Integration. Table of contents

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1 Itegrtio Tle of cotets. Defiitio of Riem Itegrtio Prtitio of itervls Upper/lower Riem sum Upper/lower Riem itegrls Riem itegrtio Criteri Properties Checkig itegrility Properties of Riem itegrls The Fumetl Theorem of Clculus The theorems Applictios of FTC Improper Riem Itegrtio Prtitio of itervls.. Defiitio of Riem Itegrtio Defiitio. (Prtitio) Let, R with <. A prtitio of the itervl, is the set of poits P = 0,,, } such tht = 0 < < < =. () A refiemet of prtitio P is prtitio Q P. I this cse we sy Q is fier th P. The most useful property of prtitios is the followig. Lemm. Let P, Q e prtitios. The P Q is fier th oth P Q. Proof. This is ovious sice prtitios re efie s sets... Upper/lower Riem sum. Defiitio 3. (upper/lower Riem sum) Let, R with <, let P = 0,, } e prtitio. Suppose f:, R is oue. i. The upper Riem sum of f over P is the umer U(f,P) 8 M j (f) ( j j ) ()

2 Itegrtio where M j (f) = j, j ii. The lower Riem sum of f over P is the umer f() (3) L(f,P) 8 m j (f) ( j j ) (4) where m j (f) = if j, j f(). (5) Remrk 4. Note tht sice we ssume f to e oue, m j (f) M j (f) re rel umers. Sice m j (f) M j (f), we clerly hve Lemm 5. U(f,P) L(f,P)..3. Upper/lower Riem itegrls. Defiitio 6. (Upper lower itegrl) We efie the upper itegrl U(f) of f over, y U(f): =if U(f,P): P is prtitio of, }. (6) We efie the lower itegrl L(f) of f over, y L(f) 8 L(f,P): P is prtitio of, }. (7) Theorem 7. Upper lower itegrls eist for every oue fuctio f. Proof. All we ee to show is tht U(f,P) is oue from elow L(f, P) is oue from ove. Tke y oue fuctio f. There is M > 0 such tht f M, tht is M f M for ll,. Now y efiitio U(f,P) = i= f() ( l l ) M ( l l ) = M ( ). (8) i, i i= Similrly we c prove L(f,P) M ( ). Theorem 8. U(f) L(f).

3 3 Nïvely, we oly ee to tke if of left/right sie of U(f,P) L(f,P). However this is wrog. If we imgie the process of tkig if s fiig P such tht U(f, P ) U(f), the the fct tht it s the sme P o oth sies les to U(f) = lim U(f,P ) lim L(f,P ) (9) which, eve if it eists, is usully ot L(f) = if P L(f, P). A relte issue is, f() g()if f() g(), for emple tke f = g. The uless f is costt, we wo t hve if f() g() = f(). To prove Theorem 8, the followig Lemm is eee. Lemm 9. Let P Q e prtitios. The U(f, Q) U(f,P) L(f, Q) L(f,P). Proof. We prove the first clim. The proof for the seco is similr. Let Q= 0,,, }. The s P Q, we hve P = 0,, k } with k, 0 =0, k =. Now cosier f() ( i i ) = i, i Therefore we hve k U(f,P) = i= k i= = l= f() i, i + f() i, i + + f() i, i f() = i, i + i, i + + i, i + i j= i + f() i, i i j= i + l, l j, j ( i i ) ( i i ) ( i + i ) ( i i ) f() ( i i ) f() ( i i ) f() ( i + i ). ( j j ). (0) f() ( j j ) j, j ( l l )= U(f, Q). () f() Proof. (of Theorem 8) We first show tht U(f,P) L(f,Q) for y prtitios P, Q. By Lemms 5 9 we hve U(f,P) U(f,P Q) L(f,P Q) L(f, Q). ()

