MATHEMATICAL ANALYSIS

Transcription

1 Mri Predoi Trdfir Băl MATHEMATICAL ANALYSIS VOL II INTEGRAL CALCULUS Criov, 5

2 CONTENTS VOL II INTEGRAL CALCULUS Chpter V EXTENING THE EFINITE INTEGRAL V efiite itegrls with prmeters Problems V 5 V Improper itegrls 9 Problems V 9 V Improper itegrls with prmeters Problems V Chpter VI LINE INTEGRALS VI Curves Problems VI 7 VI Lie itegrls of the first tpe 9 Problems VI 4 VI Lie itegrls of the secod tpe 44 Problems VI 5 Chpter VII MULTIPLE INTEGRALS VII Jord s mesure 56 Problems VII 6 VII Multiple itegrls 6 Problems VII 77 VII Improper multiple itegrls 8 Problems VII 88 V

3 Chpter VIII SURFACE INTEGRALS VIII Surfces i R 9 Problems VIII 97 VIII First tpe surfce itegrls 99 Problems VIII VIII Secod tpe surfce itegrls 4 Problems VIII VIII4 Itegrl formuls Problems VIII4 7 Chpter IX ELEMENTS OF FIEL THEORY IX ifferetil opertors 9 Problems IX 7 IX Curvilier coordites Problems IX 9 IX Prticulr fields 4 Problems IX 5 Chpter X COMPLEX INTEGRALS X Elemets of Cuch theor 55 Problems X 66 X Residues 68 Problems X 85 INEX 88 BIBLIOGRAPHY VI

4 CHAPTER V EXTENING THE EFINITE INTEGRAL V EFINITE INTEGRALS WITH PARAMETERS We cosider tht the itegrl clculus for the fuctios of oe rel vrible is kow Here we iclude the idefiite itegrls (lso clled primitives or ti-derivtives s well s the defiite itegrls Similrl, we cosider tht the bsic methods of clcultig (ectl d pproimtel itegrls re kow The purpose of this prgrph is to stud etesio of the otio of defiite itegrl i the sese tht beod the vrible of itegrtio there eists other vrible lso clled prmeter efiitio Let us cosider itervl A R, I = [, b] R d f : A IR If for ech A ( is clled prmeter, fuctio t f(, t is itegrble o [, b], the we s tht F : A R, defied b b F( = f(, tdt is defiite itegrl with prmeter (betwee fied limits d b More geerll, if isted of, b we cosider two fuctios φ, ψ : A [, b] such tht φ( ψ( for ll A, d the fuctio t f(, t is itegrble o the itervl [φ(, ψ(] for ech A, the the fuctio G( = ( ( f(, tdt is clled defiite itegrl with prmeter (betwee vrible limits The itegrls with vrible limits m be reduced to itegrls with costt limits b chgig the vrible of itegrtio: Lemm I the coditios of the bove defiitio, we hve: G( = [ψ( φ(] f(, φ( + θ[ψ( φ(]d θ Proof I the itegrl G( we mke the chge t = φ( + θ [ψ( φ(], dt for which = ψ( φ( } d Reltive to F d G we'll stud the properties cocerig cotiuit, derivbilit d itegrbilit i respect to the prmeter Theorem If f : A I R is cotiuous o A I, the F : A R is cotiuous o A

5 Chpter V Etedig the defiite itegrl Proof If A, the either Å, or is ed-poit of A I cse there eists η > such tht K η = {(, t R : η, A, t[, b]} is compct prt of A I Sice f is cotiuous o A I, it will be uiforml cotiuous o K η, ie for ε > there eists δ > such tht f(', t' f(", t" < ( b wheever (', t', (", t" K η d d((', t', (", t" < δ Cosequetl, for ll A for which < mi { η, δ } we hve b F( F( f(, t f(, t dt (b < ε, ( b which mes tht F is cotiuous t } 4 Corollr If the fuctio f : A I R is cotiuous o A I, d φ, ψ : A [, b] re cotiuous o A, the G : A R is cotiuous o A Proof Fuctio g : A [, ] R, defied b g(, θ = f(, φ( + θ[ψ( φ(], which ws used i lemm, is cotiuous o A [, ], hece we c ppl theorem d lemm } 5 Theorem Let A R be rbitrr itervl, I = [, b] R, d let us ote f : A I R If f is cotiuous o A I, d it hs cotiuous prtil derivtive f, the F CR (A, d F'( = Proof We hve to show tht t ech A, there eists F( F( f lim (, t dt b b f (, tdt For this purpose we cosider the followig helpful fuctio f (, t f (, t if h(, t = f (, t if O the hpothesis it is cler tht h is cotiuous o A I, hece we c use theorem for the fuctio b H( = b h(, tdt = f (, t f (, t F( F( dt

6 O this w, the equlit H( =, d b F'( = lim V efiite itegrls with prmeters H( shows tht F is derivble t f (, tdt The cotiuit of F' is cosequece of the cotiuit of f, b virtue of the sme theorem } 6 Corollr If, i dditio to the hpothesis of the bove theorem, we hve φ, ψ C R (A, the G C R (A d the equlit G'( = ( ( f (, tdt + f(, ψ( ψ'( f(, φ( φ'( holds t A Proof Let us cosider ew fuctio L : A I I R, epressed b v L(,u,v = u we hve f(, tdt Accordig to the bove theorem, for fied u d v v L f (, u, v (, t dt O the other hd, the geerl properties u L L of primitive led to (, u, v = f(, u d (, u, v = f(, v u v Becuse ll these prtil derivtives re cotiuous, L is differetible o A I I Applig the rule of derivig composite fuctio i the cse of G( = L(, φ(, ψ(, we obti the ouced formul The cotiuit of G' follows b usig theorem } 7 Theorem If f : A I R is cotiuous o A I, the F : A R is itegrble o compct [α, β] A, d b F ( d f (, t ddt Proof Accordig to theorem, F is cotiuous o [α, β], hece it is lso itegrble o this itervl It is well kow tht the fuctio Ф( = F(d is primitive of F o [α, β] We will show tht

7 Chpter V Etedig the defiite itegrl b Ф( = For this purpose let us ote U(, t = The, '( = f (, t ddt f(, td d ( = U (, t = f(, t, hece ccordig to theorem 5, we hve b b U(, tdt f(,tdt Cosequetl, the equlities '( = F( = Ф'( hold t [α, β], hece Ф( ( = c, where c is costt Becuse Ф(α = (α =, we obti c =, ie Ф = I prticulr, Ф(β = (β epress the required equlit } 8 Corollr If, i dditio to the coditios i the bove theorem, the fuctios φ, ψ : A[, b] re cotiuous o A, the G( d g(, dd where g(, θ = f(, φ( + θ[ψ( φ( ] [ψ( φ( ] (s i corollr 4 Proof Accordig to Lemm, we hve G( = g (, d, so it remis to use theorem 7 } 9 Remrk The formuls estblished i the bove theorems d their corollries (especill tht which refers to derivtio d itegrtio re frequetl useful i prctice for clcultig itegrls (see the problems t the ed of the prgrph I prticulr, theorem 7 gives the coditios o which we c chge the order i iterted itegrl, ie b b f (, t dtd f (, t ddt 4

