6.1 Definition of the Riemann Integral

Transcription

1 6 The Riemnn Integrl 6. Deinition o the Riemnn Integrl Deinition 6.. Given n intervl [, b] with < b, prtition P o [, b] is inite set o points {x, x,..., x n } [, b], lled grid points, suh tht x =, x n = b, nd x k < x k or ll k =,..., n. The points x k subdivide [, b] into n subintervls, [, x ], [x, x 2 ],..., [x k, x k ],..., [x n 2, x n ], [x n, b], with the length o the kth subintervl being x k = x k x k. The mesh o P, denoted by P, is deined to be the length o the longest subintervl: P = mx k n x k. The set o ll possible prtitions o [, b] we denote by P[, b]. Nottion. It is onvenient to denote prtition {x, x,..., x n } more omptly by {x k } n k=, nd write {x k } n k= P[, b] to speiy tht {x k} n k= is prtition o the intervl [, b]. Deinition 6.2. Let : [, b] R be bounded, nd let P = {x k } n k= P[, b]. Deine M k = sup{(x) : x k x x k } nd m k = in{(x) : x k x x k } or eh k n. Then U(P, ) = n M k x k nd L(P, ) = k= n m k x k re the upper sum o with respet to P nd lower sum o with respet to P, respetively. Note tht sine is bounded on [, b, ], it must be bounded on eh [x k, x k ] [, b] nd hene M k nd m k must be rel numbers s onsequene o the Completeness Axiom o R. Deinition 6.3. Let : [, b] R be bounded. The upper Riemnn integrl o over [, b] is k= = in{u(p, ) : P P[, b]},

2 2 nd the lower Riemnn integrl o over [, b] is = sup{l(p, ) : P P[, b]}. Proposition 6.4. I : [, b] R is bounded, then exist in R. nd Proo. Suppose tht : [, b] R is bounded. Then there exists some M R suh tht (x) M or ll x [, b]. Let P = {x k } n k= P[, b]. For eh k n we hve (x) M or ll x [x k, x k ], nd so or eh k n. Thus L(P, ) = m k = in{(x) : x k x x k } M n m k x k k= n M x k = M k= n x k = M(b ), nd sine P P[, b] is rbitrry we onlude tht the rel number M(b ) is n upper bound or {L(P, ) : P P[, b]}. Thereore by the Completeness Axiom the lest upper bound o {L(P, ) : P P[, b]} is rel-vlued, whih is to sy k= = sup{l(p, ) : P P[, b]} exists in R. The proo o the sttement onerning the upper Riemnn integrl o over [, b] is similr nd so let s n exerise. Proposition 6.5. I P, P 2 P[, b] re suh tht P P 2, then Proposition 6.6. I P, P 2 P[, b], then L(P, ) L(P 2, ) nd U(P 2, ) U(P, ). L(P, ) U(P 2, ). Deinition 6.7. A bounded untion : [, b] R is Riemnn integrble on [, b] i We ll the rel number I symbol = = I R. the Riemnn integrl o over [, b], nd denote it by the or (x) dx. The set o ll untions tht re Riemnn integrble on [, b] is denoted by R[, b].

3 The Riemnn integrl, lso known in these notes s the deinite integrl, is just one o mny dierent kinds o integrls deined in mthemtis. In the symbol, whih in prtie my be red s the integrl o rom to b, we ll the lower limit o integrtion, b the upper limit o integrtion, nd the integrnd. The x in the symbol (x) dx () given in Deinition 6.7 is lled the vrible o integrtion. It is lso lled dummy vrible, sine we ould substitute other letters or x nd the mening o the symbol would be unhnged. Thus (x) dx, (t) dt, (u) du nd so on re ll onsidered identil Riemnn integrls. As the simpler symbol suggests, only the integrnd nd the limits o integrtion nd b uniquely determine Riemnn integrl. Using the symbol () is preerred espeilly when the untion [, b] R in the integrnd hs no designtion. Thus we my write 9 x 2 dx to denote the Riemnn integrl 9 with integrnd (x) = x2. The symbol () is lso useul when there is more thn one independent vrible present in n nlysis, s will oten be the se rom Chpter 3 onwrd. Proposition 6.8. Let R. I on [, b], then R[, b] nd = (b ). I on [, b] then it s ommon prtie to write depending on one s preerene. = (b ) or dx = (b ), Theorem 6.9. Let : [, b] R be bounded untion. Then R[, b] i nd only i or every ɛ > there exists some P P[, b] suh tht U(P, ) L(P, ) < ɛ. 3