4 4 Itegrtio Now we tke ifimum over P : U(f) = Filly we tke reme over Q: if P is prtitio U(f,P) ifl(f, Q) =L(f, Q) (3) P L(f)= L(f, Q) U(f). (4) Q Thus es the proof..4. Riem itegrtio. Defiitio 0. A oue fuctio o, is si to e Riem itegrle if L(f) = U(f). I this cse we eote this commo vlue y f(). I the followig emples we set, R, <. Emple. The costt fuctio f() =c is itegrle with f() =c( ). Proof. Let P =, }. The Therefore f() = if f()= c. (5),, U(f,P) = L(f,P) = c( ). (6) As L(f,P) L(f) U(f) U(f,P) (7) we hve L(f) = U(f) =c( ) the coclusios follow. Emple. The fuctio f() = is itegrle with f() =. Proof. This time we cot epect to fi P such tht U(f,P)=L(f,P). But we c try to fi P such tht U(f,P ) L(f,P ) 0. } Let P 8, +,,+ ( ),, tht is we hve i = +i for i= 0,,,,. (8) f()= i=+i i, i ; if f() = i =+(i ) i, i. (9)

5 5 Thus Similrly These ow give U(f,P ) = +i ) i= ( ( ) = ( ) + L(f,P ) = i= ( i= i ) ( ) ( +) = ( ) + ( ) ( +) = ( ) +. (0) ( = ( ) + +(i ) ) ( ) ( ). () ( ) + ( ) ( ) L(f) U(f) ( ) + Tkig limit, y Compriso Theorem we hve Therefore ( ) + the coclusios follow. ( ) L(f)= U(f) =( ) + L(f) U(f) ( ) + ( ) = ( ) ( + ). () ( ). (3) Remrk 3. Note tht similr to the proof of Theorem 8, the fct tht it s the sme P i mkes it icorrect to coclue irectly if P (4) U(f,P)+ L(f,P)= (5) U(f,P)+ L(f,P) =. (6) P rtiol Emple 4. Cosier the Dirichlet fuctio: f() =, the f() is ot itegrle. 0 irrtiol Proof. Let P e ritrry prtitio. We kow tht i every i, i, there is t lest oe rtiol umer oe irrtiol umer. Tht is i, i f =, if i, i f =0. Cosequetly U(f,P) =, L(f,P) =0 (7) for every prtitio. Therefore U(f) = 0 =L(f), tht is f is ot itegrle.

6 6 Itegrtio.. Checkig itegrility.. Criteri Properties Theorem 5. A oue fuctio o, is itegrle if oly if for ech ε > 0, there eists prtitio P of, such tht U(f,P) L(f,P)< ε. (8) More prcticlly, if we c fi prtitios P such tht U(f,P ) L(f,P ) 0, the f is itegrle f() = lim U(f,P ) = lim L(f,P ). (9) Proof. Oly if. Let f e itegrle. The U(f)=L(f). By efiitio there re prtitios P P such tht U(f,P )< U(f) +ε/; L(f,P ) >L(f) ε/. (30) Now set P = P P. We hve, y Lemm 9, U(f,P) U(f,P )< U(f) +ε/; L(f,P) L(f,P ) > L(f) ε/. (3) As U(f) =L(f), U(f,P) L(f,P), it follows tht If. Tke ε = /. The there is prtitio P such tht U(f,P) L(f,P) <ε. (3) U(f,P ) L(f,P ) <ε</. (33) As U(f,P ) U(f) L(f) L(f,P ), we hve 0 U(f) L(f) </ for y N. This c hol oly whe U(f) =L(f). Proof of the more prcticlly prt. Assume tht there re prtitios P such tht U(f, P ) L(f,P ) 0. First we show tht f is itegrle. For y ε > 0, there is N N such tht U(f, P N+ ) L(f,P N+ ) <ε. Therefore f is itegrle. Now we show tht f() = lim U(f, P ) = lim L(f, P ). Note tht the eistece of the two limits is ot give, tht is we ee to show this eistece. Cosier the prtitio Q =P P. The we hve L(f, Q ) L(f, Q ) L(f, Q ) U(f, Q ); (34) U(f, Q ) U(f, Q ) U(f, Q ) L(f, Q ). (35) Thus lim U(f, Q ) lim L(f, Q ) eist. Sice we hve, y Squeeze Theorem U(f,P ) L(f,P ) U(f, Q ) L(f, Q ) (36) U(f, Q ) L(f, Q ) 0 (37)