8 / PROBLEMS V Clculte l( si t dt, where > Hit eotig the itegrl b F(, we obti F'( = Usig the substitutio tg t = u, we obti F'( = F( = π l( + + c I order to fid c, we write V efiite itegrls with prmeters / si, d so c = F( π l( + = / = si t l l dt πl( + = / = si t l dt l Tkig here, it follows c = πl Clculte I = f( = f(d, where f : [, ] R hs the vlues Hit Notice tht f( = eists l t if (,, if, dt t dt t [,, d t the ed poit, there lim f (, so ol t this poit f differs form cotiuous fuctio o [, ] Cosequetl I = [ t dt ]d = t ddt l Clculte si t tg t e e dt lim dt 5

9 Chpter V Etedig the defiite itegrl Hit This is idetermitio; i order to use L'Hospitl rule we eed the derivtives reltive to, which is prmeter i the upper limits of itegrls, so the limit reduces to lim si cos cos si tg e e tg t ( t e t e dt t dt d 4 Clculte I =, where < b <, d deduce the vlues of b cos I = d cos, K = b cos ( b cos d d L = l( b cos d Hit The substitutio tg = t is ot possible i I becuse [, π is crried ito [, Sice the itegrl is cotiuous o R, we hve I = l d lim, b cos l d this lst itegrl c be clculted usig the metioed substitutio More ectl, l d l tg dt rctg tg b cos b t ( b b hece I = b b b To obti K, we derive I reltive to b Fill, rctg rctg 5 Clculte I = d b derivig I( = d, Hit Substitutio = cos θ gives / l L =I d d I'( = ( cos Becuse the substitutio tg θ = t crries [, ito [,, d the substitutio tg = t leds to complicted clcultio, we cosider 6

10 d I'(= lim cos l l V efiite itegrls with prmeters If we replce tg θ = t i this lst itegrl, the we obti l d cos = Cosequetl I'( = tgl dt t rctg tgl, hece I( = l( + + c Becuse I( = it follows tht c =, hece I = I( = l(+ rctg rctg d 6 Clculte I = d usig the formul Hit Chgig the order of itegrtio we obti I = d d d d ( so the problem reduces to I'( from problem 5 bsi d 7 Clculte K= l, > b > bsi si Hit Usig the formul obti Sice si bsi l bsi K = b d b b d b si d si d b K = b rcsi b b b b b d it follows tht d si si d we 7

11 Chpter V Etedig the defiite itegrl 8 Show tht I + ( = d Usig this result, clculte ( Hit erive I ( reltive to I' (, where I ( = ( d, N *, 9 Use Theorem 7 to evlute I = f ( d, where si(l, if d f ( l, if or d >, > Hit Itroduce prmeter t d remrk tht I = t dt si(l d Chge the order of itegrtio to obti I = t dt si(l d dt = ( t The result is I = rctg ( ( 8

12 V IMPROPER INTEGRALS I the costructio of the defiite itegrl, oted b f ( t dt, we hve used two coditios which llow us to write the itegrl sums, mel: (i d b re fiite (ie differet from + ; (ii f is bouded o [, b], where it is defied There re still m prcticl problems, which led to itegrls of fuctios ot stisfig these coditios Eve defiite itegrls reduce sometimes to such "more geerl" itegrls, s for emple whe chgig the vribles b tg = t, the itervl [, π] is crried ito [, ] The im of this prgrph is to eted the otio of itegrl i the cse whe these coditios re o loger stisfied efiitio The cse whe b = If f : [, R is itegrble o [, β] for ll β >, d there eists L = lim f ( t dt, the we m s tht f is improperl itegrble o [,, d L is the improper itegrl of f o [, I this cse we ote f ( t dt = lim f ( t dt, d we s tht the improper itegrl is coverget Similrl we discuss the cse whe = The cse whe f is ubouded t b Let f : [, b R be ubouded i the eighborhood of b, i the sese tht for rbitrr δ > d M > there eists t (b δ, b such tht f (t > M If f is itegrble o [, β ] for ll < β < b, d there eists L = lim b f ( t dt, the we s tht f is improperl itegrble o [, b, d L is clled improper itegrl of f o [, b If L eists, we ote b f ( t dt = lim f ( t dt, d we s tht the b improper itegrl is coverget We similrl tret the fuctios which re ubouded t Remrks I prctice we ofte del with combitios of the bove simple situtios, s for emple f ( t dt f ( t dt lim ot R f ( t dt, 9

13 Chpter V Etedig the defiite itegrl b f ( t dt lim f ( t dt, where < α < β <b b b The itegrl f ( t dt c be improper becuse f is ubouded t some poit c (, b, i which cse we defie b f t dt ( = ( t dt lim lim f f ( t dt c c b c c b From the geometricl poit of view, cosiderig improper itegrls m be iterpreted s mesurig res of ubouded subsets of the ple The eistece of the bove cosidered limits shows tht we c spek of the re of ubouded set, t lest for sub-grphs of some rel fuctios c I spite of the diversit of tpes of improper itegrls, there is simple, but essetil commo feture, mel tht the itegrtio is relied o o-compct sets I fct, compct set i R is bouded d closed, hece [,, (, b], (,+ re o-compct becuse the re ot bouded, while [, b, (, b], etc re o-compct becuse of o-closeess Obviousl, other combitios like (,, (, c (c, b], etc re possible Becuse improper itegrl is defied b limitig process, whe provig some propert of such itegrls it is sufficiet to cosider ol oe of the possible cses dt Emples The itegrl I(λ = t (λ R is coverget for λ >, whe I(λ = (λ, d diverget for λ I fct, ccordig to the bove defiitio, I(λ = lim t ( t dt l Fill, it remis to remember tht lim dt, where if if if if if

14 V Improper itegrls dt b The itegrl I(μ = t (μ > is coverget for μ <, whe it equls I(μ = ( μ, d it is diverget for μ Figures V, respectivel b, suggest how to iterpret I(λ d I(μ s res of some sub-grphs (htched portios t Fig V The usul properties of the defiite itegrls lso hold for improper itegrls, mel: 4 Propositio The improper itegrl is lier fuctiol o the spce of ll improperl itegrble fuctios, ie if f, g : [, b R re improperl itegrble o [, b, d λ, μ R, the λf + μg is improperl itegrble o [, b d we hve: b ( g( t dt f ( t dt b f g( t dt b The improper itegrl is dditive reltive to the itervl, ie b c f ( t dt = f ( t dt + b c b f ( t dt c The improper itegrl is depedet o the order of the itervl, mel b f ( t dt = t b f ( t dt 5 Theorem (Leibi-Newto formul Let f : [, b R be (properl itegrble o compct [, β ]icluded i [, b, d F be the primitive of f o [, b The ecessr d sufficiet coditio for f to be improperl itegrble o [, b is to eist the fiite limit of F t b I this cse we hve: b

15 Chpter V Etedig the defiite itegrl b f ( t dt = lim F( F( b 6 Theorem (Itegrtio b prts If f, g stisf the coditios: (i f, g C R([, b] (ii there eists d is fiite lim ( fg( b (iii b b b f ( t g' ( t dt is coverget the f '( t g( t dt is coverget too, d we hve b f '( t g( t dt = lim ( fg( b b b f(g( f ( t g' ( t dt 7 Theorem (Chgig the vrible Let f : [, b R be cotiuous o [, b, d let φ : [', b' [, b be of clss C R([', b'], such tht φ(' = d lim ( b b ' ' b b If f ( t dt is coverget, the the itegrl b' ' is lso coverget, d we hve b' ' f ( ( ' ( d f ( ( ' ( d b = f ( t dt The bove properties (especill theorems 5 7 re useful i the cses whe primitives re vilble If the improper itegrl c't be clculted usig the primitives it is still importt to stud the covergece For developig such stud we hve severl tests of covergece, s follows: 8 Theorem (Cuch's geerl test Let f : [, b R be (properl b itegrble o [, β] [, b The ε > there eists δ > such tht b', b" (b δ, b implies f ( t dt is coverget iff for ever b" b' f ( t dt