4 4 6.2 Riemnn Sums Given prtition P = {x k } n k=, smple point rom [x k, x k ] is ny point x k the intervl, so tht x k x k x k or eh k n. hosen rom Deinition 6.. Given untion : [, b] R, prtition P = {x k } n k= smple points x k [x k, x k ] or k =,..., n, the sum n S(P, ) = (x k) x k is lled the Riemnn sum or with respet to P on [, b]. k= P[, b], nd In this deinition s well s the next one it is importnt to ber in mind tht the vlue o the integer n depends on the hoie o prtition P. We ould write n P insted o n to emphsize this, but will rerin rom doing so to minimize lutter. Deinition 6.. Let L R. Then we deine lim S(P, ) = L P to men the ollowing: or every ɛ > there exists some δ > suh tht i P = {x k } n k= P[, b] with < P < δ, then S(P, ) L < ɛ or ll hoie o smple points x k [x k, x k ], k n. The ollowing theorem provides, t lest theoretilly, mens o lulting Riemnn integrl by evluting limit. Further improvements re orthoming. Theorem 6.2. Let : [, b] R be bounded. Then R[, b] i nd only i or some L R, in whih se = L. lim S(P, ) = L P

5 5 6.3 Properties o the Riemnn Integrl In this setion we estblish vrious generl properties o the Riemnn integrl. Most o these properties will mke the tsk o evluting Riemnn integrls esier, nd ertinly ll o them will be useul lter on to prove urther theoretil results. Deinition 6.3. For ny R[, b] we deine b = For ny untion or whih () R we deine =. Proposition 6.4 (Linerity Properties o the Riemnn Integrl). Suppose, g R[, b] nd R. Then ± g R[, b] nd R[, b] suh tht. 2. = ( ± g) = ± Theorem 6.5. Suppose, g R[, b]. I g on [, b], then g Theorem 6.6. Suppose (, b). I R[, b], then R[, ], R[, b], nd = A result mounting eetively to onverse o the sttement o Theorem 6.6 is the ollowing proposition. Proposition 6.7. Suppose (, b). I R[, ] nd R[, b], then R[, b]. Proo. Suppose R[, ] nd R[, b]. Then is bounded on [, ] nd [, b], whih implies tht it is bounded on [, b]. Let ɛ >. By Theorem 6.9 there exist prtitions P P[, ] nd P 2 P[, b] suh tht + g.. U(P, ) L(P, ) < ɛ/2 nd U(P 2, ) L(P 2, ) < ɛ/2 Deine P P[, b] by P = P P 2, so tht U(P, ) = U(P, ) + U(P 2, ) nd L(P, ) = L(P, ) + L(P 2, )..

6 6 Relling Proposition 6.5, we hve U(P, ) L(P, ) = U(P, ) L(P, ) = [U(P, ) + U(P 2, )] [L(P, ) + L(P 2, )] U(P, ) L(P, ) + U(P 2, ) L(P 2, ) = U(P, ) L(P, ) + U(P 2, ) L(P 2, ) < ɛ/2 + ɛ/2 = ɛ. Thus there exists some prtition in P[, b] or whih U(P, ) L(P, ) < ɛ. Thereore R[, b] by Theorem 6.9. Theorem 6.8. Suppose R[, b]. I is inite set, then g R[, b] nd {x [, b] : g(x) (x)} g =.