7 7 which, together with the efiitio of U(f), L(f), gives us f() =U(f) =L(f) = lim U(f, Q ) = lim L(f, Q ). (38) Filly we show tht lim U(f, P ) lim L(f, P ) eist equl the itegrl. Write U(f) U(f,P ) L(f, Q ) +U(f,P ) L(f,P ) (39) Tkig limits o oth sies pply Squeeze Theorem gives the esire result. As we hve see i Emple, estlishig itegrility through efiitio is surprisigly tricky, eve for very simple fuctios. Also ote tht wht we essetilly i i Emple is just provig specil cse of Theorem 5. Theorem 6. If f is cotiuous over,, the f is itegrle. Proof. (Not require) Sice f is cotiuous, f is uiformly cotiuous. For y ε > 0, there is δ >0 such tht wheever y <δ, we hve f() f(y) <ε/( ). } Now tke N such tht < δ, let P =, +, +,,. The we hve It follows tht Thus we hve So f is itegrle. f() f(y) <ε/( ), y + j, +(j +). (40) M j (f) m j (f) <ε/( ). (4) U(f) L(f) U(f,P) L(f,P) < Whe f is ot cotiuous, it my or my ot e itegrle. <0 Emple 7. Let f =. Prove tht f is itegrle over,. 0 0 ε ( ) = ε. (4) Proof. Thks to Theorem 5, ll we ee to show is tht, for every ε, there eists prtitio P = 0,, } such tht U(f,P) L(f,P) <ε. } Tke N such tht / < ε. Set P =,, 0,. The we hve Thus O the other h, if f() =, if, / f() =0, if /,0 f()= 0. (43) 0, (( L(f,P) = ) ) ( ( ( ) +0 0 )) + 0 ( 0) = ; (44) f() =,, / f() =, /,0 f()= 0. (45) 0,

8 8 Itegrtio Thus (( U(f,P) = ) ) ( ( ( ) + 0 )) + 0 ( 0) =. (46) Now we hve U(f,P) L(f,P)= < ε (47) itegrility follows. Remrk 8. Note tht oce f is uoue, it is ever itegrle ccorig to Defiitio 0, ecuse the reme of f over t lest oe of the suitervl woul e, cusig U(f,P) L(f, P) = for every prtitio P. Similrly, Defiitio 0 oes ot cover the cse whe the itervl, is of ifiite size (tht is = or = or oth). We will see lter tht there is turl etesio (improper itegrls) of itegrility to these situtios. Eercise. Let f, g e itegrle o,. Prove the followig.. If f g, the f g.. f f... Properties of Riem itegrls. Theorem 9. (Arithmetics) Let f g e itegrle fuctios o, let c e rel umer. The ) c f is itegrle (c f)() =c f(). ) f + g is itegrle (f + g)() = f() + g(). c) f g is itegrle. ) If g is oue elow y positive costt, the f/g is itegrle. Proof. We prove ), c) leve ) s eercise (ote tht you ee to cosier the cses c>,<,=0). ). We try to pply Theorem 5. For y ε>0, sice f,g re itegrle, let P, Q e prtitios such tht U(f,P) L(f,P) <ε, U(g, Q) L(g, Q) < ε. (48) Now cosier the prtitio P Q. We hve U(f,P Q) L(f,P Q)< ε/, U(g, P Q) L(g, P Q) < ε/. (49) Sice we hve L(f, P Q) + L(g, P Q) L(f + g, P Q) U(f + g, P Q) U(f, P Q) + U(g, P Q) We coclue tht U(f + g, P Q) L(f + g, P Q) <ε. (50)

9 9 c). Sice f, g re oue, there is M > 0 such tht f, g M. Now for y ε > 0, let P e prtitio such tht U(f, P ) L(f, P ) < ε M let P e prtitio such tht U(g,P ) L(g,P ) < ε M. Now set P =P P = 0 =,, = }. We hve U(f,P) L(f,P),U(g, P) L(g,P) < ε M. (5) Let y, z e two poits i j, j. We hve f(y) g(y) f(z) g(z) f(y) (g(y) g(z)) + g(z) (f(y) f(z)) As this hols for every pir y,z, we coclue Thus M g(y) g(z) + f(y) f(z) M ( g if g)+ ( f if f) = M M j (g) m j (g)+m M j (f) m j (f). (5) M j (f g) m j (f g) M M j (g) m j (g) +M M j (f) m j (f). (53) U(f g, P) L(f g, P) = M j (f g) m j (f g) j j M M j (f) m j (f) j j + M j (g) m j (g) j j Itegrility ow follows from Theorem 5. M U(f,P) L(f,P) U(g,P) L(g, P)} < ε. (54) We will ot spe time ivetig tricks to prove ) here. But otice tht ll the ove clims ecome trivil to prove i light of Remrk 8. For emple, if f, g re itegrle, the their sets of iscotiuity A f,a g hve mesure 0. But the clerly the set of iscotiuity for f + g, f g, f/g re ll just susets of A f A g so they ll must hve mesure 0 too. Remrk 0. Note tht there is o cler reltio etwee ( f() g() )( ) f() g(). This is wht mkes my prolems i sciece egieerig hr to stuy mthemticlly. The most fmous mog these prolems is turulece i flui mechics. Theorem. (Composite fuctio) Let f e itegrle over, g e cotiuous. The g f is itegrle of,. Proof. The proof requires uiform cotiuity is omitte.. Or more precisely, etwee the scle-free qutities f() g() f() g().

