7.8 Improper Integrals

Transcription

1 Improper Integrls The Completeness Axiom of the Rel Numers Roughly speking, the rel numers re clled complete ecuse they hve no holes. The completeness of the rel numers hs numer of importnt consequences. For exmple, the Intermedite Vlue Property (of continuous function) depends on the completeness of the rel numers. The following result is nother consequence. A ounded nondecresing function hs finite limit s x More precisely, we hve Theorem. Let F e n nondecresing function defined on the intervl [, ). Then either () () lim F(x) = L R x lim F(x) = x or Notice tht F need not e continuous. Stted nother wy: If for ll x (3) F(x) M < then lim x F(x) = L < In generl, the most we cn sy out the limit L is tht L M. Also, see section. of the text.

2 7.8 Type I - Infinite Limits of Integrtion Consider the following exmple. Exmple. Ares under infinite curves? Suppose tht >. Evlute the following definite integrl. (4) x dx Since the integrnd is positive, this mounts to finding the re under curve /x from x = to x =. Now y = /x dx = x x (5) = Now we evlute (5) s. Tht is, lim dx x = In prctice, we dopt the following convenient nottion. (6) ˆ dx x x dx The left-hnd side (6) is clled n improper integrl nd is defined more precisely elow.

3 7.8 Definition. Improper Integrls - Type I Definite integrls with infinite limits of integrtion re clled improper integrls of type I. They re defined s follows.. If f(x)dx exists for ll, then (7) f(x)dx f(x)dx ˆ provided the right-hnd limit exists s finite numer.. If f(x)dx exists for ll, then (8) f(x)dx f(x)dx provided the right-hnd limit exists s finite numer. In oth cses we sy the improper integrl converges whenever the right-hnd side is finite. Otherwise, the improper integrl diverges. 3. If c f(x)dx nd c f(x)dx converge, then (9) f(x)dx = ˆ c f(x)dx+ c f(x)dx for ny rel numer c. Notice tht this lst cse is hndled y comining the first two. Remrk. So the improper integrl in the first exmple converged nd its vlue ws. Also, since f(x) = /x > we cn sy tht the re under the curve is finite...in fct, the re is (defined to e).

4 7.8 3 As we might hope, improper integrls lso stisfy some nice properties. Proposition. Suppose tht f is continuous on [, ) nd f(x)dx converges. Then. kf(x)dx lso converges for ny k R nd () kf(x)dx = k f(x)dx. For c >, we hve () f(x)dx = ˆ c f(x)dx+ c f(x)dx Proof. Let > nd k R. Then y the sic properties of the definite integrl It now follows y the properties of limits tht kf(x)dx = k f(x)dx kf(x)dx kf(x)dx ˆ = k lim = k ˆ f(x)dx f(x)dx < This estlishes (). We leve the proof of () s n esy exercise.

5 7.8 4 Proposition 3. Suppose tht the continuous functions f nd g stisfy the inequlity f(x) g(x) for ll x [, ). Then () f(x)dx g(x) dx provided oth improper integrls exist s extended rel numers. For exmple, if f nd g re continuous functions nd g(x) f(x), then oth improper integrls exist s extended rel numers. Thus () holds. See Exmple 6. Also, compre with Corollry 5 elow. Exmple. Evlute the improper integrl x dx Notice tht the integrnd is continuous on [, ) so y (7) we hve ˆ dx x x dx Now we cn use the usul techniques (prtil frctions in this cse) to evlute the integrl. ˆ x x+ dx (ln x ln x+ ) ln x x+ = ln3 ( ln + ln 3 )

6 7.8 5 The next exmple is referred to often in chpter. Exmple 3. p-integrls Consider the improper integrl (3) x p dx, p > Notice tht /x p is continuous on the intervl [, ). We clim tht the integrl converges if p > (s we sw in the first exmple for p = ) nd the integrl diverges when p. The integrl in (3) is clled p-integrl. Note: These conclusions remin vlid if the lower limit of integrtion is replced y ny positive.

7 7.8 6 Cse. Suppose p <. Then since p >. ˆ dx xp x dx p x p p = p lim =, ( p ) Cse. Suppose p =. Then ˆ dx x x dx ln x (ln ln) = Cse 3. The cse p > is left s n esy exercise. (Hint: see Exmple.)