7 7 6.4 Integrble Funtions In this setion we develop tools to help determine whether given untion is Riemnn integrble on n intervl [, b]. Theorem 6.9. I ϕ R[, b], Rn(ϕ) [α, β], nd ψ : [α, β] R is ontinuous, then ψ ϕ R[, b]. Proposition 6.2. I is ontinuous on [, b], then R[, b]. Proo. Suppose tht : [, b] R is ontinuous. Let ϕ : [, b] R be the identity untion on [, b], so tht ϕ(x) = x or ll x [, b]. Sine Rn(ϕ) = [, b], nd ϕ R[, b] by Proposition??, it ollows by Theorem 6.9 tht ϕ R[, b]. Now, or ny x [, b], ( ϕ)(x) = (ϕ(x)) = (x), so we see tht ϕ = nd thereore R[, b]. Proposition 6.2. I, g R[, b], then. 2 R[, b], 2. g R[, b], 3. R[, b] with, 4. g, g R[, b]. Proo. Proo o Prt (). Suppose tht R[, b]. Then : [, b] R is bounded untion, so tht Rn() [α, β] or some < α < β <. Deining the untion ψ by ψ(x) = x 2, it is ler tht ψ is ontinuous on [α, β]. By Theorem 6.9 we onlude tht ψ R[, b]. Now, sine (ψ )(x) = ψ((x)) = [(x)] 2 = 2 (x) or ny x [, b], we see tht ψ = 2 nd thereore 2 R[, b]. Proo o Prt (2). Suppose tht, g R[, b]. By Proposition 6.4 we hve +g, g R[, b], nd then by Prt () we hve ( +g) 2, ( g) 2 R[, b]. Applying Proposition 6.4 one more, it ollows tht g = 4 [( + g)2 ( g) 2 ] R[, b] s ws to be shown. Proo o Prt (3). Suppose tht R[, b]. One gin Rn() [α, β] or some < α < β <. I ψ(x) = x, then ψ is ontinuous on [α, β]. By Theorem 6.9 we onlude tht ψ R[, b]. Now, sine (ψ )(x) = ψ((x)) = (x) = (x) or ny x [, b], we see tht ψ = nd thereore R[, b].

8 8 6.5 The Fundmentl Theorem o Clulus Rell tht, ording to the Extreme Vlue Theorem, ontinuous untion on losed intervl [, b] will ttin n bsolute mximum vlue nd n bsolute minimum vlue. Tht is, there will be some x, x 2 [, b] suh tht (x ) (x) or ll x [, b] nd (x 2 ) (x) or ll x [, b]. Then we n write (x ) = mx{(x) : x [, b]} nd (x 2 ) = min{(x) : x [, b]}. This is essentil in wht ollows. Lemm Let be ontinuous on [, b] nd [, b]. Deine α, β : [, b] R by { mx{(t) : t [, x]}, i x α (x) = mx{(t) : t [x, ]}, i x < nd β (x) = Then lim x α (x) = lim x β (x) = (). { min{(t) : t [, x]}, i x min{(t) : t [x, ]}, i x < Remrk. Note α nd β re indeed rel-vlued untions or ny b sine is bounded on [, b] by the Extreme Vlue Theorem. Proo. We will ssume tht (, b). I = or = b only right- or let-hnd limit, respetively, would need to be onsidered, but otherwise the rgument would be the sme. Let ɛ >. Sine is ontinuous t, there exists some δ > suh tht x < +δ implies (x) () < ɛ/2. Suppose tht < x < + δ. For ny t [, x] we hve (t) () < ɛ/2, whene () ɛ/2 < (t) < () + ɛ/2 obtins nd thus From this it n be seen tht () ɛ/2 < α (x) = mx{(t) : t [, x]} () + ɛ/2. α (x) () ɛ/2 < ɛ, nd so lim α (x) = (). x + Next, ontinuity o t implies tht there is some δ 2 > suh tht (x) () < ɛ/2 whenever δ 2 < x. Suppose tht δ 2 < x <. For ny t [x, ] we hve (t) () < ɛ/2, whih gives () ɛ/2 < (t) < () + ɛ/2 nd thus () ɛ/2 < α (x) = mx{(t) : t [x, ]} () + ɛ/2.

9 9 One gin we re led to onlude tht α (x) () ɛ/2 < ɛ, nd so lim α (x) = (). x So lim x α (x) = () or ny (, b). The proo runs long similr lines or the untion β nd so it omitted. Theorem 6.23 (The Fundmentl Theorem o Clulus, Prt ). I is ontinuous on [, b], then the untion Φ : [, b] R given by Φ(x) = (t) dt, x b is dierentible on [, b], with Φ (x) = (x) or eh x b. Remrk. As is our ustom, it is intended tht Φ () nd Φ (b) be tken s signiying the one-sided derivtives Φ +() nd Φ (b), respetively. Proo. Suppose tht is ontinuous on [, b], so tht R[, b] by Proposition 6.2. Let (, b). Using Theorem 6.6, we hve ( Φ(x) Φ() x ) x lim = lim = lim (2) x + x x + x x + x Let α, β : [, b] R be s deined in Lemm For ny ixed x (, b) we hve β (x) (t) α (x) or ll t [, x]. Thus by Theorem 6.5 we obtin whene Proposition 6.5 gives nd thereore β (x) β (x) (x ) α (x), α (x) (x ) β (x) x α (x). (3) Sine the inequlity (3) holds or ll < x < b nd lim α (x) = lim β (x) = () x + x + by Lemm 6.22, by (2) nd the Squeeze Theorem we obtin A similr rgument shows tht Φ(x) Φ() lim x + x Φ(x) Φ() lim x x = lim x + x = lim x x = (). = (),