16 V Improper itegrls Proof Let F : [, b R be defied b F( = f ( t dt The f is improperl itegrble o [, b if F hs fiite limit t b, which mes tht for ever ε > we c fid δ > such tht b', b" (b δ, b implies F(b' F(b" < ε It remis to remrk tht F(b' F(b" = b" b' f ( t dt } The bove Cuch's geerl test is useful i reliig logies with bsolutel coverget series s follows: 9 efiitio If f : [, b R, the we s tht the itegrl b bsolutel coverget iff b f ( t dt is f ( t dt is coverget, ie f is improperl itegrble o [, b Remrk I wht cocers the itegrbilit of f d f, the improper itegrl differs from the defiite itegrl: while f itegrble i the proper sese implies f itegrble, this is ot vlid for improper itegrls I fct, there eist fuctios, which re improperl itegrble without beig bsolutel itegrble For emple, let f : [, R be fuctio of ( vlues f ( =, d f (t = if t (, ], where N * This fuctio is improperl itegrble o [,, d f ( t dt ( l, but it is ot bsolutel itegrble sice f ( t dt The et propositio shows tht the opposite implictio holds for the improper itegrls: Propositio Ever bsolutel coverget itegrl is coverget Proof Usig the Cuch's geerl test, the hpothesis mes tht for ever ε > there eists δ > such tht for β', β" (b δ, b we hve " ' f ( t dt < ε Becuse f is properl itegrble o compct from [, b, d

17 Chpter V Etedig the defiite itegrl " ' f ( t dt " ' f ( t dt 4 " = ' f ( t dt it follows tht f is improperl itegrble o [, b } Theorem (The compriso test Let f, g : [, b R be such tht: f, g re properl itegrble o compct from [, b for ll t [, b we hve f(t g(t b g ( t dt is coverget b The f ( t dt is bsolutel coverget " Proof Becuse ( t dt ' " f g( t dt holds for ll, b, b, ', we c ppl the Cuch's geerl test } Remrk Besides its utilit i estblishig covergece, the bove theorem c be used s divergece test I prticulr, if f(t g(t for b ll t [, b, d f ( t dt is diverget, the g ( t dt is diverget too b I prctice, we relie compriso with fuctios like i emple, ie o [,, t o [, b, q ( b t t o [,, etc The compriso with such fuctios leds to prticulr forms of Theorem, which re ver useful i prctice We metio some of them i the followig theorems Theorem specil form # I of the compriso test (Test bsed o lim t t f ( t Let f : [, R + be itegrble o compct from [, d let us ote = lim t f ( t t If λ > d <, the If λ d <, the b f ( t dt is coverget f ( t dt is diverget Proof If (,, the for ever ε > there eists δ > such tht t > δ implies < ε < t λ f(t < + ε, ie

18 V Improper itegrls f ( t t t If, the the itegrl of o [δ, is diverget, so the first iequlit from bove shows tht λ >, the t the itegrl t f ( t dt is diverget too Similrl, if is itegrble o [δ,, d the secod iequlit shows tht f ( t dt is coverget The cses = d = re similrl discussed usig sigle iequlit from bove } 5 Theorem specil form # II of the compriso test (Test bsed o lim( b t tb f ( t Let f : [, b R + be itegrble o compct from [, b, d let us ote = lim( b t f ( t, where λ R tb If λ < d <, the If λ d <, the b b f ( t dt is coverget, d f ( t dt is diverget The proof is similr to the bove oe, but uses the testig fuctio ( b t o [, b } The bove two tests hve the icoveiet tht the refer to positive fuctios The followig two theorems re cosequeces of the compriso test for the cse of o-ecessril positive fuctios 6 Theorem specil form # III of the compriso test (Test of (t itegrbilit for f(t = t o [, Let f : [, R, where >, (t be fuctio of the form f(t = t where: φ is cotiuous o [, There eists M > such tht ( tdt M for ll α > 5

19 Chpter V Etedig the defiite itegrl The f ( t dt is coverget, wheever λ > Proof B hpothesis, for Φ( = ( tdt we hve α [, Sice λ + >, it follows tht d ( M for ll is coverget So, ( ccordig to theorem, d is bsolutel coverget Itegrtig b prts we obti ( t dt t '( t t dt ( t dt t which shows tht f is itegrble o [, } 7 Theorem specil form # IV of the compriso test (Test of itegrbilit for f(t = (b t λ φ(t o [, b Let f : [, b R, where b R, be fuctio of the form f(t = (b t λ φ(t If φ is cotiuous o [, b there eists M > such tht ( tdt M for ll α [, b, b the the itegrl f ( t dt is coverget for λ > Proof Let us remrk tht Φ( = ( tdt verifies the iequlit b ( ( b M ( b d Sice λ <, b is coverget, hece ( bsolutel coverget It remis to itegrte b prts b ( b t ( t dt ( b t ' ( t dt b b ( b t b ( t ( ( b dt d d use the form of f } is 6

20 7 V Improper itegrls The followig test is bsed o the compriso with the prticulr fuctio g : [, R, of the form g( = q, where q > d > (see lso problem V 8 Theorem specil form # V of the compriso test (The Cuch's root test Let f : [, R, where >, be itegrble o compct from [,, d let us suppose tht there eists = If <, the If >, the f ( t dt is bsolutel coverget, d f ( t dt is ot bsolutel coverget / t lim f ( t t Proof B the defiitio of, we kow tht for ever ε > there eists δ > such tht t > δ implies f(t /t < ε, ie ε < f(t /t < + ε If <, let us ote q = + ε < If t > δ, we hve f(t < q t So, it remis to see tht q t is itegrble o [δ, sice q < Becuse f is itegrble o the compct [, δ ], it will be itegrble o [, too The secod cse is similrl led b otig q = ε >, whe q t diverget, d f(t > q t } The covergece of some improper itegrls c be reduced to the covergece of sequeces d series 9 Theorem (Test of reductio to series If f : [, R + is decresig fuctio, itegrble o [, b] [,, the the followig ssertios re equivlet: f ( t dt is coverget b The sequece of terms u = f ( t dt, N, is coverget c The series N f ( is coverget Proof implies b becuse if there eists = lim f ( t dt, the lim f ( t dt = too b b The writte itegrls eist becuse decresig fuctios re itegrble o compct itervls dt is

21 Chpter V Etedig the defiite itegrl b c follows from the iequlit f (t f ( + o [ +, + ], which leds to b k f ( k f ( t dt Fill, c becuse from k k k f ( t dt f( + k it follows tht f ( t dt f ( k for ll b [, + ] } Remrks Betwee improper itegrls d series there re still sigifict differeces For emple, the covergece of f ( t dt does ot geerll impl lim f (t = (see problem 6 t b The otio of improper itegrl is sometimes used i more geerl sese, mel tht of "priciple vlue" (lso clled "Cuch's pricipl vlue", deoted s pv B defiitio, pv pv b f ( t dt lim f ( t dt, d c b f ( t dt lim f ( t dt f ( t dt c where c (, b is the poit roud where f is ubouded Of course, the coverget itegrls re lso coverget i the sese of the pricipl vlue, but the coverse implictio is geerll ot true (see problem 7 8