10 0 Itegrtio Eercise. Try to prove this theorem with wht we hve lere so fr. Wht ifficulty rises? C you esig slightly stroger otio of cotiuity to overcome this ifficulty? Compre your stroger cotiuity with the efiitio of uiform cotiuity (c e fou i y lysis tetook or just wiki uiform cotiuity ). Eercise 3. Fi f:, c, g:c, R oth itegrle, ut g f:, R is ot itegrle. Theorem. Let f e oue fuctio o,. Let c (,). The f is itegrle o, if oly if f is itegrle o oth, c c,. Proof. Oly if. If U(f,P) L(f,P) <ε (55) the we c refie ( c) to get P = P Q. U(f,P ) +U(f, Q ) <L(f,P )+ L(f, Q ) + ε (56) which is eough sice we hve U(f,P ) L(f,P ). This prt is esy s y prtitios P, Q of,c, c, c e comie ito prtitio of,. Applictio of Theorem 5 the yiels the esire result. If. This prt is esy s prtitio P c e refie to prtitios P, Q. Emple 3. f() = si 0 0 =0 is itegrle o every fiite itervl,. Proof. By Theorem, we oly ee to prove the itegrility of f over 0, for every >0,0 for every <0. We prove the first cse here, the proof for the seco cse is similr. Thks to Theorem 5, ll we ee to show is tht, for every ε, there eists prtitio P = 0 =0,, =} such tht U(f,P) L(f,P)<ε. Let δ = ε. Sice f is cotiuous o δ,, it is itegrle there. Thus there is prtitio P = δ,,} such tht Now set P = 0} P. The we hve U(f,P) = Thus U(f,P ) L(f,P )< ε. (57) 0,δ f δ +U(f,P ) =δ + U(f,P ); (58) L(f,P) = if f δ + L(f,P ) = L(f,P ). (59) 0,δ U(f,P) L(f,P) =δ + U(f,P ) L(f,P ) <ε. (60) Thus es the proof.

11 3. The Fumetl Theorem of Clculus As we hve see i Emple, clcultig f() y efiitio is very complicte eve for simple fuctios like f() =. Fortutely, the followig theorems mkes this tsk much esier through coectig itegrtio to (the much simpler) ifferettio. Before presetig it, let s suffer through the clcultio of oe more itegrl so tht we c pprecite the theorem etter. Emple 4. Use efiitio to clculte cos. First sice cos is oue cotiuous o,, it is itegrle. For N let P = 0 =,, =} with j j =. The we clculte cos j = = = = cos + j ( ) si si cos + j ( ) si + si si ( j + ) ( ) j ( ) si + ( ) si ( + ) ( si + Tkig we coclue tht (recll tht si s 0) lim ). (6) cos j = si si. (6) Now cosier ritrry prtitio Q. Set P = Q P. The o ech suitervl isie j, j we hve (cos) cos j, if(cos) cos j (63) which les to Tkig limit we coclue U(cos, Q) U(cos, P) L(cos, Q) L(cos, P) cos j cos j + ( ) (64) ( ). (65) U(cos, Q) si si L(cos, Q). (66) As this hols for ll prtitios Q, we hve U(cos) si si L(cos). Filly ecuse cos is itegrle, we must hve cos =U(cos)=L(cos)=si si. Remrk 5. I the ove rgumet we use the fct tht cos is ifferetile. It c e prove tht i geerl, if f() is itegrle over,, the U(f,P ) } f() whe. Here P =, +,, + ( ), s i the ove emple. More geerlly, y prtitio with m j j j 0 woul o.

