied: () f(x) nd g(x) re differentible ner x =, but not necessrily t x =. (b) g (x) 6= for x ner but x 6=. (c) x! f (x) g (x) exists. }
11 7.6. L Hôpitl s rule 99 (d) nd either (i) f(x) = g(x) =, or x! x! (ii) f(x) =±, g(x) =±. x! x! Then, l Hôpitl s rule sttes tht f(x) x! g(x) = f (x) x! g (x). (7.4) Note: vlues of the it cn include ±. In loose terms, l Hôpitl s rule cn be used if f(x) x! g(x) is of type or ±. Let us now explore the power of l Hôpitl s rule through severl exmples. Exmple : Clculte the following it: x! sin x e x. Solution: x! (e x In this exmple l Hôpitl s rule cn be used becuse x! sin x =nd ) =. Thus, x! sin x e x = x! (sin x) (e x ) = cos x x! e x cos x x! = (becuse of nonzero denomintor) x! ex = cos =. } Exmple 2: Clculte the following it: t! t ln t t 2 +.
12 2 Chpter 7. Improper integrls Solution: In this exmple, we cn use l Hôpitl s rule becuse both the numertor nd the denomintor diverge: t! (t ln t) = nd t! (t 2 + ) =. Thus, t! t ln t t 2 + = t! (t ln t) (t 2 + ) = t! ln t +. 2t Unfortuntely, the new it tht results from pplying l Hôpitl s rule remins tricky. However, we cn simply pply l Hôpitl s rule gin becuse the prerequisites re still stisfied. In this cse, the numertor nd denomintor still diverge: t! (ln t + ) = nd t! 2t =. Using l Hôpitl s rule gin, we obtin ln t + (ln t + ) = t! 2t t! (2t) = t! t 2 = t 2 t! =. Note: As long s ll prerequisites of l Hôpitl s rule remin stisfied, the rule cn be pplied repetedly. } Exmple 3: Clculte the following it: sin x x!+ x 2. Note: plussign (+) dded to the it mens tht x pproches the it from bove (or from the right). In the present cse, x>lwys holds s x pproches zero. Lter in this exmple we will see why this subtle point is importnt. In nlogy, minussign ( ) dded to the it mens tht the it is pproched from below (or from the left). If no sign is dded, it does not mtter whether the it is pproched from bove, below or even in n lternting mnner. Solution: Agin, l Hôpitl s rule cn be used becuse x! sin x =nd x! x 2 =. Thus, sin x (sin x) cos x x!+ x 2 = x!+ (x 2 ) = x!+ 2x. Check it! Check it! In order to evlute this new it, we might be tempted to pply l Hôpitl s rule gin. Is this permissible? Stop for moment nd think bout why or why not. Of course, it is not permissible becuse the numertor x!+ cos x = nd hence violtes the prerequisites for pplying l Hôpitl s rule. Ignoring the prerequisites nd blindly pplying l Hôpitl s rule gin would yield n incorrect it of zero. Insted, we get cos x x!+ 2x = cos x x!+ x!+ 2x =+. Hence, the it does not exist, it diverges to +. In order to see why it ws importnt to pproch the it from bove, x! +, clculte the it s x pproches zero from below:
13 7.7. Summry 2 x! sin x x 2. In contrst to the bove result, the it is now pproches zero. becuse x<lwys holds s x } Exmple 4: Clculte the following it: ln.!+ Note: The it of zero cn only be pproched from bove,! +, becuse ln is undefined for <. Solution: At first, this it does not seem to hve the correct form to pply l Hôpitl s rule. However, we cn use net little trick nd rewrite the it s quotient of two functions: ln =!+!+ ln. Once rewritten, it becomes pprent tht l Hôpitl s rule cn indeed be pplied becuse gin both the numertor nd the denomintor diverge:!+ ln = nd!+ / =. Thus, ln (ln )!+ =!+ ( = )!+ 2 =!+ ( ) =. } 7.7 Summry The min points of this chpter cn be summrized s follows:. We reviewed the definition of n improper integrl (type one) over n infinite domin: Z f(x) dx = b! f(x) dx. 2. We computed some exmples of improper integrls nd discussed their convergence or divergence. We reclled (from erlier chpters) tht wheres Z Z e rt dt x dx converges, diverges.
14 22 Chpter 7. Improper integrls 3. More generlly, we showed tht Z dx converges if p>, diverges if p pple. xp 4. We reviewed the definition of improper integrls (type two) for integrnds tht re unbounded t either end of the domin of integrtion, sy x = : f(x) dx = f(x) dx.! + 5. If the integrnd is not defined t one (or more) point(s) in the interior of the domin of integrtion, then the integrl is split into two (or more) prts nd we proceed s bove. 6. In prticulr, we showed tht Z dx converges if p<, diverges if p. xp 7. L Hôpitl s rule is powerful tool to evlute tricky its tht my rise for improper integrls of both kinds. It sttes tht f(x) x! g(x) = f (x) x! g (x) if f() =g() =or x! f(x) = x! g(x) =± s well s some, more subtle prerequisites.