10 nd thus Φ Φ(x) Φ() () = lim = (). x x Sine < < b is rbitrry, we onlude tht Φ is dierentible on (, b) with Φ (x) = (x) or ll x (, b). Now let =. We hve nd Lemm 6.22 gives α (x) = mx{(t) : t [, x]} nd β (x) = min{(t) : t [, x]}, lim α (x) = lim β (x) = (). (4) x + x + Given ny x (, b) we hve β (x) (t) α (x) or ll t [, x], implying nd thus β (x) α (x), β (x) α (x). (5) x Sine (5) holds or ll < x < b, by (4) nd the Squeeze Theorem we obtin Φ +() = lim x + Φ(x) Φ() x Φ(x) = lim x + x = lim x + x = (), where Φ() = by Deinition 6.5. A similr rgument shows tht Φ (b) = (b). Thereore Φ is dierentible on [, b] with Φ (x) = (x) or ll x b. Exmple Given tht ind F. F (x) = sin x 2 ( t 2 ) 7 dt, Solution. I we deine Φ(x) = 2 ( t2 ) 7 dt, then nd the Chin Rule gives F (x) = Φ(sin(x)) = (Φ sin)(x) F (x) = (Φ sin) (x) = Φ (sin(x)) sin (x) = Φ (sin(x)) os(x) or ll x R. Now, by Theorem 6.23 we hve Φ (x) = ( x 2 ) 7, nd so F (x) = Φ (sin x) os x = ( sin 2 x) 7 (os x) = (os 2 x) 7 (os x) = os 5 (x) or ll x R. Relling Deinition 4.3, wht Theorem 6.23 sys is tht Φ is n ntiderivtive or on [, b]. We mke use o this t to prove the ollowing.

11 Theorem 6.25 (The Fundmentl Theorem o Clulus, Prt 2). Let be ontinuous on [, b]. I F is ny ntiderivtive or on [, b], then (t)dt = F (b) F () Proo. F be n ntiderivtive or on [, b]. By Theorem 4.33 there exists some onstnt suh tht F = Φ +, where Φ is the ntiderivtive or on [, b] tht is given in Theorem Tht is, or x [, b], whih gives F () = Thereore = F (b) F (). F (x) = Φ(x) + = F (b) = + =, nd so + = + + F (). The irst prt o the Fundmentl Theorem o Clulus shows how deinite integrl n be used to determine n ntiderivtive or untion on losed intervl [, b], while the seond prt shows how n ntiderivtive n be used to determine deinite integrl or on [, b]. The symmetry is something to behold, the two prts tken together eetively uniting the dierentil nd integrl brnhes o lulus. The ollowing proposition is modest generliztion o Theorem 6.23 nd will prove useul in the study o dierentil equtions. Proposition I is ontinuous on (, b) nd (, b), then the untion Φ : (, b) R given by Φ(x) = (t) dt, < x < b is dierentible on (, b), with Φ (x) = (x) or eh < x < b. Proo. Let d [, b). Sine is ontinuous on [, d], by Theorem 6.23 the untion Φ is dierentible on [, d] suh tht Φ (x) = (x) or ll x [, d]. Sine d [, b) is rbitrry we onlude tht Φ (x) = (x) or ll x (, b), with Φ +() = () in prtiulr. Now let d (, ]. Sine is ontinuous on [d, ], by Theorem 6.23 the untion Ψ : [d, ] R given by Ψ(x) = d (t) dt or ll x [d, ] is dierentible suh tht Ψ (x) = (x) or eh d x <, nd Ψ () = (). Now, by Theorem 6.6 nd Deinition 6.5 we obtin nd thus d (t) dt = d (t) dt + x (t) dt = d Φ(x) = Ψ(x) (t) dt d (t) dt. (t) dt = Ψ(x) Φ(x),