8 7.8 7 Type II - Integrnds with Verticl Asymptotes Definition. Type II - Improper Integrls Integrls of functions tht ecome infinite t point within the intervl of integrtion re clled improper integrls of Type II.. If f(x) is continuous on (,] nd discontinuous t x = then (4) f(x)dx f(x)dx c + c. If f(x) is continuous on [,) nd discontinuous t x = then (5) ˆ c f(x)dx f(x)dx c 3. If f(x) is discontinuous t c (,) nd continuous on (,c) (c,) then (6) f(x)dx = ˆ c f(x)dx+ c f(x)dx In ech cse, the improper integrl is sid to converge if the right-hnd side exists. Otherwise, the improper integrl diverges.

9 7.8 8 Exmple 4. Evluting Type II Integrls Evlute the following (improper) integrls.. ˆ 6 dx 6 x dx 6 6 x x 6 4 ( 6 6 ) = 4 6. ˆ lnxdx On the limits hndout, we showed lim x +xlnx = Thus ˆ ˆ lnxdx lnxdx + +x(lnx ) +(( ) (ln )) =

10 7.8 9 Exmple 5. Evlute the following integrl. ˆ xdx x 4 This isn t so d. Suppose tht < < nd we let y = x. Then dy = xdx nd It follows tht ˆ xdx = x 4 dy y = rcsiny = rcsin ˆ xdx xdx x 4 x 4 rcsin = ( π ) Testing for Convergence It is frequently the cse tht one cnnot directly compute improper integrls. And it is wste of time to use numeric techniques to estimte n improper integrl only to discover tht the integrl diverges. In such cses we would like to know priori whether the integrl converges. Consider the next exmple.

11 7.8 Exmple 6. Direct Comprison Does the improper integrl /(x+) dx converge? Of course, we could evlute this integrl directly ut, s we will see lter, tht might not lwys e possile. Erlier we sw tht the p-integrl /x dx converged. y = /x y = /(x+) The sketch suggests (7) < ˆ (x+) dx < x dx < Why is this importnt? Consider the (re) function (8) F(x) = ˆ x dt (+t) Since the integrnd is continuous for ll t, we my differentite F with respect to x to otin (9) F (x) = (+x) >

12 7.8 It follow tht F(x) is n incresing function. So y the completeness xiom of the rel numers, there re two possiilities (s we discussed in the introduction). Either () () dx (x+) F(x) = L R or x dx (x+) x F(x) = We will now eliminte the second possiility y showing tht F(x) is ounded from ove. Suppose tht t. Then () (t+) t > nd hence (3) < (t+) t So for ll x we hve (4) F(x) = ˆ x ˆ x (t+) dt ˆ t dt t dt = The lst pir of inequlities follow from (3) nd Theorem 5.3. of the text (see the dditive nd domintion properties of the definite integrl in Tle 5.4). Thus (5) dx (x+) F(x) = L < x In other words, the improper integrl converges.

13 7.8 We cn mke the lst exmple more precise with the following. Theorem 4. Direct Comprison Test Let f nd g e continuous on [, ) with f(x) g(x) for x. Then.. f(x)dx converges whenever g(x)dx diverges whenever g(x)dx converges. f(x)dx diverges. Proof. We rgue s we did in Exmple 6. Suppose tht >. Then g(x)dx g(x)dx+ g(x) dx = } {{ } g(x) dx Now the lst integrl is either rel numer or positive infinity. Since f(x) g(x) it follows tht for ll Hence (6) f(x)dx g(x)dx f(x)dx g(x) dx g(x) dx Now if the right-hnd side of (6) is finite, so is the left-hnd side (gin ecuse of the erlier remrks concerning completeness). This estlishes prt (). On the other hnd, if the left-hnd side is infinite then the right-hnd side must lso e infinite. Corollry 5. Let f nd g e continuous on [, ) with f(x) g(x) for x. Then (7) f(x)dx g(x) dx Note: We do not ssume the convergence of either integrl in (7). Compre with Proposition 3.

14 7.8 3 Now consider the following Exmple 7. Direct Comprison Does the improper integrl /(x ) dx converge? y = /(x ) y = /x Notice tht we cn drw no conclusion in this cse ecuse the inequlity is not working in our fvor. However, we do not give up so esily. By Proposition, /x dx < for ny >. Now choose lrge enough ( = 6 will do) so tht /x > /(x ) for x. It follows tht (x ) dx < y the direct comprison test. y = /(x ) y = 6/x