12 Itegrtio 3.. The theorems. Defiitio 6. Let f e rel fuctio o itervl I. A fuctio F o I is clle tierivtive of f o I if F ()=f() for ll I. F is lso clle iefiite itegrl of f, eote y f(). Lemm 7. If F is tierivtive so is F + C. A ll tierivtives re of this form. Proof. If F is tierivtive the y efiitio F () = f(). This mes (F +C) = f() therefore F +C is lso tierivtive. O the other h, if G is lso tierivtive, the G ()= f()=f () (G F) =0 G F is costt. Cll this costt C we rech G() =F()+ C. Theorem 8. (FTC st Versio) Let f e itegrle o,. If F is cotiuous o, is tierivtive of f o (, ), the f() =F() F(). (67) Remrk 9. Note tht if F is tierivtive of f o (,) the F is ifferetile o (,) therefore cotiuous o (,). So wht the coitio F is cotiuous o, relly is out re cotiuity t the e poits,. Proof. Tke y ε > 0. As f is itegrle there is prtitio P = 0,, } such tht U(f,P) L(f,P)< ε which les to f() ε<l(f,p) U(f,P) < f() +ε Now sice F is cotiuous o, F = f o (,), y me vlue theorem we hve F( i ) F( i )= f(ξ i ) ( i i ) (68) for ξ i ( i, i ). Recllig the efiitios of U(f,P),L(f,P) we hve L(f,P) i= F( i ) F( i ) U(f,P) (69) which les to f() ε F() F() f() +ε. (70) This hols for y ε > 0, cosequetly we must hve f() =F() F(). Remrk 30. The ssumptios re ll ecessry. I prticulr we hve to ssume f to e itegrle sice otherwise f() is ot efie. It c iee hppe tht f()=f () is ot itegrle. For emple F() = si f() = si cos which is ot oue therefore is ot itegrle. Emple 3. We compute cos gi. Sice cos is oue cotiuous o, it is itegrle. We kow tht si is oe tierivtive of cos. As si is cotiuous o, we c pply FTC st versio to get cos=si si. (7)

13 3 Theorem 3. (FTC Versio) Let f e itegrle o,. The G() 8 f(t) t (7) is cotiuous o,. Furthermore, if f is cotiuous t poit 0 (,), the G is ifferetile t 0 G ( 0 ) = f( 0 ). (73) Proof. First show G cotiuous. Sice f is itegrle, it is oue. Let M > 0 e such tht f() M for ll,. Now set 0,. For y ε >0, tke δ = ε M. We hve, for ll 0 < 0 +δ, G() G( 0 ) = 0 f(t) t M ( 0) <ε. (74) Similrly we c show tht the sme hols for 0 δ < 0. Thus for ll 0 < δ we hve G() G( 0 ) <ε tht is G is cotiuous t 0. Now we prove G ( 0 ) = f( 0 ) whe f is cotiuous t 0. For y ε > 0, sice f is cotiuous t 0 there is δ >0 such tht whe 0 <δ, f() f( 0 ) <ε. Now for such we hve either > 0, i this cse G() G( 0) f( 0 ) 0 = (f(t) f( 0 ))t 0 < ε, (75) 0 or < 0, i this cse G() G( 0) f( 0 ) 0 = 0 Therefore the proof es. 0 (f(t) f(0 )) t <ε, (76) G() G( lim 0 ) = f( 0 ) (77) 0 0 Defiitio 33. Whe, efie f() = f(). (78) Eercise 4. Prove tht the ove efiitio ecessrily les to f() = 0 (79) for y oue fuctio f(). Eercise 5. Prove tht this efiitio is cosistet with ll the properties of itegrtio we erive so fr. Tht is these properties cotiue to hol for y,, o mtter >, <, or =. Remrk 34. Ielly oe woul like to hve theorem sttig f() itegrle G() 8 f(t) t ifferetile G ()= f(). (80)

