Chapter 10. Newton Integral Primitive Function. Strongly Primitive Function

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1 Chpter 0 Newton Integrl The derivtive of function of single vrible ws introduced in chpter 8, bsed on motivtion from both geometry (construction of the tngent line to the grph of the function) nd physics (determining the instntneous velocity of prticle whose position s function of time is known). One cn think of two other problems common in geometry nd physics: nding the surfce re nd clcultion of the prticle's pth from the given velocity function tht, despite being t rst glnce very dierent, led to mthemticl tsks closely relted to the concept of dierentibility of function. The gol of the net two chpters is to develop the formlism tht is needed to ddress this type of problems. Hence in this chpter the concepts nd bsic properties of the primitive function nd Newton nd indenite integrls of rel-vlued functions of single vrible will be discussed followed by some methods how to clculte them. Further ppliction nd etension of the results will be presented in the net chpters on the Riemnn nd generlised integrls. 0. Primitive Function Finding the primitive function to given function is in some sense n inverse problem to tht of nding the function's proper derivtive. (A function f is primitive to its derivtive f.) However, contrry to the clcultion of the function's derivtive no similrly suitble formlism eists for the clcultion of its primitive function in terms of the elementry functions. Strongly Primitive Function One of the bsic types of the primitive function is introduced by the following denition: Denition 0... Let f : (J R) R nd I J be n intervl (of ny type). We sy tht function F : I R is strongly primitive (or is strong potentil) to function f on

2 0. Primitive Function intervl I if I : F () = f(). Note From denition.9. it follows directly tht function F is dierentible on intervl I. According to theorem 4.8. tht implies tht F is continuous on I, F C(I).. The sine function is strongly primitive to the cosine function on intervl (, + ). It cn be seen from emple tht the strongly primitive function on R to discontinuous function is function f : sin cos ; R \ {0} 0, = 0 F : sin, R \ {0} 0, = 0. Similrly, there eists strongly primitive function on (0, ] to function f : /3, (0, ], nmely F : 3 /3, (0, ]. Smple Problem 0... Prove tht no strongly primitive function on R eists to function sgn : 8>< > :, R+ 0, = 0, R. Solution: If the strongly primitive function F on R to function sgn eisted then ccording to the corollry to theorem the derivtive F () would not hve points of discontinuity of the rst kind on R. This is, however, in contrdiction with the requirement tht F () = sgn, R. (In the denition of the sgn function point 0 is point of discontinuity of the rst kind.) The sucient conditions for the eistence of the strongly primitive function re listed in chpter. Now let us focus on the problem of the correctness (uniqueness) of the strongly primitive function. THEOREM 0... (i) Let F be the strongly primitive function to function f on intervl I. Then for ny C R the function F () + C, I, is lso strongly primitive to function f on I. (ii) Let F nd G be strongly primitive functions to function f on intervl I. Then there eists unique constnt C R such tht I : F () = G() + C. Proof. (i) The sttement follows directly from denition.9.. (ii) If both F nd G re strongly primitive functions to f on I then I : (F G) () = 0. According to corollry (prt e ) to theorem.8.4 function F G is constnt function on I. The uniqueness of this constnt cn be shown indirectly.

3 0. Primitive Function 3 Note The strongly primitive function on n intervl is not dened uniquely. If it eists then there re innitely mny of them, diering from ech other by constnt.. It is, however, esy to see tht if function f hs strongly primitive function on intervl I then to every point ( 0, y 0 ) I R there eists one nd only one strongly primitive function F to f on I such tht ( 0, y 0 ) F. (The grph of F psses through the point ( 0, y 0 ).) Primitive Function In emple we sw tht even such simple function s the sgn function does not hve strongly primitive function on R. Hence it is sensible to etend the concept to wider clss of functions. Denition 0... Let f : (J R) R nd I J is gin n intervl. We sy tht function F : I R is primitive (or is potentil) to function f on intervl I if (i) F C(I), (ii) nite set M I eists such tht I \ M : F () = f(). Note The concept of the primitive function is the generlistion of the concept of the strongly primitive function. In fct, denition.9. is obtined from denition.9. for M =.. Function f :, R is primitive function to the sgn function on R, with M = {0}, nd function F :, 0 is primitive to unbounded function f : ( 5 for = 0, for (0, + ). There is similr sttement on the uniqueness of the primitive function s there ws for the strongly primitive function. THEOREM 0... (i) Let F be the primitive function to function f on intervl I nd C R. Then function F () + C, I is lso primitive function on I to function f. (ii) Let F nd G be primitive functions to function f on intervl I. Then there eists one nd only one constnt C R such tht I : F () = G() + C. (iii) If function f hs primitive function on I then for ny 0 I, y 0 R there eists one nd only one primitive function F to f on I such tht F ( 0 ) = y 0. 3