12 or eh d x. Sine is onstnt we hve d Φ (x) = Ψ (x) = (x) or x [d, ], nd sine d (, ] is rbitrry we onlude tht Φ (x) = (x) or ll x (, ), with Φ () = () in prtiulr. Finlly, rom Φ +() = () = Φ () we hve Φ () = (), nd thereore Φ (x) = (x) or ll < x < b. 2

13 3 6.6 Improper Riemnn Integrls Suppose is Riemnn integrble on [, x] or ll x. By deinition we hve R or eh x. This observtion leds us to sk whether tends to some limiting vlue L R s x ; tht is, does the limit lim x exist in R? Suh question rises requently in pplitions, nd so motivtes the ollowing deinition. Deinition I R[, b] or ll b, then we deine = lim b nd sy tht onverges to L i = L or some L R. Otherwise we sy tht diverges. I R[, b] or ll b, then we deine nd sy tht onverges to L i diverges. = lim = L or some L R. Otherwise we sy tht I n improper integrl onverges to some rel number L then it is ustomry to sy simply tht the integrl onverges or is onvergent. An integrl tht diverges is lso sid to be divergent. Any integrl o the orm,, or (see below) is lled n improper integrl o the irst kind. The next proposition estblishes linerity properties speiilly or integrls o the orm tht re identil in orm to the linerity properties o the Riemnn integrl given in 5.3. There re similr linerity properties or ll types o improper integrls. Proposition I re onvergent suh tht. 2. = ( ± g) = nd ± g g re onvergent nd R, then nd ( ± g) The proo is routine pplition o relevnt lws o limits estblished bk in Chpter 2, nd so let s n exerise.

14 4 Exmple Determine whether onverges or diverges. Evlute i onvergent. ln(x) x 2 Solution. It will be esier to irst determine the indeinite integrl ln(x) dx. x 2 We strt with substitution: let w = ln(x), so tht dw = (/x)dx nd e w = e ln(x) = x; now, ln(x) dx = we w dw. x 2 Next, we employ integrtion by prts, letting u (w) = e w nd v(w) = w to obtin we w dw = we w + e w dw = we w e w + C. dx Hene, ln(x) dx = ln(x) x 2 x x + C = ln(x) + + C. x Now we turn to the improper integrl, ln(x) b [ dx = lim ln(x) + ] b x 2 b b x ] ln(x) dx = lim x 2 [ = lim ln(b) + b b ( ) LR /b = lim =, b + ln() + = lim b using L Hôpitl s Rule where indited. Thereore the improper integrl is onvergent, nd its vlue is. ( ) b ln(b) + b Proposition 6.3. Suppose tht R[s, t] or ll < s < t <. I nd onverge or some R, then or ny the integrls nd lso onverge, nd + = + Proo. Suppose nd onverge or some R, mening the limits both exist. Let <. For ll b > we hve lim = nd lim + b,

15 where, R sine is integrble on [, ] nd [, b], nd so ( ) = lim = lim + = + lim b b Observing tht, R, we redily onlude tht For ll < we hve =, b where, R sine is integrble on [, ] nd [, ], nd so ( ) = lim = lim = lim = + 5. (6) R nd hene onverges. =. (7) Observing tht, R, we redily onlude tht R nd hene onverges. Finlly, ombining (6) nd (7), we obtin s desired. + ( = ) + ( + ) = + Due to Proposition 6.3 we n unmbiguously deine n improper integrl o the irst kind whose intervl o integrtion is (, ). Deinition 6.3. Suppose tht R[s, t] or ll < s < t <. I nd both onverge or some < <, then we deine = + nd sy tht onverges. Otherwise we sy diverges. It should be stressed tht limit s the next exmple illustrtes.. n not be relibly evluted simply by omputing the lim, b b, Exmple Show tht diverges, nd yet 2x + x 2 dx 2x lim dx =. b b + x2