15 7.8 4 The lst exmple seems like too much work. For strters, we would need to prove 6/x > /(x ) for x. Fortuntely there is n esier pproch. Theorem 6. Limit Comprison Test (LCT) Suppose tht f nd g re positive continuous functions on [, ) with (8) f(x) lim x g(x) = L, < L < then (9) f(x)dx nd g(x) dx oth converge or oth diverge. Notice tht, using the lnguge of section 7.5p, (8) sys tht f nd g grow t the sme rte! So if f nd g grow t the sme rte, nd one of the ove integrls converges then so does the other. On the other hnd, if one of the integrls diverges then so does the other. Remrk. In Proposition we sw tht f(x)dx = f(x)dx+ f(x)dx Now since the integrnds in this section re ssumed to e continuous (or, t lest piecewise continuous), the first integrl on the right-hnd side exists. Since we my choose > s lrge s we wish, Proposition implies tht, when estlishing convergence or divergence of n improper integrl, the only thing tht is importnt is the function s (right) end ehvior. Tht is, how fst does the function decy to s x. Theorem 6 confirms this oservtion. In fct, we hve the following. Theorem 7. Suppose tht f is continuous on [, ) nd. Then ech of the following sttements implies the other. (3) (3) lim f(x)dx converges f(x)dx =

16 7.8 5 Py close ttention to the precise detils of the following exmples. Exmple 7. (cont) Using the LCT Does the improper integrl /(x ) dx converge? Let g(x) = /x nd notice tht }{{} f(x) f(x) lim x g(x) /(x ) x /x x /x x (x ) /x ( x = = ( ) x ) Tht is, f nd g grow t the sme rte. Now y exmple 3, /x dx converges since it is p-integrl with p = >. So y the Limit Comprison Test /(x ) dx lso converges.

17 7.8 6 Notice tht we could hve nswered the question ove y evluting the integrl. Unfortuntely, tht is not lwys the cse. Consider the following exmple. Exmple 8. (3) Does the improper integrl elow converge or diverge? Justify your clim. x+ dx 5x 3 +6 We clim the integrl converges. Let f(x) = x+ 5x 3 +6 nd let g(x) = /x 5/. Then f(x) lim x g(x) x x x = 5 x+ 5x 3 +6 /x 5/ x+ 5x /x 5+6/x 3 x 5/ /x 3 /x 3 Tht is, f nd g grow t the sme rte. But /x 5/ dx is convergent p-integrl since p = 5/ >. It follows y the Limit Comprison Test tht the integrl in (3) must lso converge.

18 7.8 7 Wht were the essentil ingredients in the lst two exmples? Exmple 9. (33) Does the improper integrl elow converge or diverge? Justify your clim. x 3x5 +x+ dx We clim the integrl diverges. Let f(x) = x 3x 5 +x+ nd let g(x) = / x. Then f nd g grow t the sme rte (exercise). But the p-integrl / xdx diverges since p = / <. One cnnot lwys find fmilir function tht grows t the sme rte s given integrnd. Fortuntely we hve Theorem 8. Limit Comprison Test - Modified Suppose tht f nd g re positive continuous functions on [, ) nd (34) lim x f(x) g(x) = L i. If L < nd g(x)dx converges then so does f(x)dx. ii. If < L nd g(x)dx diverges then so does f(x)dx. Compre with Theorem 6.

19 7.8 8 Exmple. (35) Does the following improper integrl converge or diverge? Justify your clim. e x / dx We clim the integrl converges. Let f(x) = e x / nd g(x) = e x. By direct computtion e x dx e ˆ x dx ( e ) = Now f(x) lim x g(x) e x / x e x x e x /+x = since x /+x s x. So y Theorem 8 (which we will lso refer to s the Limit Comprison Test) the integrl in (35) converges. Remrk. The integrnd in (35) is clled proility density function. More precisely, the proility density function ssocited with norml distriution (or ell curve) is given y h(x) = σ e(x µ) σ π Here µ nd σ re clled the men nd vrince, respectively. Lter we will show tht (36) h(x)dx =

20 7.8 9 We conclude with few more useful results. Proposition 9. Suppose tht g(x) is continuous function on [, ) nd tht the integrl g(x) dx converges. Then g(x)dx lso converges. Note: In such cses we sy tht g is solutely integrle (on [, )). Proposition 9 sys tht solutely integrle functions re (improperly) integrle. Proof. Recll tht for ny rel numer we hve. It follows tht for ll x [, ) (37) g(x) g(x) g(x) or g(x) +g(x) g(x) So y Proposition nd the Direct Comprison Test, ( g(x) +g(x))dx < Hence g(x)dx g(x) dx ˆ = ˆ (g(x)+ g(x) g(x) )dx ( (g(x)+ g(x) )dx (g(x)+ g(x) )dx lim (g(x)+ g(x) )dx ) g(x) dx g(x) dx g(x) dx <