4 0. Primitive Function 4 Proof. Sttement (i) is obvious. (ii) There eist nite sets M I, M I such tht I \ M : F () = f(), I \ M : G () = f(). The set M = M M is nite set in I nd I \ M : F () = f() = G (). Since F C(I), G C(I), ccording to prt b) of emple.8.4 there eists constnt C R such tht F () G() = C for ll I. The uniqueness is obvious. (iii) It is seen immeditely tht if G is primitive function to f on I then F : G() G( 0 ) + y 0, I is the only primitive function to f on I with the required properties. Smple Problem 0... Let c 0, c,..., c m ; l,..., l m be rel numbers nd let 0 = < < <... < m < m = b. On intervl [, b] we dene the "step-like" function Prove tht function f : cj, = j, j = 0,,..., m l i, ( i, i ), i =,..., m. F : 8>< > : l ( 0 ) [ 0, ) l ( ) + l ( 0 ) [, ) l m ( m ) + l m ( m m ) l ( 0 ), [ m, m ] is primitive to function f on intervl [,b]. Solution: Let M be the nite set of points { 0,,..., m } [, b]. Then nd the sttement is proved. [, b] \ M : F () = f() Note Note tht the concepts of the strongly primitive function nd primitive function hve been dened on n intervl. Both concepts my be etended to more generl sets on R. However, the crucil ssumption tht I is n intervl should be kept in the uniqueness theorems (prt (ii) of theorem.9. nd prts (ii) nd (iii) of theorem.9.). For emple, functions f :, (0, ) +, (, ) nd g : +, (0, ), (, ) re primitive (lso strongly primitive) to function, (0, ) (, ), on (0, ) (, ). Nevertheless the dierence f g is not constnt.. Further generlistion of the concept of the primitive function cn be obtined when set M in denition.9. is countble. In this cse the primitive function to the Dirichlet function is ny rel constnt (since Q, the rtionl number set, is countble). (See, for emple, tetbook [?], p. 359.) 4

5 0. Primitive Function 5 Note It is obvious tht function dened on I R my not hve primitive function on I. For instnce, function f : 5, for = 0, for (0, ] hs no primitive function on [0, ]. On intervl (0, ], the primitive function to f is function ln, (0, ]. The set of ll functions f : I R whose primitive function on intervl I R eists will be denoted s P (I). Smple Problem Let f : (, b) R hve point of discontinuity of the rst kind t point 0 (, b). Furthermore, let F (F ) be the primitive function to restriction f (, 0 ) (f (0, b)) such tht proper limits lim F () nd lim F () eist. Show then tht primitive function F to function f on intervl (,b) eists nd =8>< > : F () + k, (, 0 ), (k R) F () F (), ( 0, b) lim 0 F () + k = lim + 0 F (), = 0. Solution: From the continuity of functions F nd F on intervls (, 0 ) nd ( 0, b), respectively, nd from the choice of constnt k = lim F () lim F () it follows tht F is continuous on (, b). Tht mens tht nite sets M (, 0 ) nd M ( 0, b) eist such tht F () = f() for ll (, 0 ) \ M nd F () = f() for ll ( 0, b) \ M. Hence F () = f() for ll (, b) \ (M M { 0 }). This emple shows how to construct primitive function to discontinuous function with points of discontinuity of the rst kind. This construction will be used frequently nd will be referred to s "gluing". Problems. Show tht the eistence of the primitive function to f + g on intervl I does not gurntee the eistence of the primitive functions to f nd g, seprtely, on I.. Let functions f nd f + g hve primitive functions on J R. Find out if primitive function to g on J eists. 3. Find out if primitive functions to f nd g on J R eist provided the primitive functions to f + g nd f g on J eist. 4. Prove tht there eists no primitive function to function χ(), I, where χ is the Dirichlet function nd I R is n rbitrry intervl in R. 5