22 V Improper itegrls PROBLEMS V Show tht q t dt, where >, q > is coverget for q < d it is diverget for q Hit If q =, the d is diverget Otherwise q si Stud the covergece of the itegrls d Hit Use theorems 4 d 5 for b si d l d [q b q ] l q d l d si Show tht d is coverget but ot bsolutel coverget si Hit Becuse lim, the itegrl is improper ol t the upper limit We c ppl theorem 6 (specil form # III to φ ( = si, for λ = The itegrl is ot bsolutel coverget becuse for > we hve si si, d which is diverget si d d d 4 Estblish the covergece of (cos cos d, for λ (, Hit Appl theorem 7 (specil form # IV for φ ( = cos dt t t 5 Ale the covergece of the itegrls si si cos, sice 9

23 Chpter V Etedig the defiite itegrl I = where N * d, d J = d, ( Hit Use theorem 8 (specil form # V For I, lim, hece I is (bsolutel coverget For the (positive fuctio i J we hve lim, so J is diverget for > I the cse =, we hve lim, hece J is diverget 6 Show tht t cost dt is coverget eve if lim cos does't eist Is this situtio possible for positive fuctios isted of cos? Hit Use theorem 6 for φ ( = cos d λ =, sice t cost dt si si Accordig to theorem 4, the swer to the questio is egtive, ie positive fuctios which re itegrble o [, must hve ull limit t ifiit I fct, o the cotrr cse, whe lim f ( does't eist or is differet from ero, we hve i the metioed test, it would follow tht lim f (, hece tkig λ = d f ( t dt is diverget 7 Stud the pricipl vlues of the itegrls I = e t si tdt, J = t dt, where [] is the etire prt of,

24 V Improper itegrls dt K = cos tdt, d L = t Solutio I is (bsolutel coverget; J is diverget, but pvj = ; K is diverget i the sese of pv; L is diverget, but pvl = l 8 Stud the covergece of the itegrls I = d K = cos d, where N Hit lim e e d, J = si d, for N, hece pplig theorem 4, I is coverget J, J, K, K re diverget ccordig to the defiitio I J d K, for we m replce = t, d use theorem 6 9 Show tht the followig itegrls hve the specified vlues: I = e d! b J = e d! Hit Estblish the recurrece formul I = I b Replce = t i the previous itegrl Usig dequte improper itegrls, stud the covergece of the series:, R * ; b l, R ; c, R (l Hit Use theorem 9 I itegrl d (l b l d we c itegrte b prts I the we c chge l = t All these itegrls (d the correspodig series re coverget iff α >

25 V IMPROPER INTEGRALS WITH PARAMETERS We will recosider the topic of V i the cse of improper itegrls efiitio Let A R, I = [, b R, d f : A I R be such tht for ech A, the fuctio t f(, t is improperl itegrble o [, b The F : AR, epressed b b F( = f (, t dt ; f (, t dt ; f (, t dt ; etc is clled improper itegrl with prmeter Remrk Accordig to the defiitio of improper itegrl, F is defied s poit-wise limit of some defiite itegrls, ie F( p lim f (, t dt b More ectl, this mes tht for A d ε >, there eists δ(, ε > such tht for ll β (b δ, b, we hve f (, t dt F( M times we eed stroger covergece, like the uiform oe, which mes tht for ε >, there eists δ(ε > such tht for ll A d β (b δ, b, we hve the sme iequlit: f (, t dt F( I this cse we s tht the improper itegrl uiforml coverges to F, d we ote F( u b lim f (, t dt The followig lemm reduces the covergece of the itegrl to the covergece of some fuctio sequeces d series Lemm Let us cosider A R, I = [, b R, d f : A I R fuctio, such tht for ech A, the mp t f(, t is itegrble o ech compct from I The followig ssertios re equivlet: (i The improper itegrl b f (, t dt, with prmeter, is uiforml (poit-wise coverget o A to F ; (ii For rbitrr icresig sequece (β N for which β = d lim b, the fuctio sequece (F N, where F : A R hve the vlues F (= f (, t dt, is uiforml (poit-wise coverget o A to F

26 V Improper itegrls with prmeters (iii For rbitrr icresig sequece (β N such tht β = d lim b, the fuctio series u, of terms u : A R, where u ( = f (, t dt, is uiforml (poit-wise coverget o A to F The proof is routie d will be omitted, but we recommed to follow the scheme: (i (ii (iii 4 Theorem (Cuch's geerl test Let A R, I = [, b R, d f : A I R be such tht the mp t f(, t is itegrble o ech b compct from I, for rbitrr A The the improper itegrl f (, t dt with prmeter, is uiforml coverget o A iff for ever ε >, there eists δ(ε > such tht for rbitrr A d b', b" (b δ, b, we hve Proof If F( u b" b' lim b b' f (, t dt b" b' f (, t dt f (, t dt, the we evlute f (, t dt F( b" f (, t dt F( s we usull prove Cuch coditio Coversel, usig the bove lemm, we show tht the sequece (F N, where F ( = f (, t dt, β =, β < β +, d lim b, is uiforml Cuch o A I fct, for ε > we hve F ( F m ( = f (, t dt, wheever β, β m (b δ, b, ie m, > (δ N } m Usig this geerl test we obti more prcticl tests: 5 Theorem (Compriso test Let A, I d f be like i the bove theorem Let lso g : I R + be such tht:

27 Chpter V Etedig the defiite itegrl f(, t g(t for ll (, t A I b g ( t dt is coverget b The f (, t dt is uiforml coverget o A Proof I order to ppl the bove geerl test of uiform covergece we evlute b" b' b" b" f (, t dt f (, t dt g( t dt The lst itegrl c be b' mde rbitrril smll for b', b" i pproprite eighborhood of b, sice g is itegrble o [, b } 6 Remrk If compred to theorem,, we see tht the uiform boudedess reltive to, f(, t g(t, leds to the uiform covergece o A Cosequetl, prticulr tests similr to theorems 48 i V re vlid, if the hpothesis re uiforml stisfied reltive to A As i V, we re iterested i estblishig the rules of opertig with prmeters i improper itegrls 7 Theorem (Cotiuit of F Let f : A I R be cotiuous o A I, b where A R, d I = [, b R If the itegrl coverget o A, the F : A R, epressed b F( = cotiuous o A Proof Accordig to Lemm, F u b' lim F f (, t dt is uiforml b f (, t dt is O the other hd, F re cotiuous o A (see theorem i Cosequetl, F is cotiuous s uiform limit of cotiuous fuctios } 8 Theorem (erivbilit of F Let A R, I = [, b R, d f : A I R be such tht: f is cotiuous o A I f is cotiuous o A I b f (, t dt is poit-wise coverget o A to F : A R bf 4 (, t dt is uiforml coverget o A 4

