6.5 Improper integrals

Transcription

1 Eerpt from "Clulus" 3 AoPS In IMPROPER INTEGRALS 6.5 Improper integrls As we ve seen, we use the definite integrl R f to ompute the re of the region under the grph of y = f () long the intervl [, ]. By definition, these integrls n only e used to ompute res of ounded regions. In some situtions, however, we re interested in unounded regions these re regions tht etend towrds infinity in t lest one diretion. Yet, mny unounded regions still hve finite re. We strt with si emple of this phenomenon: Prolem 6.6: Wht is the re of the region ordered y the urve y =, the line =, nd the -is? Solution for Prolem 6.6: We sketh piture of this region t right. Notie tht this region is unounded: the region etends towrds + s grows lrge. Even though this region is unounded, we n ttempt to determine its re. We ertinly n ompute the re of the portion of the region to the left of = (for ny > ) s the definite integrl y d. As grows lrger, we epet tht the re under the urve on [, ] pprohes the re of the entire region under the urve on [, +). Speifilly, this re is y =! d. The integrl is esy to evlute: d = = Thus, when we tke the it, we get tht the re of the region is. d =!! = =. Note the prdo here: even though the region is unounded, it hs finite re. Prolem 6.6 suggests logil definition: Definition: integrl Let f e ontinuous funtion nd R suh tht (, ) Dom( f ). We define the improper f () d = f () d,! provided the it is defined. If the it is defined nd is not ±, we sy tht the improper integrl onverges. Otherwise, we sy tht the improper integrl diverges. There s n oviously similr definition for improper integrls in the other diretion: 7 Copyrighted Mteril

2 Eerpt from "Clulus" 3 AoPS In. CHAPTER 6. INFINITY Definition: integrl Let f e ontinuous funtion nd R suh tht (, ) Dom( f ). We define the improper f () d = f () d,! provided the it is defined. If the it is defined nd is not ±, we sy tht the improper integrl onverges. Otherwise, we sy tht the improper integrl diverges. Let s generlize Prolem 6.6: Prolem 6.7: Let r e rel numer. Compute r d. Solution for Prolem 6.7: hve: This equls By definition, we ompute the improper integrl y writing it. If r,, then we d =! r!! r. (r ) r. r If r >, then the term pprohes s pprohes. Thus, in this se, the improper integrl onverges r to r. If r <, then the term grows without ound s pprohes. Thus, the integrl diverges. We might lso r write d = if r <. r Our originl integrtion ws not vlid for r =, so we hve to do tht se seprtely:! d =! (log ) =! (log ). As goes towrds infinity, this grows without ound, so the integrl diverges. In summry: r d = 8 >< r if r >, >: diverges if r pple. The net prolem is nother ommon emple of n improper integrl: Prolem 6.8: Compute e d, where is rel numer. Solution for Prolem 6.8: We ompute, for, (we ll investigte = t the end):! e d =! (e ) =! (e ). 8 Copyrighted Mteril

3 Eerpt from "Clulus" 3 AoPS In IMPROPER INTEGRALS If is positive, then e =, so the integrl diverges. If is negtive, then e =, so the integrl equls!!. (Note this is positive numer when is negtive, so this nswer mkes sense.) Finlly, if =, then the integrl is R d, whih lerly diverges. Thus, the integrl diverges for nonnegtive eponents, nd onverges for negtive eponents. The result of Prolem 6.8 is typilly written s follows: if r >, then e r d = r. Prolem 6.9: Suppose f nd g re ontinuous funtions on [, ) nd f () pple g() for ll. () Show tht, if R f nd R g oth onverge, then Show tht if oth funtions re positive, nd R () Show tht if oth funtions re positive, nd R Solution for Prolem 6.9: () For ny, we hve (g f )() for ll [, ], thus Therefore, f pple g. g onverges, then R f onverges. f diverges, then R g diverges. (g f )() d. f () d pple g() d, nd sine its preserve non-strit inequlities, we onlude tht f () d = f () d pple g() d = g() d.!! Define funtion F() = f (t) dt. Note tht F is n inresing funtion (sine f () for ll ), nd tht R eists. Also, sine pple f () pple g() for ll, we hve f () d =! F(), if this it pple F() = f (t) dt pple g(t) dt pple g(t) dt. Thus F is inresing nd hs n upper ound (nmely, R g(t) dt, whih y ssumption onverges), so y the result of Prolem 6.5, the it F() = f (t) dt! eists, so the integrl onverges. 9 Copyrighted Mteril