14 4 Itegrtio 0 <0 0 < 0 However this is ot true. A emple is f() =, for which G() = which is 0 0 ot ifferetile t =0. It my lso hppe tht f(t) t is ifferetile t some poit 0 ut the erivtive is ot 0 0 f( 0 ). For emple let f() =. Set =. The =0 for ll ut G (0) = 0 f(0)=. 3.. Applictios of FTC. G()= f(t) t=0 (8) Theorem 35. (Itegrtio y prts) If u, v re cotiuous o, ifferetile o (, ), if u,v re itegrle o,, the u()v () =u() v() u() v() u ()v(). (8) Proof. Let F =u v. The F =u v +u v. Sice u,v re cotiuouso,, they re lso itegrle o,. Together with itegrility of u, v we coclue F is itegrle (Theorem 9) o,. Applictio of the first versio of FTC gives the esire result. Eercise 6. Prove the itegrtio y prts formul usig efiitio of Riem itegrl oly. Emple 36. Clculte 0 e. (83) Solutio. We set v =e, u=. The we hve 0 e = u v 0 = u()v() u(0)v(0) = e e 0 = 0 v u = e e P =0 =. (84) Emple 37. Tylor epsio with itegrl remier.. Clerly this couteremple is very rtificil. There is reso ehi this. It turs out tht, if f() is itegrle over,, the the erivtive of f(t) t must coicie with f() for lmost ll (, ). This is the so-clle Leesgue s Differetitio Theorem.

15 5 We c oti Tylor epsio usig itegrtio y prts. f() f() = f (t) t = t f (t) tf (t)t = f () ( ) +f () f () tf (t)t = f () ( ) + ( t) f (t) t = f ( t) () ( ) f (t) t } = f () ( ) f t= ( t) ( t) (t) + f t= (t) t = f () ( ) + f () ( ) ( t) + f (t)t = = f () ( ) + f () ( ) + f () ( ) ( t) + 3 f (4) (t)t 3! 6 = f (m) () ( ) m ( t) + f (+) (t) t. (85) m! m=0! ( t) The remier f (+) (t) t is clle itegrl remier. Oe c show tht if! f (+) (t) is cotiuous, the there is ξ (, ) such tht ( t) f (+) (t) t= f(+) (ξ) ( ) + (86)! ( +)! which is ectly the Lgrge remier. The isvtge of the Lgrge remier is tht. We hve o kowlege of where ξ ectly is;. The epeece of ξ o my e rough. For emple, we c ifferetite the itegrl remier ut ot the Lgrge remier (ue to ξ() my ot e ifferetile). Therefore i lysis oe shoul try to use itegrl remier wheever possile. Theorem 38. (Chge of vriles) Let u e cotiuous o,, ifferetile o (, ) ssume u is itegrle o,. If f is cotiuous o I 8 u(, ), the f(u(t)) u u() (t)t = f(). (87) Remrk 39. I the cse u()< u(), the itegrl is uerstoo s u() u() f() = f(). (88) u() u() u() Proof. We otice tht, if we efie F() = f(t) t, the F () = f() it follows from u() FTC Versio tht u() f(t) t=f(u()) F(u()); (89) u()

16 6 Itegrtio O the other h, if we set the y Chi rule G(t) 8 F(u(t)) (90) G (t) = t F(u(t))=F (u(t))u (t)= f(u(t)) u (t). (9) Note tht the lst equlity is result of FTC Versio oly hols ecuse f is cotiuous t every u(t). Net we check tht f(u(t)) u (t) is itegrle: f(), u(t) cotiuous f(u(t)) cotiuous f(u(t)) itegrle f(u(t))u (t) itegrle thks to Theorem 9 sice u (t) is itegrle. Filly pplyig FTC Versio to G we hve f(u(t))u (t) t=g() G() =F(u()) F(u()). (9) the proof es. Note tht i this lst step we ee G to e cotiuous, which follows from the cotiuity of f of u. Eercise 7. Prove the chge of vriles formul usig efiitio of Riem itegrtio oly. Remrk 40. Note tht we o t ee u to e oe-to-oe! 3 Emple 4. Clculte Solutio. Set y = si. We see tht 0,π/4 is mppe to si 0, si (π/4) = Now clculte π/4 0 cos 0 π/4 = cos cos π/4 cos = 0 si / y = 0 y = / y y y y + 0 π/4 cos. (93) 0,. 0 y= / = l y l y + y=0 = ( ) y + l y= / y y=0 ( = l + ). (94) 3. The reso for this is quite eep. I fct, i higher imesios, we o ee the chge of vrile fuctio to e oeto-oe. To fully uerst this issue, check out egree theory.