28 V Improper itegrls with prmeters f The F is derivble o A, its derivtive is F'( = (, t dt, d F' is cotiuous o A Proof Let us ote F ( = sequece for which b = d lemm, F= lim F b b f (, t dt, where (b N is icresig lim b b Accordig to the previous poit-wise O the other side F is derivble s defiite itegrl with prmeter (see theorem 5,, d b F '( = f (, t dt Now, usig the sme lemm for uiforml coverget itegrls, we obti ll the climed properties of F } The opertio of itegrtio m be relied either i the proper sese (s i defiite itegrls, or i the improper sese 9 Theorem (The defiite itegrl reltive to the prmeter Let us cosider A = [α, β] R, I = [, b R, d f : A I R be such tht: f is cotiuous o A I b f (, t dt is uiforml coverget o A = [α, β] to F b The F is itegrble o [α, β] d F( d f (, t d dt Proof Let (b N be icresig sequece such tht b = d lim b b Accordig to Lemm, F u lim F, where F : [α, β] R re epressed b F ( = b f (, t dt O the other hd, ccordig to theorem, V, F re cotiuous fuctios, hece F is cotiuous too So, we deduce tht F is itegrble o [α, β], d F( d lim F ( d Now it remis to use theorem 7, V, i order to clculte b F ( d f (, t d dt, d to ppl lemm gi } 5

29 Chpter V Etedig the defiite itegrl Theorem (The improper itegrl reltive to the prmeter Let us cosider A = [α, β R, I = [, b R, d f : A I R be such tht: f is positive d cotiuous o A I b b f (, t dt is uiforml coverget to F: AR o compct from A f (, t d is uiforml coverget to G : I R o I 4 G ( t dt is coverget The F is improperl itegrble o [α, β, d F ( d = G ( t dt Proof Accordig to the previous theorem, for ech η [α, β, the fuctio b F is itegrble o [α, η], d F( d f (, t d dt Let us ote b φ : [α, β] [, b R the fuctio of vlues φ(η, t = f (, t d if t, G( t if t The third hpothesis of the theorem shows tht φ is cotiuous o the set [α, β] [, b O the other hd, if we ote b Φ: [α, β] R the fuctio Φ(η = b (,tdt, we obti Φ(η = F ( d for ll η [α, β Now, the problem reduces to etedig this reltio for η = β I fct, becuse f is positive, for ll η [α, β d t [, b we hve b f (, t d f (, t d, b ie φ(η, t G(t Sice G ( t dt is coverget, the compriso test shows tht b (,tdt is uiforml coverget to Φ Addig the fct tht φ is cotiuous, theorem 7 shows tht Φ is cotiuous o [α, β], hece there 6

30 eists lim Φ(η = Φ(β, ie Φ(β = 7 V Improper itegrls with prmeters F ( d Replcig Φ d φ b their vlues, we obti the climed formul } Remrks Theorems 9 d estblish the coditios whe we c chge the order of itegrtio, ie b b f (, t dtd f (, t ddt b The coditio f to be positive i theorem is essetil For emple, if t f : [, [, R is epressed b f(, t =, the f(, t ( t s well s f(, t t for ll (, t [, [,, hece f is itegrble o [, reltive to t, d lso reltive to B direct clcultio we fid F( = d G(t = Cosequetl, F d G re lso ( ( t itegrble o [,, but G ( t dt F( d Eceptig the coditio of beig positive, f stisfies ll coditios of theorem The itegrls with prmeter re useful i defiig ew fuctios The Euler's Γ d B fuctios re tpicl emples i this sese: efiitio The fuctio Γ : (, (, epressed b Γ( = t is clled Euler's gmm fuctio The fuctio B: (, (, (, of vlues B(, = t e t dt ( t dt is clled Euler's bet fuctio This defiitio mkes sese becuse: Propositio The itegrls of Γ d B re coverget Proof The itegrl which defies Γ is improper both t d Becuse t e t t for t [, ], d t is itegrble if >, it follows tht the itegrl of Γ is coverget t This itegrl is coverget t becuse t e t is itegrble o [, for ll N

31 Chpter V Etedig the defiite itegrl The itegrl which defies B is lso improper t d t, d, i dditio, it depeds o two prmeters The covergece of this itegrl follows from the iequlit t ( t [t + ( t ], which holds for t [, ], > d > (see the compriso test This iequlit m be verified b cosiderig two situtios: If t [/,, d >, the t, so tht i this cse t ( t ( t [t + ( t ]; b If t (, /], the ( t [/,, d sice > too, we hve ( t, d similr evlutio holds } 4 Theorem Fuctio Γ hs the followig properties: (i it is cove d idefiitel derivble fuctio; (ii Γ( + = Γ( t > ; (iii Γ( + =! for ever N, ie Γ geerlies the fctoril Proof (i It is es to see tht f(, t = t e t stisfies the coditios i theorem 8, hece t Γ'( = t e l tdt B repetig this rgumet we obti Γ (k ( = t 8 e t k l tdt for k N *, ie Γ is idefiitel derivble Its coveit follows from Γ"( > for ll > (ii Itegrtig b prts we obti we obti Γ(+ = t e t dt = lim t e t + t t e t dt = Γ( (iii Accordig to (ii, Γ( + = Γ( = ( Γ(, d Γ( = e t dt = 5 Theorem Fuctio B hs the properties: (i B(, = B(,, ie B is smmetric; (ii For (, (, (, we hve B(, = (iii It hs cotiuous prtil derivtives of order Proof (i Chgig t = θ, B(, becomes B(, v v (ii Replcig t = i B, we obti B(, = v ( v other hd, chgig t = ( + vu i Γ, it follows tht ( ( ; ( dv O the

32 Γ( = ( + v u 9 e V Improper itegrls with prmeters u( v Writig this reltio t + isted of, we hve du u( v Γ(+ u e du ( v Amplifig b v d itegrtig like i B, we obti u( v Γ(+B(, = v u e dudv Usig theorem we chge the order of the itegrls d we obti u uv Γ( + B(, = u e v e dvdu = u = u e u ( d = = Γ( u e u du = Γ( Γ( (iii This propert results form the similr propert of Γ, tkig ito ccout the bove reltio betwee Γ d B } e t 6 Remrkble itegrls Γ( = dt d t (lso clled Euler-Poisso itegrl I fct, B(, = Γ ( = replcig = si t e u du d, which turs out to be π, if ( The secod itegrl follows from Γ( b tkig t = u m b The biomil itegrl I = d, >, b >, p > m + > p ( b m be epressed b elemetr fuctios ol if p is iteger m is iteger (positive p m is iteger (positive

33 Chpter V Etedig the defiite itegrl I fct, otig b = u d k = I = k u m Aother chge of vribles, mel I = kv m m p p b p m ( u du u u ( v dv = k B(, we obti = v, leds to m m, p = m m p = k ( p This formul shows tht i geerl, I is epressed b Γ; i the metioed cses Γ reduces to fctorils, so I cotis ol elemetr fuctios m We recll tht i the cse whe is iteger, we mke the substitutio + b = t s, where s is the deomitor of the frctio p m Similrl, if p is iteger, the evlutio of the itegrl m be mde b the substitutio + b = t s

34 V Improper itegrls with prmeters PROBLEMS V Show tht F( = si t e t dt is coverget for [, d t F(=rctg Hit The itegrl is improper t ; the covergece is cosequece of si t the compriso test, if g(t=, t (see lso theorem 6, V B t theorem 8, F'( =, hece F( = rctg + C Tke = Clculte I(r = ( r cos r d, where r < Hit The substitutio t = tg i I'(r gives r cos I'(r = r cos r r r d 4 r ( t t ( t dt A where = > Brekig up ( t ( t t t where A = B =, we obti 4 dt I'(r = ( r t ( t Cosequetl, I(r = C, but I( =, hece I(r = too B, Show tht Φ( = si t dt (Poisso t e t si t dt rctg, d deduce tht t Hit Usig the result of problem, Φ( = F( = rctg = rctg Aother method cosists i itegrtig two times b prts i Φ'(, d obtiig Φ'( = Φ'(, wherefrom it follows tht Φ( = rctg + C