6 0. Newton Integrl nd Its Properties 6 5. Find ll primitive functions to the functions dened below together with their denition domins: f :, (, 0); f :, (0, + ); f 3 :, (, 0) (0, + ); f 4 : [], [0, n], n N; f 5 :, R; f 6 :, R \ {0} 0, = Prove tht tn + cot = π for ny R. Answers. Yes, it does. 3. Yes, it does. 5. F : ln( ) + c, (, 0); F : ln + c, (0, + ); F 3 : ln( ) + c, (, 0) ln + c, (0, + ) ( specil cse: F 3 : ln + c, R \ {0}); F 4 : [] []([]+), [0, n]; F 5 : ( F 6 does not eist. + c, [0, + ) + c, (, 0); 0. Newton Integrl nd Its Properties Inspired by the preceding section we re redy to clim tht determining the trjectory of moving prticle out of its velocity given s function of time eectively mens nding the primitive function to the velocity function. In this section, besides the properties of the Newton integrl we shll show tht when clculting the surfce re one is gin fced with the issue of nding the primitive function. Newton Integrl Before ddressing the surfce re question we shll study the following concept nd its properties: Denition 0... Let I be n intervl in R nd f P (I). Let F be primitive function to f on I nd let [, b] I. Then the dierence F (b) F () is clled Newton (denite) 6

7 0. Newton Integrl nd Its Properties 7 integrl (N-integrl) of function f from to b. It is denoted s (N)Rb f() d. We lso sy tht f is Newton integrble on [, b]. (The dierence F (b) F () is denoted s [F ()] b or F () b.) Note 0... The denition of the Newton integrl is correct, i.e. its vlue does not depend on the choice of the primitive function. Smple Problem 0... Clculte the Newton integrl from to b of the following functions: ) f = sgn ; b) f : sgn, R \ {, b}, f () = c R, f (b) = c R; c) the function dened in smple problem.0.; d) n unbounded (in the neighbourhood of point 0) function for =, b =. Solution: For the rst two cses c) d) It is esy to see tht f 3 : sin π π cos π for 0 0 for = 0, sgn d = [ ] b = b = f () d. f() d = [F ()] b = mx i= F 3 : sin π for 0 0 for = 0 l m ( i i ). is strongly primitive function to function f 3 on R, nd hence (N) f 3 () d = F 3 () = sin π sin π = 0. Note 0... In the cse when l i > 0 for i =,..., m, the Newton integrl in emple c) gives very intuitive geometricl interprettion the surfce re of the histogrm-shped object in gure 0.. Note Note tht the N-integrl is dened on closed intervl. The etension of this concept to include n rbitrry (lso unbounded) intervl will be introduced in chpter 3. Bsic Properties of the N-integrl THEOREM 0... Let J R be n intervl, J, b J, f P (J) nd M J is nite set. (i) If f J\M = g J\M then g() d = f() d. 7

8 0. Newton Integrl nd Its Properties 8 (ii) (N) f() d = 0, f() d = (N) f() d. (iii) For function k : c R, J \ M the following is true: k() d = c(b ). Proof. (i) Tking the ssumptions into ccount it follows tht the primitive functions to functions f nd g on intervl J dier by constnt nd the sttement is then obvious. (ii) Let F be primitive function to f on J. Then (N) f() d = F () F () = 0, f() d = F (b) F () = [F () F (b)] = (N) f() d. (iii) According to emple.0. the primitive function to function k on intervl J is F : c( ), J. The result then follows trivilly. b b THEOREM 0... Let nd b be numbers from intervl J R, C R, C R, nd f P (J) nd g P (J). Then (C f + C g) P (J) nd [C f() + C g()] d = C f() d + C g() d. Proof. Let F nd G be primitive functions to f nd g, respectively, both on intervl J. Then C F + C G is primitive function to C f + C g nd the following reltions hold: [C f() + C g()] d = C F (b) + C G(b) C F () C G() = nd the proof is done. = C [F (b) F ()] + C [G(b) G()] = = C f() d + C g() d, THEOREM Let, b nd c be numbers from intervl J R, nd let f P (J). Then f() d = (N)c f() d + f() d c 8