4 Eerpt from "Clulus" 3 AoPS In. CHAPTER 6. INFINITY () This is just the ontrpositive sttement to prt, so there is nothing dditionl to prove. WARNING!! j We n only use the omprison tests in prts nd () of Prolem 6.9 if oth funtions re positive. As trivil emple, if f () = nd g() =, then for ny R, R g =, so R g onverges, ut R f diverges. Thus fr in this setion, we hve looked t improper integrls tht ompute res of regions tht re unounded in the -diretion. There is nother type of improper integrl tht ours when the region tht we re emining is unounded in the y-diretion, s in the following emple: Prolem 6.: Compute p d. Solution for Prolem 6.: Skething the grph will immeditely show the issue. We hve! + p =. So the re under y = p is potentilly infinite (nd in ft the funtion is not even defined t ). We n do essentilly the sme thing we did for improper integrls with it of integrtion of ±. We define y y = p p d =! + p d. Note the + sine we only re out the intervl (,], we only re out wht hppens to the right of. This integrl is now esy to ompute: p d = p = p. As! +, this pprohes. Hene p d =. One gin, seemingly infinite re turns out to e finite. We n generlize the definition from Prolem 6.: Definition: integrl Suppose f is funtion, ontinuous on (, ], suh tht f () = ±. We define the improper! + f () d = f () d,! + provided this it is defined. If the it is defined, we sy tht this improper integrl onverges, nd if it is undefined, we sy tht the improper integrl diverges. Of ourse, we n do the sme thing if the funtion hs it of ± t the end of [, ). (We will omit writing out the forml definition.) Copyrighted Mteril

5 Eerpt from "Clulus" 3 AoPS In IMPROPER INTEGRALS Sidenote: Note tht the ove definition is onsistent with our usul (non-improper) integrls. In prtiulr, if R f is defined, then y the Fundmentl Theorem of Clulus, the funtion g() = is di erentile, hene ontinuous, nd thus f (t) dt f (t) dt = g() = g() = f (t) dt.! +! + We know tht for regulr (not improper) integrls, we n rek them prt t ny point into two seprte integrls. Speifilly, if (, ), then f = f + f. This is lso how we evlute integrls tht re improper t oth ends, s in the following emple: Prolem 6.: Compute d for ll r > (or determine when it diverges). r Solution for Prolem 6.: The orret thing to do with n integrl tht is improper t oth ends is to split it somewhere in the middle. For emple, we n write r d = r d + r d. (We didn t hve to pik = s the point t whih to split them, ut it seems onvenient sine r is niely ehved t =.) We lredy know y Prolem 6.7 tht R d onverges if nd only if r >. The other integrl is r! d = d = = r!+ r!+ (r ) r r.!+ r If r >, then the frtion gets ritrrily lrge, so the it is infinite. Thus R d diverges for r >. r Hene our originl douly-improper integrl is never onvergent: the integrl on (, ] diverges for r >, nd the integrl on [, ) diverges for r pple. Importnt: If (, ) Dom( f ) nd f (t) dt is improper t oth ends of (, ), then for ny (, ). f (t) dt = f (t) dt + f (t) dt,! +! As noted in the solution to Prolem 6., it doesn t mtter t whih point we rek up the douly-improper integrl. Copyrighted Mteril

6 Eerpt from "Clulus" 3 AoPS In. CHAPTER 6. INFINITY Conept: We n rek n integrl prt s f = f + f t ny (, ) tht we hoose. Thus, hoose to e s onvenient s possile. We will leve it s n eerise to prove this. Also, it is not orret to try to tke shortut nd del with oth ends of doule-improper integrl t one. In prtiulr: WARNING!! j f () d is not the sme s f () d.! The orret wy to evlute n integrl over ll of R is to hoose R, nd then ompute f () d = f () d + f () d = f () d + f () d.!!+ We will leve it s n eerise to eplore this further. We lso hve to e it utious when deling with funtions with domins tht re not ll of R. Integrls of suh funtions might e improper ut not immeditely pper so. For emple: 3 Prolem 6.: Compute d. Solution for Prolem 6.: If you weren t pying lose ttention, you might do this: Bogus Solution: 3 d = 3 = 3 + = 6. We n t do this, euse the funtion is not defined t! To e little more preise, the funtion does not hve n ntiderivtive on the intervl [, 3], euse it is not defined t =, so we nnot pply the Fundmentl Theorem of Clulus. In order to evlute the integrl, we need to rek it up into sum of two improper integrls t the point t whih the funtion is undefined: 3 d = 3 d + d. As we sw in Prolem 6., oth of these diverge. Thus, the originl integrl itself diverges. d More generlly, when omputing something like, it might e tempting to sy is n odd funtion, so the integrl from to will nel out the integrl from to, nd thus the overll integrl is. This is lso the result tht nive lultion will give: Bogus Solution: d = log = log() log() =. Copyrighted Mteril

7 Eerpt from "Clulus" 3 AoPS In. REVIEW PROBLEMS But this is not orret! The only wy leglly to evlute this integrl is to rek it up into its improper prts. Neither prt onverges, so the integrl diverges. Eerises 6.5. Compute the following improper integrls: () 6.5. () 3 d ( ) Compute + d. Compute Compute + d. e d. (log ) d d = d + d. () e d (d) 4 d Show tht it doesn t mtter t whih point we rek up douly-improper integrl. Speifilly, show tht, for ny, d (, ), if R f nd R f onverge, then R d f nd R f lso onverge, nd d d f + f = f + f. d Hints: 3, ? () Show tht if f () d onverges, then f () d = f () d. Hints: 65! Show tht the onverse of prt () is not true; tht is, it is possile tht f () d diverges. Hints: 8, 5! f () d onverges ut tht Review Prolems 6.3 Compute the following: ()! log os! 3! () e 4 (Soure: Rie) 6.4 sin t tn t Suppose nd re nonzero rel numers. Find nd. Hints: 3 t! sin t t! tn t 6.5 Compute () e d 3 d () d Hints: Copyrighted Mteril