35 Chpter V Etedig the defiite itegrl For we deduce C = Fill, the Poisso's itegrl is Φ( b e e cos cosb 4 Clculte I = d, d J = d, where < < b b b b t Hit I = t b e dt d e ddt dt l t b b t si si t J = tdt d d dt si = (b, where d = is the Poisso's itegrl (see problem idepedetl of t > 5 Let f : (, ] (, ] R be fuctio of vlues Show tht dd ( ( wh these itegrls hve differet vlues Hit Theorem does ot work sice f chges its sig, d d 6 Use the fuctios bet d gmm to evlute the itegrls I = b J = p p e m q ( d, p, q, m > ; q d, p > -, q > Hit Chge the vrible m = t, d evlute I = m t p m b Replce q = t, d clculte J = q t p q ( t e q t dt dt p = B, q m m p = q q t f (, t ( t, d epli

36 CHAPTER VI LINE INTEGRAL We will geerlie the usul defiite itegrl i the sese tht isted of fuctios defied o [, b R we will cosider fuctios defied o segmet of some curve There re two kids of lie itegrls, depedig of the cosidered fuctio, which c be sclr or vector fuctio, but first of ll we must precise the termiolog cocerig curves (there re plet mterils i the literture VI CURVES We le the otio of curve i R, but ll the otios d properties c be obviousl trsposed i R p, p N \ {, }, i prticulr i R efiitio The set γ R is clled curve iff there eists [, b] R d fuctio φ : [, b] R such tht γ = φ ([, b] I this cse φ is clled prmeteritio of γ Tpes of curves The poits A = φ( d B = φ(b re clled edpoits of the curve γ ; if A = B, we s tht γ is closed We s tht γ is simple (without loops iff φ is ijective Curve γ is sid to be rectifible iff φ hs bouded vritio, ie there eists b V sup ( ti ( ti, i where δ = {t = < t < < t = b} is divisio of [, b] The umber b L = V is clled legth of γ We s γ is cotiuous (Lipschite, etc iff φ is so Let us ote φ(t = ((t,(t,(t for t [, b] If φ is differetible o [, b], d φ' is cotiuous d o-ull, we s tht γ is smooth curve This mes tht there eist cotiuous derivtives ', ' d ', d ' (t + ' (t + ' (t, t [, b] The vector t ( / (t, / (t, / (t is clled tget to γ, t M ((t,(t,(t For prcticl purposes, we frequetl del with cotiuous d piecewise smooth curves, ie curves for which there eists fiite umber of itermedite poits C k γ, k =,, where C k = φ(c k for some c k (, b, such tht φ is smooth o ech of [, c ], o [c k, c k+ ] for ll k =,,, d o [c, b], d φ is cotiuous o [, b] The imge of restrictio of φ to [c, d] [, b] is clled sub-rc of the curve γ, so γ is piece-wise smooth iff it cosists of fiite umber of smooth sub-rcs

37 Chpter VI Lie itegrl Remrks The clss of rectifible curves is ver importt sice it ivolves the otio of legth Geometricll spekig, the sum i ( t, i ( t i from the bove defiitio of the vritio b V, represets the legth of broke lie of vertices φ(t i Pssig to fier divisios of leds to loger broke lies, hece is rectifible iff the fmil of these iscribed broke lies hs u upper boud for the correspodig legths Without goig ito detils, we metio tht fuctio f :[, b] R hs bouded vritio if it hs oe of the followig properties: mooto, Lipschit propert, bouded derivtive, or it is primitive, ie f ( ( t dt, [, b] (for detils, icludig properties of the fuctios with bouded vritio, see [FG], [N--M], etc The bove defiitio of the rectifible curves is bsed o the followig reltio betwee bouded vritio d legth of curve: 4 Theorem (Jord Let = (, : [, b]r be prmeteritio of ple curve The curve is rectifible if d ol if the compoets, d of hve bouded vritio We omit the proof, but the reder m cosult the sme bibliogrph 5 Corollr If is smooth curve, the it is rectifible, d its legth is L ( t ( t dt b / A similr formul holds for curves i R d R Becuse ll the otios from bove re bsed o some prmeteritio, it is importt to kow how c we chge this prmeteritio, d wht hppes whe we chge it These problems re solved b cosiderig the followig otio of "equivlet" prmeteritios of smooth curve 6 efiitio The fuctios φ : [, b] R d ψ : [c, d] R re equivlet prmeteritios iff there eists diffeomorphism σ : [, b] [c, d] such tht σ'(t for ll t [, b], d φ = ψ σ I this cse we usull ote φ ψ, d we cll σ itermedite fuctio 7 Remrks (i Reltio from bove is rell equivlece I dditio, this equivlece is pproprite to prmeteritios of curve becuse equivlet fuctios hve ideticl imges Whe we re iterested i studig more geerl th smooth curves, the "itermedite" fuctio σ (i defiitio stisfies less restrictive coditios, s for emple, it c ol be topologicl homeomorphism / 4

38 VI Curves (ii Becuse σ : [, ] [, b] defied b σ(t = tb + (t, is emple of itermedite (eve icresig fuctio i defiitio, we c lws cosider the curves s imges of [, ] through cotiuous, smooth or other fuctios Aother useful prmeteritio is bsed o the fct tht the fuctio σ : [, b] [, L], defied b σ(t = t ' ( ' ( ' ( d stisfies the coditios of beig itermedite fuctio I this cse s = σ(t represets the legth of the sub-rc correspodig to [, t], d L is the legth of the whole rc γ If s is the prmeter o curve, we s tht the curve is give i the coicl form (iii From pure mthemticl poit of view curve is clss of equivlet fuctios I other words we must fid those properties of curve, which re ivrit uder the chge of prmeters More ectl, propert of curve is itrisic propert iff it does ot deped o prmeteritio i the clss of equivlet fuctios (the sese of the cosidered equivlece defies the tpe of propert: cotiuous, smooth, etc For emple, the properties of curve of beig closed, simple, cotiuous, Lipschite, d smooth re itrisic Similrl, the legth of curve should be itrisic propert, so tht the followig result is ver useful: 8 Propositio The propert of curve of beig rectifible d its legth do ot deped o prmeteritio Proof Beig mootoic, σ relies : correspodece betwee the divisios of [, b] d [c, d], such tht the vritio of the equivlet fuctios o correspodig divisios re equl It remis to recll tht the legth is obtied s supremum } The fct tht either σ' > or σ' < i defiitio llows us to distiguish two subclsses of prmeteritios which defie the oriettio of curve 9 Orietted curves To oriette curve mes to split the clss of equivlet prmeteritios ito two subclsses, which cosist of prmeteritios relted b icresig itermedite fuctios, d to choose which of these two clsses represet the direct oriettio (sese, d which is the coverse oe B covetio, the direct (positive sese o closed, simple d smooth curve i the Euclide ple is the ti-clockwise oe More geerll, the closed curves o orietted surfces i R re directl orietted if the positive orml vector leves the iterior o its left side whe ruig i the sese of the curve Altertivel, isted of cosiderig two seses o curve, we c cosider two orietted curves More ectl, if γ is orietted curve (ie the itermedite diffeomorphism i defiitio is lso 5