21 7.8 Propositions nd 9 (nd the inequlity in (37)) imply Corollry. Suppose tht g is solutely integrle (on [, )). Then or g(x) dx g(x)dx g(x) dx g(x) dx g(x) dx The previous proof lso contined nother importnt (improper) integrl property. Do you see it? Proposition. Suppose tht f(x) nd g(x) re continuous functions on [, ) nd f(x)dx nd g(x)dx oth converge. Then (f(x)+g(x))dx lso converges nd (38) f(x)dx+ g(x)dx = (f(x)+g(x))dx

22 7.8 Exmple. Evluting the Are Under the Bell Curve The stndrd method for evluting the integrl in (35) requires techniques from third semester clculus. However, creful serch of MthSci Net yielded the following eutiful lterntive. The ide is to compute the volume generted y revolving the function f(x) = e x / out the y-xis using two different methods nd equte the results. y = e x / y x t We leve it s n exercise to show tht the volume is π. (Hint: Use the method of disks nd integrte in the the y-direction from to. You might find one of the results from Exmple 4 useful.)

23 7.8 y (t,x,y) r (t,x) x t Now let A(t ) e the re of cross-section for some fixed distnce, sy t, long the t-xis. Let r = x +t nd oserve tht y = e r / = e (x +t )/ = e t / e x / It follows tht re of the cross-section shown is A(t ) = e t / e x / dx / = e t e x / dx }{{} constnt

24 7.8 3 Now let t e ritrry. Then A(t) = e t / e x / dx }{{} I Using the volume technique we discovered in section 6., we hve Hence Volume = π = = = I = I e x / dx = I = π Since f is n even function, we immeditely otin e x / dx = A(t) dt Ie t / dt e t / dt }{{} I e x / dx = π We cn now use simple chnge-of-vriles (u-sustitution) to verify (36). Exercise: Show tht σ π e (x µ) σ dx =

25 7.8 4 Exmple. A Chllenging Exmple Evlute the following type I improper integrl. (39) sinx x dx y = sinx x y = /x π x The integrl si(x) = sint/tdt is clled the sine integrl nd is very importnt in mny res of mthemtics (c.f. exercise ). At first glnce (39) my pper to e type II integrl t the lower limit, however, this is n illusion since lim x + sinx/x =. It is nontrivil to show tht sine integrl converges. For exmple, it is nturl to try the comprison test with /x since sinx x x, x But the p-integrl dx/x diverges so the inequlity ove doesn t help us. It turns out tht sin x/x is not solutely integrle ut the improper integrl in (39) does converge. We will postpone the proof of oth sttements until chpter. Insted we consider the following.

26 7.8 5 Oserve tht ( ) sinx x x, x y = /x y = sin x x π It follows tht the improper integrl (sinx/x) dx converges y the Comprison Test. We cn ctully sy more. Proposition. (4) ( sinx x ) dx = π We postpone the proof for the moment. To tke dvntge of Proposition we try integrtion y prts on (39). Since x = (x/) we my rewrite the integrl s Now let Then sin (x/) x dx = u = /x nd sin(x/) cos(x/) x dx dv = sin(x/) cos(x/) du = dx/x nd v = (sinx/) Thus sin (x/) x dx = sin(x/) cos(x/) x dx = (sinx/) x/ + ( sinx/ x/ ) dx = + ( sinx x ) dx = π Remrk. It is worth oserving tht the integrnd in (4) is solutely integrle while the integrnd in (39) is not!

27 7.8 6 Wrning: The proof of Proposition is not for the fint of hert. We need few preliminry results. First we introduce the Dirichlet kernel. For ny nonnegtive integer n, let n (4) D n (x) = coskx (4) k= n = + n coskx Since the cosine function is even, the second formul is n immedite consequence of the first. The Dirichlet kernel stisfies the following Lemm 3. Let D n e the Dirichlet kernel. Then (43) (44) k= D n (x) = sin(n+/)x sin x/ ˆ π D n (x)dx = π π Proof. To prove (43) we comine (4) with the product-to-sum trig identities discussed t the end of section 8.4. n sin(x/)d n (x) = sinx/+ sin(x/) cos kx, etc. This yields telescoping sum. We leve the detils s n exercise (lso, see the proof of (46) elow). To prove (44) oserve tht for ny non-zero integer k Thus ˆ π π ˆ π π coskxdx = sinkx k D n (x)dx = ˆ π π = π + π π dx+ k= nd = (sinkπ sin( kπ)) = k n k= ˆ π π coskxdx Next we define the Fejer kernel. For ny nonnegtive integer n, let (45) K n (x) = n D k (x) n+ Here D k is the Dirichlet kernel. k=