16 6 Solution. Letting u = + x 2 gives du = 2x dx. Then nd so Thus 2x + x 2 dx = diverges, nd thereore +b 2 du = [ln u ]+b2 = ln( + b 2 ) ln() = ln( + b 2 ), u 2x b 2x dx = lim dx = lim + x2 b + x ln( + 2 b b2 ) =. 2x + x 2 dx 2x + x 2 dx diverges s well. On the other hnd, gin employing the substitution u = + x 2 we ind tht 2x +b 2 b + x dx = du =, 2 +b u 2 nd so 2x lim dx = lim b b + x2 b () =. An improper integrl o the seond kind is n integrl o the orm where < < b <, or whih there exists some p [, b] suh tht p / Dom(). The ollowing deinition estblishes how suh n integrl is to be evluted, i it n be evluted t ll, in the se when p = or p = b. Deinition I R[, b] or ll (, b] nd / Dom(), then we deine, = lim + nd sy tht onverges to L i = L or some L R. Otherwise we sy tht diverges. I R[, ] or ll [, b) nd b / Dom(), then we deine = lim b nd sy tht onverges to L i = L or some L R. Otherwise we sy tht diverges.

17 Very oten i is ontinuous on, sy, (, b] nd / Dom(), then hs vertil symptote t ; tht is, lim x + (x) = ±. However, it ould just be tht vlue or is simply not speiied t by onstrution. For exmple or the untion { 3x 2, i x < 5 ϕ(x) = 4 8x, i x > 5 it s seen tht ϕ(5) is let undeined, nd so the integrl 9 ϕ is n improper integrl o the 5 seond kind. By Deinition 6.33 we obtin 9 5 ϕ = lim (4 8x) dx = lim 5 + [ 4x 4x 2 ] 9 = lim 5 + [( 4(9) 4(9) 2 ) ( 4 4 2)] = ( 4(9) 4(9) 2) ( 4(5) 4(5) 2) = 28, whih shows tht 9 ϕ is onvergent. 5 Exmple Determine whether onverges or diverges. Evlute i onvergent. x 2 dx Solution. The untion (x) = /x 2 being integrted hs vertil symptote t x =, whih is the right endpoint o the intervl o integrtion [, ]. By Deinition 6.33 we obtin [ dx = lim dx = lim ] = lim ( ) x2 x2 x =, whih shows tht the improper integrl is divergent. Exmple Determine whether 2 onverges or diverges. Evlute i onvergent. x 4 x 2 dx Solution. Here x/ 4 x 2 hs vertil symptote t x = 2, the right endpoint o the intervl o integrtion [, 2]. By Deinition x dx = lim 4 x 2 2 nd so, letting u = 4 x 2 so tht x dx = du, we obtin 2 lim 2 x dx = lim 4 x x 4 x 2 dx, /2 u du = lim 2 ( 2 ) [ ] u ( = lim 2 ) 4 2 = = 2. 2 Hene the improper integrl is onvergent, nd its vlue is

18 The next deinition ddresses the irumstne when untion is not deined t some point p in the interior o n intervl o integrtion. Agin, this is ommonly due to hving vertil symptote t p, so tht lim x p but other senrios re possible. x p (x) = or lim (x) =, + Deinition Suppose tht R[, ] or ll [, p), R[, b] or ll (p, b], nd p / Dom(). I p nd both onverge, then we deine p = p + nd sy tht onverges. Otherwise we sy diverges. Exmple Determine whether 3 onverges or diverges. Evlute i onvergent. 2 x 4 dx Solution. Here /x 4 hs vertil symptote t x =, n interior point o the intervl o integrtion [ 2, 3]. Now, by Deinition [ dx = lim dx = lim ] 3 = lim ( 27 x4 + x4 + x + ) =, whih shows tht 3 x 4 dx is divergent. Thus, sine 3 x 4 dx nd p 2. x 4 dx nnot both be onvergent, by Deinition 6.36 it s onluded tht 3 2 x 4 dx is divergent. The integrl treted in Exmple 6.37, like ll improper integrls o the seond kind, does not look improper t irst glne. I one is reless nd undertkes to evlute the integrl by onventionl mens, one is likely to rrive t resonble-looking nswer without ever suspeting tht something is miss: 3 2 x 4 dx = [ ] 3 = x = 35 26, whih is inorret! So, beore ttempting to evlute deinite integrl, it is neessry to hek tht the integrl is not improper in some wy. It is possible to hve n integrl tht is improper in more thn one sense, suh s x 2 dx. 8