9 0. Newton Integrl nd Its Properties 9 Proof. Let F be primitive function to f on intervl J. Then nd the proof is done. f() d = F (b) F () = [F (b) F (c)] + [F (c) F ()] = (N)c f() d + f() d Using the method of mthemticl induction nd theorem.0.3 one cn stte Corollry 0... Let 0,,..., n be numbers from intervl J R, nd let f P (J). Then n (N) f() d = (N) f() d (N)n f() d. n 0 0 THEOREM Let J R be n intervl, J, b J, < b, nd f P (J) nd g P (J) be such tht f g on intervl [, b]. Then f() d g() d Proof. Let F ( G ) be primitive function to f ( g ) on intervl J. There eist nite sets M J nd M J such tht nd Thus for ny J \ (M M ) J \ M : F () = f(), J \ M : G () = g(). (G F ) () = (g f)() 0. Tht mens tht the dierence G F is non-decresing function on J (compre with smple problem 8.4. prt )) nd hence (g f)() d = [G(b) F (b)] [G() F ()] 0. The proof is then esily completed using theorem.0.. Corollry 0... Let, b, c nd d be numbers from intervl J R such tht c d b nd let g P (J) be non-negtive function on intervl [, b]. Then the following reltions hold: nd g() d 0 (N)d g() d g() d. c 9 c

10 0. Newton Integrl nd Its Properties 0 Proof. The rst inequlity follows from theorem.0.4 for f : 0, [, b]. The second inequlity is obtined using the corollry to theorem.0.3 for this specic cse: g() d = (N)c g() d + (N)d g() d + g() d (N)d g() d. c d c The following theorem discusses the integrl representtion of the primitive function: THEOREM Let nd be numbers from intervl J R nd let f P (J). Then function F : (N) f(t) dt, J, is the primitive function to function f on J with the property F () = 0. Proof. Let us denote primitive function to f on J s G. We then get Thus nite set M J eists such tht which completes the proof. J \ M : (N) f(t) dt = G() G(). d d (N) Surfce Are s Primitive Function f(t) dt = dg () = f(), d The numericl evlution of the surfce re of two-dimensionl objects hs been one of the oldest mthemticl tsks. The re's iomtic denition for so-clled elementry regions is introduced in section.. There it is lso shown tht this denition of the surfce re is correct (tht mens the surfce re of n elementry region eists nd is dened uniquely) nd the surfce re of elementry regions cn be epressed using the Riemnn integrl. Here we shll focus on the reltion between the surfce re P (f; c, d) of curved trpezoid M(f; c, d) = {(, y) R ; [c, d] [, b], 0 y f()}, where function f : [, b] R is continuous on intervl [, b] (gure 0.), nd the primitive function to f. THEOREM Let function f : [, b] R be continuous nd non-negtive on intervl [, b] nd let for every (, b] surfce re of curved trpezoid M(f;, ) eist. Then function p f : P (f;, ) for (, b] 0 for = is the strongly primitive function to function f on [, b]. 0

11 0. Newton Integrl nd Its Properties Proof. Let 0 [, b). Continuity of function f implies ɛ > 0 0 < δ < b 0 [ 0, 0 + δ) : f( 0 ) ɛ < f() < f( 0 ) + ɛ. From this observtion nd due to well-known (nturl) properties of surfce res tht gree with our intuition, nd lso bsed on the non-negtivity of function f we obtin [f( 0 ) ɛ] δ m{[f( 0 ) ɛ] δ, 0} P (f; 0, 0 + δ) [f( 0 ) + ɛ] δ, which mens tht the surfce re of the lrgest rectngle plced inside curved trpezoid M(f; 0, 0 + δ) is not greter thn the surfce re of the trpezoid, nd the ltter is in turn not greter thn the surfce re of the smllest rectngle contining the trpezoid. Furthermore we hve P (f;, 0 ) + P (f; 0, 0 + δ) = P (f;, 0 + δ), which mens tht the sum of the surfce res of curved trpezoids M(f;, 0 ) nd M(f; 0, 0 + δ) (the two pieces we brek trpezoid M(f;, 0 + δ) into) is equl to the surfce re of trpezoid M(f;, 0 + δ). From these reltions we get ) p f( 0 + δ) p f ( 0 ) f( 0 δ < ɛ. If in this inequlity one considers the limits δ 0 + nd then ɛ 0 +, then the reltion p f +( 0 ) = f( 0 ) emerges. The nlogous sttement, p f ( 0 ) = f( 0 ) for 0 (, b], for the left derivtive is shown in the sme wy. Thus we cn conclude tht p f() = f() for rbitrry [, b] nd the proof is done. Corollry Let J R be n intervl nd function f : J R be continuous nd non-negtive on J. If for ny intervl [, b] J ( R, b R) the surfce re P (f;, b) of curved trpezoid M(f;, b) eists then the strongly primitive function to f on J eists nd the following reltion holds: P (f;, b) = f() d. Proof. Bsed on theorem.0.6 strongly primitive function p f (), [, b] to function f eists on ny intervl [, b] J, nd therefore it eists on the whole intervl J. Furthermore, from the denition of the N-integrl it follows tht since p f () = 0. f() d = p f (b) p f () = P (f;, b) Note In section.5, we shll prove n even stronger sttement thn the one in the corollry bove. Here we just quote the stronger sttement: Every continuous function on n intervl in R hs strongly primitive function on this intervl.