17 7 4. Improper Riem Itegrtio Recll tht Defiitio 0 cot el with uoue fuctios or uoue itervls. However for some fuctios there is resole wy to efie itegrls. Defiitio 4. Let < where, R, }. Let f:(,) R. The f is si to e improperly itegrle o (,) if oly if ) f is loclly itegrle: For y c, R, < c < <, f is itegrle o c,, ( ) ) lim c + lim c f() eists is fiite. Cll this limit the improper Riem itegrl of f over (, ), eote it y f(). (95) Remrk 43. The est wy to uerst this efiitio is through FTC. Fi c (, ). For y t (, ), efie t F(t)= f(). (96) c The we see tht F(t) is cotiuous fuctio o (, ). The if the left limit (right limit) lim t F(t) (lim t +F(t)) eists, it is resole to ete the omi of F(t) to iclue the e poit(s) y settig F() 8 ( lim F(t) F() 8 t ) lim F(t). (97) t + We stte without proof the followig theorems which gurtees the cosistecy of the ove efiitio withi itself lso with the efiitio of itegrility. Theorem 44. ) If f is itegrle o,, the it is improperly itegrle its improper itegrl equls its itegrl. ) If f is itegrle o, for every (, ), the its improper itegrl c e efie s f() = lim f() (98) Similrly, if f is itegrle o c, for every c (, ), the its improper itegrl c e efie s f() = lim f() (99) c + c c) If f is improperly itegrle o (,), the the orer of limit tkig oes ot mtter: ( ) ( ) f() = lim lim f() = lim lim f(). (00) c + c + c c Proof. Not require. Emple 45. Show tht f() = / is improperly itegrle o (0,.

18 8 Itegrtio Proof. For y 0 <c < < f is cotiuous o c, so f is loclly itegrle. We clculte f() = / P =c = = / c /. (0) The lim c 0+ c ( lim c f() ) = lim c 0+ c / =. (0) So f is improperly itegrle o (0, its improper itegrl over (0,) is. Emple 46. Show tht f() = is ot improperly itegrle o (0, ). Proof. For y 0 <c < < f is cotiuous o c, so f is loclly itegrle. We hve =l lc. (03) The lim c 0+ ( c lim c f() ) = lim lc= (04) c 0+ The limit is ot fiite so f is ot improperly itegrle o (0,). rtiol Emple 47. Show tht f() = 0 irrtiol is ot improperly itegrle o (0, ). Proof. For 0<c<<, we kow tht f() is ot itegrle o (c,). So f is ot loclly itegrle cot e improperly itegrle. Theorem 48. (Arithmetics) Let f, g e improperly itegrle o (, ). The ) For y c R, c f is improperly itegrle o (, ) with c f() =c f(). ) f + g is improperly itegrle o (,) with (f()+ g()) = f() + g(). c) If g is oue loclly itegrle, f is improperly itegrle, the f g is improperly itegrle. Iste of provig the theorem, we iscuss riefly prt c). Note tht we o ot ee g to e improperly itegrle o (, ). This reltio shoul ecome turl i light of the followig emple: Tke =0,=. Tke g=. The g is ot improperly itegrle o (0, ) ut clerly if f is improperly itegrle the so is f g. The (stroger) requiremet tht f (iste of f) eig improperly itegrle is ecessry, s c e see from the followig emple: Cosier =π,= f()= si. The for y <c<<, we hve si c c = ( cos) cos = cosc cos + c. (05) c

19 9 Now the limit of the first term oviously eists. For the seco term, sice cos is cotiuous o π, for y, we hve lim cos lim c π c cos = lim π. (06) It turs out tht this limit eists: Set F()8 cos, we oly ee to show π tht lim F() eists is fiite. It suffices to show tht for y sequece, lim F( ) eists is fiite. For y ε > 0, tke N N such tht >ε for ll > N. The for y, m > N, we hve (ssume m >) m cos F( m ) F( ) = m = < m m, m } < ε. (07) Thus F( ) is Cuchy therefore coverges to fiite limit. Therefore f() is improperly itegrle over (π, ). Now tke g() = si >0 si. We hve f() g()=. We hve si 0 π f() g() = (k+)π si π k= kπ k= = π k= (k + ) π kπ (k+)π si (k +). (08) This sum tes to s. So we kow π f() g() = (09) which mes eve if π lim π f() g() (0) eists, it cot e fiite. Therefore f g is ot improperly itegrle o (π, ). Remrk 49. Recll tht the fuctio si is itegrle o (0, (0, if we further efie its vlue t 0 to e ). Therefore si is improperly itegrle o (0, ).