39 Chpter VI Lie itegrl icresig of prmeteritio φ : [, b] R, the the curve deoted γ of prmeteritio ψ : [, b] R defied b ψ(t = φ ( + b t is clled the opposite of γ Aother w of epressig the oriettio o curve is tht of defiig order o it More ectl, we s tht X = φ (t is "before" X = φ (t o γ iff t t o [, b] Usig this termiolog, we s tht A = φ ( is the first d B = φ (b is the lst poit of the curve If o cofusio is possible, we c ote γ = AB d γ = BA Cotrril to the divisio of curve ito sub-rcs, we c costruct curve b likig together two (or more curves with commo ed-poits efiitio Let γ i, i =, be two curves of prmeteritio φ i : [ i, b i ] R such tht φ (b = φ ( The curve γ, of prmeteritio φ : [, b + (b ] R, where ( t if t, b ( t is clled coctetio ( t b if t b, b ( b (uio of γ d γ, d it is oted b γ = γ γ Propositio The coctetio is ssocitive opertio with curves hvig commo ed-poits, but it is ot commuttive The proof is routie, d will be omitted If γ γ mkes sese, the the coctetio γ γ is possible, but geerll γ γ is ot Propositio The smooth curves hve tget vectors t ech M γ, cotiuousl depedig o M The directios of tget vectors do ot deped o prmeteritios I coicl prmeteritio, ech tget t = ('(s, '(s, '(s is uit vector Proof If fuctio φ : [, b] R, of vlues φ(t = ((t, (t, (t is prmeteritio of γ, the M M = ((t (t, (t (t, (t (t Sice φ is differetible, M M ('(t (t t, '(t (t t, '(t (t t, with equlit whe t t Cosequetl the directio of t is give b ('(t, '(t, '(t B chgig the prmeter, t = σ(θ, this vector multiplies b σ'(θ, hece it will keep up the directio For the coicl prmeteritio we hve Δ s = Δ + Δ + Δ, hece the legth of the tget vector is ' (s + ' (s + ' (s = } 6

40 VI Curves PROBLEMS VI Is the grph of fuctio f : [, b] R curve i R? Coversel, is curve i R grph of such fuctio? Hit Ech fuctio f geertes prmeteritio φ : [, b] R of the form φ(t = (t, f(t The circle is curve, but ot grph Show tht the coctetio of two smooth curves is cotiuous piecewise smooth curve, but ot ecessril smooth Hit Use defiitio 7 of coctetio Iterpret the grph of, where [, +], s coctetio of two smooth curves Let γ i, i =, be two curves of prmeteritio φ i : [ i, b i ] R with commo ed-poits, ie φ ( = φ ( d φ (b = φ (b Show tht both γ γ d γ γ mke sese d the re cotrril orieted closed curves 4 Fid the tget of ple curve implicitl give b F(, = I prticulr, tke the cse of the circle Hit If = (t, = (t is prmeteritio of the curve, from F((t, (t o [, b], we deduce df =, hece F' ' + F' ' = Cosequetl, we c tke t = ('(t, '(t = λ(f', F' 5 If the ple curve γ is implicitl defied b F(, =, we s tht M γ is criticl poit iff F' (M = F' (M = Stud the form of γ i the eighborhood of criticl poit ccordig to the sig of Δ = " " " F F F Emple = +, d M = (, Hit M is sttior poit of the fuctio = F(,, d γ is the itersectio of the ple O with the surfce of equtio = F(, I this istce F( + h, + k F" (, h + F" (, hk + F" (, k, hece Δ < leds to isolted poit of γ, Δ > correspods to ode (double poit, d Δ = is udecided (isolted poit I the emple, M is isolted for <, it is ode for > ; it is cusp for = 6 Fid the legth of the logrithmic spirl φ(t = (e t cos t, e t si t, e t, where t Solutio L = ' ' ' dt 7

41 Chpter VI Lie itegrl 7 Estblish the formul of the legth of ple curve which is implicitl defied i polr coordites, r = r(θ Use this formul i order to fid the legth of the crdioid r = ( + cos θ Hit Followig Fig VI, we hve Δs = (rδ θ + (Δr dr r d r r r s b Fig VI The legth of the crdioid (sketched i Fig VIb is L = r r' d cos d 4 cos d 8 8 Fid the legth of the curves defied b the followig equtios: si r, [, ]; b r si, [, ] Aswer (8 ; b 8 9 Fid the legth of the curve of equtio r, r [, ] r Hit Estblish formul similr to tht i the bove Problem 7 The legth of the curve is l 8

42 VI LINE INTEGRALS OF THE FIRST TYPE I this prgrph we cosider the lie itegrl of sclr fuctio Such itegrls occur i the evlutio of the mss, ceter of grvit, momet of ierti bout is, etc, of mteril curve with specified desit The costructio of the itegrl sums Let γ be smooth d orietted curve i R, of ed-poits A d B B divisio of γ we uderstd set δ ={M k γ : k =,,, } such tht M = A, M = B, d M k < M k+ i the order of γ, for ll k =,,, The orm of δ is δ = m M k M k k If γ k = M km k deotes the sub-rc of the ed-poits M k d M k+ o γ, we write Δs k for the legth of γ k, k =,,, O ech sub-rc γ k we choose poit P k betwee M k d M k+ i the order of γ The set S = {P k k : k =,,, } represets the so clled sstem of itermedite poits B P k M k+ B = M R M k s k M k+ Pk f A M k A = M Fig VI Now we cosider tht γ is etirel cotied i the domi o which the sclr fuctio f is defied (see Fig VI Uder these coditios, we c clculte S γ, f (δ, S = k f ( P k s k, 9

43 Chpter VI Lie itegrl which is clled itegrl sum of the first tpe of f o the curve γ, correspodig to the divisio δ, d to the sstem S of itermedite poits efiitio We s tht f is itegrble o the curve γ iff the bove itegrl sums hve (fiite limit whe the orm δ, d this limit is ot depedig o the sequece of divisios with this propert, d o the sstems of itermedite poits If this limit eists, we ote lim S γ, f (δ, S = 4 fds, d we cll it lie itegrl of the first tpe of f o the curve γ Remrk The bove defiitio of the lie itegrl mkes o use of prmeteritios, but cocrete computtio eeds prmeteritio i order to reduce the lie itegrl to usul Riem itegrl o R I fct, if φ : [, b] R is prmeteritio of γ, the to ech divisio δ of γ there correspods divisio d of [, b], defied b M k = φ(t k for ll k =,, Of course, d iff δ Similrl, to ech sstem S = {M k γ k : k =,,, } of itermedite poits of γ, there correspods sstem T = {θ k [t k, t k+ ] : k =,,, } of itermedite poits of [, b] The vlues f(p k m be epressed b (f φ(θ k, such tht k f S γ, f (δ, S = ( f ( k sk k tk ( ( k, ( k, ( k ' ( t ' ( t ' t k ( t dt Fill, usig the me theorem for the bove itegrls, we obti S γ, f (δ, S = f ( ' ( ' ( ' ( ( t t, ( k k k k k k which looks like itegrl sum of simple Riem itegrl Thus we re led to the followig ssertio: 4 Theorem Let γ be (simple smooth curve i R, d let f : R be cotiuous sclr fuctio The there eists the lie itegrl of f o γ, d for prmeteritio φ : [, b] R of γ we hve fds b = ( f ( t ' ( t dt I prticulr, the lie itegrl does ot deped o prmeteritio Proof Let us ote F(t = (f φ(t φ'(t, d let ( k k k k σ F (d, T = f ( ' ( ( t t k k