12 0. Newton Integrl nd Its Properties Problems. Find ll α R for which the Newton integrl (N)R 0 f() d eists for f : α, 0 c, = 0 where c R.. Clculte the following N-integrls provided they eist: ) (N) m (, 4 ) d b) (N)+π/ π/ sgn(sin ) d c) (N) f() d for f : ( ) /, (, ) c R, = or = +, d) (N) [ ] d. 3. Find out if function 0 f :, 0 0, = 0 is Newton integrble. Consider seprtely the following three intervls: [, ], [, 0] nd [, ]. If it is, clculte the Newton integrl. 4. Prove the corollry tht follows fter theorem Show tht the following sttement is true: Let < c < b be rel numbers, nd f P ([, c]) nd f P ([c, b]). Then f P ([, b]) nd (N)c f() d + f() d = f() d. c 6. Formulte nd prove the theorem on the N-integrl of sum of functions (nlogous to theorem.0.) for n rbitrry nite number of terms. 7. Clculte d b ) d (N) d b e d, d (N) d b e d, db (N) where R, b R. d (sin b) d (N) )/3 sin t 3 dt, for [ π/, +π/], R 0 sin Rsin t dt c) lim 0, lim 0 0 (tn t) /n dt Rtn 0 (sin t) /n dt, n N. e d, 8. Clculte the surfce re of two-dimensionl plnr objects bounded by the following curves: ) the is, grph of function 3 ( R), nd lines = nd = ; b) the grphs of functions ( R) nd + ( R).

13 0.3 Indenite Integrl 3 Answers α <. ) /5, b) π /4, c) π, d) 5 ln 3/ ln. 3 f is Newton integrble only on the lst of the three intervls. The result is ln. 7 ) 0, e, e b ; b) 3 (sin ) /3 cos sin 3 ; c),. 8 ) 6 ; b) Indenite Integrl In section we sw tht the hrd prt in the clcultion of the N-integrl consists of the determintion of the primitive function to given function on n intervl under considertion. Therefore, in this section we shll focus on dierent lterntive methods for nding the primitive function. Concept of Indenite Integrl We lredy know from theorem.9. tht the primitive function to given function is not uniquely determined. If single primitive function to f on n intervl I eists, then there re innitely mny of them, diering from ech other by n dditive constnt. R Denition The set of ll primitive functions to f : (J R) R on n intervl I J will be clled n indenite integrl of function f on I nd denoted by symbol f() d, I. Note Function f is clled n integrnd, the symbolr is referred to s the integrl sign, nd letters d denote the integrtion vrible ( in this cse). The mening of the integrtion vrible is s follows: clculting the derivtive with respect to the integrtion vrible of ny function from the setr f() d on I \ M ( M is nite subset of I) one obtins restriction f I\M.. If F is primitive function to f on I we shll write f() d c = F (), I or f() d = {F () + c} c R, I. For instncer e d c = e =: F (), R since df d () = e for R. 3. The indenite integrl will usully be sought on the so-clled miml set A \ R which does not hve to be n intervl but could lso be union of nite number of intervls tht 3

14 0.3 Indenite Integrl 4 stises the following sttement: If f hs primitive function on set B then B A. For emple, d = c ln, (, 0) (0, + ) 3 d = c 3 3, (, 0) (0, + ). Bsic Indenite Integrls The following indenite integrls cn be derived from the rules for derivtives (section 8.):. 0 d c = 0, R. d =: d c =, R 3. n d = c n+ n +, R (n N). If n R \ { }, the formul holds for R +. For some vlues of n the formul is vlid on n etended set (see, e.g. prt 3 of the lst note); 4. d = c ln, R +, d = c ln( ), R or d = c ln, R \ {0} 5. + d c = tn, R, + d c = cot, R 6. d c = sin, (, ), d c = cos, (, ) 7. e d c = e, R 8. d c = ln, R ( R+ \ {}) 9. sin d c = cos, R 0. cos d = c sin, R [. (sin ) d = c cot, (kπ, π + kπ) [ k. (cos ) d = c tn, (kπ π, kπ + π ) 3. sinh d c = cosh, R k 4