28 7.8 7 Lemm 4. Let K n e the Fejer kernel. Then (46) (47) (48) K n (x) = ( sin(n+)x/ n+ sin x/ ˆ π K n (x)dx = π π K n (x) ) Clerly (48) is n immedite consequence of (46). Proof. (47) follows immeditely from (44) nd (45) since ˆ π π K n (x)dx = ˆ π n+ π = n+ n k= = π(n+) n+ n D k (x)dx k= ˆ π D k (x)dx π }{{} π To prove (46) we once gin ppel to the product-to-sum trig identities (c.f. section 8.4 from the text). sin(x/) sin(k +/)x = coskx cos(k +)x

29 7.8 8 Thus (sinx/) K n (x) = (sinx/) n+ = n+ = n+ = n+ n D k (x) k= n sin(x/) sin(k +/)x k= n (coskx cos(k +)x) k= { (cos cosx)+(cosx cosx)+ } +(cosnx cos(n+)) = cos(n+)x n+ = n+ (sin(n+)x/) Here we hve used the hlf-ngle formul, (sinu/) = cosu. Now divide oth sides y (sinx/). Before we prove our next preliminry result, it might e useful to sketch few of the kernels π π

30 7.8 9 Notice tht K n (x) is n even function nd tht lim K n(x) = x n+ lim x = n+ (n+) = n+ ( sin(n+)x/ sin x/ So it isn t difficult to figure out tht the sketch ove includes the grphs of K,K 3,K 7 nd K 5. Also, from the sketch it ppers tht, with incresing n, ech kernel s mss (re) ecomes more concentrted ner the origin. Specificlly, we hve Proposition 5. Let < δ < π. Then ˆ δ lim K n (x)dx = π n δ )

31 7.8 3 Proof. Let < δ < π. As nottionl convenience, we write ˆ π π f(x)dx = nd, if f(x) is n even function = = ˆ δ π ˆ δ δ ˆ δ δ f(x)dx+ ˆ f(x)dx+ ˆ δ δ δ< x π f(x)dx+ f(x)dx ˆ π f(x)dx+ f(x)dx δ ˆ π δ f(x)dx Now if δ < x < π, then sin δ/ < sin x/. It follows tht Thus ( ) sin(n+)x/ (n+)k n (x) = sin x/ π = = = = ˆ π π ˆ δ δ ˆ δ δ ˆ δ δ ˆ δ δ K n (x)dx ˆ π K n (x)dx+ K n (x)dx δ K n (x)dx+ ˆ π n+ δ K n (x)dx+ n+sin δ/ K n (x)dx+ n+ sin x/ < sin δ/ dx sin δ/ ˆ π δ π δ sin δ/ }{{} independent of n dx So for ll n, π π δ n+sin δ/ ˆ δ δ K n (x)dx π Now let n nd pply the Squeeze Lw. Which of the properties from Lemm 4 were needed in the previous proof?

32 7.8 3 Proof. (of Proposition ) Once gin we let < δ < π. It is n esy (Mth 3) exercise to show tht x implies x/ sinx/. Hence for x R we hve the following string of implictions. sin x/ (x/) (x/) sin x/ ( ) ( ) sin(n+)x/ sin(n+)x/ = (n+)k n (x) x/ sin x/ Thus (49) n+ ( ) sin(n+)x/ K n (x) x/ y = ( ) sinx/ x/ π δ δ π On the other hnd, for δ x δ, ( ) ( sinδ/ sinx/ ) Hence δ/ x/ ( ) ( ) sinδ/ sinx/ K n (x) K n (x) δ/ x/ = ( ) ( sin(n+)x/ sinx/ n+ sin x/ x/ = ( ) sin(n+)x/ n+ x/ ) Together with (49) we otin ( ) sinδ/ (5) K n (x) ( ) sin(n+)x/ K n (x) δ/ n+ x/ provided δ x δ.

33 7.8 3 Integrting (5) nd pplying Lemm 4 we hve ( ) ˆ sinδ/ δ K n (x)dx δ/ δ ˆ δ δ ˆ δ δ ˆ π = π π n+ K n (x)dx K n (x)dx ( ) sin(n+)x/ dx x/ Tht is ( ) ˆ sinδ/ δ K n (x)dx δ/ δ ˆ δ δ n+ ( ) sin(n+)x/ dx π x/ Applying chnge of vriles to the second integrl yields ( ) ˆ sinδ/ δ δ/ δ ˆ δ(n+)/ K n (x)dx δ(n+)/ ( ) sinx dx π x Now let n nd pply Proposition 5 to otin ( ) sinδ/ ( ) sinx π ˆ dx π δ/ x Finlly, we let δ to conclude ( ) sinx dx = π x Since the integrnd is n even function, we hve proven (4).