44 VI Lie itegrls of the first tpe be the Riem itegrl sum of F o [, b] Becuse γ is smooth, it follows tht F is cotiuous, hece there eists F( t dt lim σ F (d, T More b d ectl, for ever ε > there eists η > such tht for ever divisio d of [, b], for which d < η, we hve σ F (d, T b F ( t dt < (* O the other hd, f φ is uiforml cotiuous o the compct [, b], hece for ε > there eists η > such tht for ll t', t" [, b] for which t' t" < η, we hve (f φ(t' (f φ(t" <, where L is the L legth of γ If d is divisio of [, b] such tht d < η, the S γ, f (δ, S σ F (d, T = = f ( k ( f ( k '( k ( tk tk sk (** L k k Cosequetl, if d is divisio of [, b] for which d < η = mi {η, η }, the usig (* d (** we obti S γ, f (δ, S b F ( t dt S γ, f (δ, S σ F (d, T + σ F (d, T b F ( t dt < ε, ie b F ( t dt is the limit of the itegrl sum of f o γ The lst sttemet of the theorem follows from the fct tht the itegrl sums S γ, f (δ, S do ot deped o the prmeteritio, d the prmeteritio used i the costructio of F is rbitrr } The geerl properties of the lie itegrl of the first tpe re summried i the followig : 5 Theorem (i The lie itegrl of the first tpe is lier fuctiol, ie for smooth curve γ, cotiuous f, g, d λ, μ R, we hve ( f g ds fds gds (ii The lie itegrl is dditive reltive to the rc, ie fds = fds + fds, wheever γ = γ γ (iii The lie itegrl of the first order does ot deped o the oriettio o the curve, ie fds = fds The proof is directl bsed o defiitio, d will be omitted 4

45 Chpter VI Lie itegrl PROBLEMS VI Clculte ( + + ds, where γ (spirl hs the prmeteritio φ : [, π] R, φ(t = (cos t, si t, t Aswer π Evlute the itegrl ( ds, where is the curve of equtio ( (, Hit Recogie the lemiscte i polr coordites r the prmeteritio cos cos, cos si, 4 4 The swer is cos, d use Clculte the mss of the ellipse of semi-es d b, which hs the lier desit equl to the distce of the curret poit up to the is Hit The recommeded prmeteritio is give b φ : [, π] R, where φ(t = (cos t, bsi t We must clculte where e = b ds = b b + rcsi e, e is the e-cetricit of the ellipse 4 etermie the ceter of grvit of hlf-rc of the homogeeous ccloid = (t si t, = ( cos t, where t [, π] Hit G = M ρ(, sds, G = M ρ(, sds, where M is the mss of the wire I this cse G = G = 4 5 Fid the momet of ierti, bout the is of the first loop of the homogeeous spirl = cos t, = si t, = bt Hit I = ( + ρ(,, ds = π b 4

46 VI Lie itegrls of the first tpe 6 A mss M is uiforml distributed log the circle + = i the ple = Fid the force with which this mss cts o mss m, locted t the poit V(,, b Mm Hit Geerll spekig, F k r I the prticulr cse F = (,, F, r where F = km ( (,, t kmmb ds r ( b 7 Let be rc of the stroid i the first qudrt, whose locl desit equls the cube of the distce to the origi Fid the force of ttrctio eerted b o the uit mss plced t the origi Hit A prmeteritio of the stroid is cos t, si t Up to costt k, which depeds o the chose sstem of uits, the compoets of the force hve the epressios: F k ds 4 k = k sit cos t dt = ; 5 F k ds = si 4 k k t cost dt = 5 8 Show tht if f is cotiuous o the smooth curve γ, of legth L, the there eists M * γ such tht the me vlue formul holds fds = Lf(M * Hit Usig prmeteritio of γ, we reduce the problem to the me vlue formul for Riem itegrl / 9 Show tht if f is cotiuous o the smooth curve γ, the fds f ds Hit Use theorem 4 4

47 VI LINE INTEGRALS OF THE SECON TYPE The mi object of this prgrph will be the lie itegrl of vector fuctio log curve i R The most sigifict phsicl qutit of this tpe is the work of force The costructio of the itegrl sums Let γ R be smooth orietted curve, d let F : R be vector fuctio We suppose tht γ, d tht F hs the compoets P,Q, R : R, ie for ever (,,, we hve F (,, = (P(,,, Q(,,, R(,, Altertivel, usig the coicl bse { i, j, k } of R (see Fig VI, we obti r = i + j + k d F = Pi + Q j + R k B T k M k+ F i k j M k A Fig VI If δ = {M k γ : k =,, } is divisio of γ, we ote r k for the positio vector of M k For ech sstem of itermedite poits S = {T k = (ξ k, η k, ζ k M km k : k =,, } we costruct the itegrl sum S, F ( δ, S = F( Tk, r k 44 k = [P(ξ k, η k, ζ k ( k+ k + Q(ξ k, η k, ζ k ( k+ k + R(ξ k, η k, ζ k ( k+ k ] k where <, > is the Euclide sclr product o R These sums re clled r itegrl sums of the secod tpe of F log the curve γ k

48 VI Lie itegrls of the secod tpe efiitio We s tht F is itegrble o γ iff the itegrl sums of the secod tpe hve (fiite limit whe the orm of δ teds to ero, d this limit is idepedet of the sequece of divisio which hve δ, d of the sstems of itermedite poits I this cse we ote the limit b lim S, F ( δ, S = < F, d r > = F d r = Pd + Qd + Rd d we cll it lie itegrl of the secod tpe of F o γ Remrk The mi problem is to show tht such itegrls re lso idepedet of the prmeteritio of γ, d to clculte them usig prmeteritios We will solve this problem b reducig the itegrl of the secod tpe to itegrl of the first tpe, which is kow how to be hdled I order to fid the correspodig sclr fuctio, we modif the form of the itegrl sums b usig prmeteritio φ : [, b] R of γ I fct, if φ (t = ((t, (t, (t, the ccordig to Lgrge's theorem, o ech [t k, t k+ ] we hve (t k+ (t k = '(θ k (t k+ t k (t k+ (t k = '(θ k (t k+ t k (t k+ (t k = '(θ k (t k+ t k, where θ k, θ k, θ k (t k, t k+ Cosequetl, S (δ, S becomes k, F [P(φ(θ k '(θ k + Q(φ(θ k '(θ k + R(φ(θ k '(θ k ](t k+ t k, (* where φ(θ k = P k, k =,,, re the itermedite poits of δ r Let us ote the uit tget vector t curret poit of γ b C r More ectl, if M = φ(θ, θ [, b], the '( i '( j '( k C( M ' ( ' ( ' ( Let us cosider the sclr fuctio f = < F, >, which hs the itegrl sums of the first tpe (see remrk i, F (f φ( θ k r '( k (t k+ t k (** S (δ, S = k B comprig the itegrl sums of F d f, we turll clim tht the lie itegrl of the secod order of F reduces to the lie itegrl of the first order of f I fct, this reltio is estblished b the followig 4 Theorem Uder the bove ottios, if F is cotiuous o γ, the F is itegrble o γ, d we hve F d r = f ds 45