19 Here we hve n integrl o over n unbounded intervl [, ), so it s n improper integrl o the irst kind, nd lso is undeined t, so it s n improper integrl o the seond kind. Suh n integrl is lled mixed improper integrl. Deinition I R[s, t] or ll < s < t <, / Dom(), nd nd both onverge or some (, ), then we deine = + nd sy onverges. Otherwise we sy diverges. I R[s, t] or ll < s < t < b, b / Dom(), nd nd some (, b), then we deine = + nd sy onverges. Otherwise we sy diverges. Exmple Determine whether the mixed improper integrl onverges or diverges. Evlute i onvergent. x( + x) dx Solution. We strt by determining the indeinite integrl dx. x( + x) Let u = x, so tht + u 2 = + x nd we reple dx with 2u du to obtin dx = x( + x) 2u u(u 2 + ) du = 2 u 2 + du = 2 rtn(u) + = 2 rtn( x ) +. 9 both onverge or Now, [ ] dx = lim dx = lim 2 rtn( x ) x( + x) + x( + x) + = lim 2[rtn() rtn()]= 2[rtn() rtn()] + ( π ) = 2 4 = π 2, nd x( + x) dx = lim b = lim b 2 [ [ ] b dx = lim 2 rtn( x ) x( + x) b rtn( ] ( π b ) rtn() = 2 2 π ) 4 = π 2.

20 2 Sine both onverge, we onlude tht by Deinition x( + x) dx = x( + x) dx nd x( + x) dx dx + dx = π x( + x) x( + x) 2 + π 2 = π

21 2 6.7 Comprison Tests or Integrls It n be diiult to determine by diret mens whether n improper integrl is onvergent or not, lrgely beuse deinite integrls themselves n be diiult to evlute. One tool to remedy this is omprison test. Beore stting the irst suh test we need to estblish lemm tht will be needed or its proo. Lemm 6.4. Suppose the untion is monotone inresing on (, ). I then (x) M or ll x >. lim (x) = M, x Proo. Suppose there exists some x > suh tht (x ) > M. Thus (x ) = M + ɛ or some ɛ >. Now, or ny β > we n let x = mx{x, β} +. Sine x > x nd is monotone inresing, we hve (x ) (x ) = M + ɛ (x ) M ɛ (x ) M ɛ. Observing tht x > β lso, we onlude tht or ny β > there exists some x > β or whih (x) M ɛ, nd thereore lim (x) M. x The ontrpositive o the sttement o the lemm is proven. Theorem 6.4. Suppose R[, x] or ll x, nd g on [, ). I onvergent, then is onvergent. Proo. Suppose g is g is onvergent. By deinition it ollows tht g R[, x] or ll x, g. Similrly we deine ϕ : [, ) R by nd so we my deine ψ : [, ) R by ψ(x) = ϕ(x) =. Now, g on [, ) implies tht y x g or ny x < y. Thus, or ny x, y [, ) suh tht x < y we hve ψ(y) = y g = g + y x g g = ψ(x), whih shows tht ψ is monotone inresing on [, ). Sine on [, ), similr rgument estblishes tht ϕ lso is monotone inresing on [, ). Sine g onverges, there exists some M R suh tht lim x Tht is, ψ is monotone inresing on (, ) nd g = M. lim x ψ(x) = M, so ψ(x) M or ll x > by Lemm 6.4.

22 Finlly, rom the hypothesis tht g on [, ), we hve ϕ(x) = g = ψ(x) M or ll x >. It is now estblished tht ϕ is both monotone inresing nd bounded bove on (, ), nd thereore by Proposition 2.2 lim x ϕ(x) exists in R. Sine it ollows tht = lim x = lim x ϕ(x), is onvergent. Proposition Suppose R[, x] or ll x. I onvergent. 22 is onvergent, then is Proo. Suppose tht is onvergent. Then R[, x] or ll x, nd sine the sme holds true or by hypothesis, we hve + R[, x] or ll x by Proposition 5.5. Now, sine + 2 on [, ), nd 2 is onvergent by Proposition 6.28, Theorem 6.4 implies tht ( + ) is onvergent. Then, beuse is onvergent by Proposition 6.28, it ollows by Proposition 6.28 tht [ ] ( + ) + ( ) is onvergent. O ourse ( + ) + ( ) = on [, ), nd thus we onlude tht is onvergent.