15 7.8. Exercises Exercises Exercise 7. Consider the integrl A = Z D x p dx () Sketch region in the plne whose re represents this if (i) p> nd (ii) p<. (b) Evlute the integrl for p 6=. (c) How does the re A depend on the vlue of D in ech of the cses (i) nd (ii). Does the re increse without bound s D increses? Or does the re pproch some constnt? (d) With this in mind, how might we try to understnd n integrl of the form Exercise 7.2 cse. () (d) Z x p dx Which of the following improper integrls converge? Give reson in ech Z Z dx x. (b) e x dx (e) Z Z xdx e 2x dx (c) (f) Z Z x 3 dx xe x dx Exercise 7.3 The grvittionl force between two objects of mss m nd m 2 is F = Gm m 2 /r 2 where r is the distnce of seprtion. Initilly the objects re distnce D prt. The work done in moving n object from position D to position x ginst force F is defined s W = Z x D F (r) dr. Find the totl work needed to move one of these objects infinitely fr wy. Exercise 7.4 Gbriel s Horn is the surfce of revolution formed by rotting the grph of the function y = f(x) =/x bout the x xis for pple x<. () Find the volume of ir inside this shpe nd show tht it is finite. (b) When we cut crosssection of this horn long the x yplne, we see flt re which is wedged between the curves y =/x nd y = /x. Show tht this crosssectionl re is infinite. (c) The surfce re of surfce of revolution generted by revolving the function y = f(x) for pple x pple b bout the xxis is given by S = 2 f(x) p +(f (x)) 2 dx Write down n integrl tht would represent the surfce re of Gbriel s Horn.
16 24 Chpter 7. Improper integrls (d) The integrl in prt (c) is not esy to evlute explicitly i.e. we cnnot find n ntiderivtive. However, we cn show tht it diverges. Set up comprison tht shows tht the surfce re of Gbriel s horn is infinite. Exercise 7.5 Z sin (x) Does the integrl x 4 + x 2 dx converge or diverge? + Exercise 7.6 Suppose tht n irborne disese is introduced into lrge popultion t time t =. At time t >, the rte t which this disese is spreding is r (t) = 4te 4t new infections per dy. () After how long is this disese most infectious? (b) How mny people in totl cquire the disese? (c) Wht eventully hppens to the rte of new infections? Exercise 7.7 Z 5 x Does the integrl x 2 + x dx converge or diverge? 2 Exercise 7.8 Suppose tht you plce $,, in n ccount tht erns 5% nnul interest, nd you wish to withdrw z dollrs from this ccount one yer from now, 2z dollrs two yers from now, 3z dollrs three yers from now, nd so on. Wht is the mximum vlue of z so tht you never run out of money? Exercise 7.9 For which vlues of p does the integrl Exercise 7. Evlute the following its: () x! xe4x (b) x sin x! x Z 2 x (ln x) p dx converge? (c) cos ( )! 2 2 Exercise 7. Determine the convergence of the following integrls. If they converge, find their vlues: () Z xe x dx (b) Z Z 2x p dx (c) x 2 x 3 dx Exercise 7.2 Convergence of n integrl of the form splitting the integrl up into two prts: Z f (x) dx = Z f (x) dx + Z Z f (x) dx is determined by f (x) dx.
17 7.8. Exercises 25 Z The integrl f (x) dx is sid to converge if both of these two integrls converge. Cn you come up with n exmple of function f (x) for which Z but the integrl f (x) dx diverges? f (x) dx < b! b Exercise 7.3 Exercise 7.4 Evlute the integrl Evlute the integrl Z Z cos p sin 3 d. e x sin xdx.
18 7.9. Solutions Solutions Solution to 7. () See Figure 7.3 (b) A = D p p (c) p>: A pproches constnt s D increses p<: A increses with D. DNE: does not exist. y y (d) p>: A = p p<: A DNE. D x D x Figure 7.3. Solution for problem?? Solution to 7.2 () convergent, p> (b) divergent, p< (c) convergent, p> (d) divergent,! (e) convergent, 2 (f) convergent, Solution to 7.3 W = Gm m 2 D Solution to 7.4 () V = (b) A! Z p p x4 + x4 + (c) S =2 x 3 dx (d) x 3 > for x> ) S! x Solution to 7.5 The integrl converges. Solution to 7.6 () fter 6 hours (b) 25 people (c) the rte tends to
19 336 Chpter 7. Improper integrls Solution to 7.7 It converges: R 5 Solution to 7.8 z = 25 dollrs x x 2 +x 2 dx = R 5 Solution to 7.9 It converges if p> nd diverges otherwise. Solution to 7. x+2 dx = ln (7) ln (2). () (b) (c) Solution to 7. () it converges to (b) it converges to 2 (c) it diverges Solution to 7.2 One possible exmple is f (x) =xe x2 ; ny odd function will do. Solution to 7.3 Solution to 7.4 Z Z cos p d =. sin 3 e x sin xdx= 2.
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