15 0.3 Indenite Integrl 5 4. cosh d c = sinh, R 5. (sinh ) d c = coth, R \ {0} 6. (cosh ) d c = tnh, R In order to speed up clcultions the following formuls re often useful: 7. + d c = ln( + + ), R 8. d = c ln( + ), (, + ) 9. d = c ln, R \ {, } + 0. Let f : (I R) R (I is n intervl) be dierentible on I. Then f () f() d = c ln f(), {z I; f(z) 0}. Let f : (I R) R (I is n intervl) be dierentible on I. Then f () È f() d c = È f(), {z I; f(z) > 0} Formuls 7- cn be esily veried by performing the derivtive on the right-hnd sides, or with the use of forthcoming theorems 3 or 4. Bsed on the formuls listed bove one cn clculte indenite integrls for just nrrow clss of functions. Hence in the following tet we shll develop integrtion methods llowing us to trnsform integrls of wide clss of functions in such wy tht the formuls bove cn R R be pplied. We wrn the reder right here tht such trnsformtion my not be possible for every function f P (I). For instnce, indenite integrls cos d, R; e d, R, lthough they do eist (see note 4.0.), cnnot be epressed in terms of the elementry R functions (nd, therefore, cnnot be trnsformed into ny of the formuls - for n elementry function f). Similrly, neither of the integrlsrsin d, R \ {0} nd d, R \ { } belongs to the set of the elementry functions. (A comple 3 + proof of the lst sttement cn be found in tetbook [].) There is no generl solution to this problem. Decomposition Method THEOREM Let k,..., k m be rel numbers nd indenite integrlsr fi () d c = F i () eist on intervl I R for i =,..., m. Then the following indenite integrl eists "mx i= k i f i ()# d c = 5 mx i= k i F i (), I.

16 0.3 Indenite Integrl 6 Proof. The ssumptions imply tht nite sets M i I, i =,..., m eist such tht Set M m i= M i I \ M i i =,..., m F i () = f i () ( R). is lso nite subset mx of I nd the mx following reltion mx holds I \ M k i F i! () = k i F i () = k i f i () i= which mens tht the proof is done. "mx Note The sttement in theorem.. is usully written in the following form i= k i f i ()# d = mx i= i= i= k i f i () d, J. Smple Problem Clculte the following indenite integrls on their respective miml sets )R R d ; b) 4 Pm () d, where P m is polynomil of degree m m m , R nd i R for i = 0,,..., m; c)r d. sin Solution: ) Using formuls 5 nd 9 one obtins 4 d = + + ( )( + ) d = d + d c = ln + tn = "mx i# for R \ {, }. b) P m () d i d = i=0 c) Using formul 0 one nds d sin sin = [ + cos sin cos for (kπ, kπ + π). k mx i i d = i=0 d = sin cos Method of Integrtion By Prts mx i=0 d+ cos sin i i + i+, R. d c = ln sin ln cos = ln tn Let F nd G be the primitive functions to functions f nd g on intervl I R. In generl, it is not true tht product F G is the primitive function to function f g. Tht is becuse (F G) = F G + F G = fg + F g on I. However, the following theorem holds: THEOREM Let f P (I) nd g P (I) where R f() d c = F (), R g() d c = G(), I nd furthermore let fg P (I) (or F g P (I) ). Then lso F g P (I) (or fg P (I) ) nd f() G() d + F () g() d c = F () G(), I. (0.) 6

17 0.3 Indenite Integrl 7 Proof. Let fg P (I) nd R f()g() c = H(), I. Then nite set M I eists such tht I \ M : (F G H) () = f()g() + F ()g() f()g() = F ()g(). Tht mens tht function F G H is primitive to F g on I nd F g P (I). The second prt of the sttement follows from theorem.. nd reltion (F G) () d = c F ()G(), I which completes the proof. Note Theorem.. on integrtion by prts is of lrge prcticl signicnce. However, when pplied we shll use R f()g() d = F ()G() R F ()g() d, I, insted of (.9). Smple Problem Clculte the following indenite integrls on their miml sets ) cos d; b) sin d; Solution: ) If we choose f : cos, R nd G :, R then F : sin, R nd g :, R nd ccording to theorem.. cos d = sin sin d c = sin + cos for R. b) Let us choose f :, (, ) nd G : sin, (, ). Then F :, (, ) nd g :, (, ). Now using formul we hve sin d = sin + d = c sin + for (, ). According to theorem (see lso problem 6.8.4) the primitive function tht hs been found bove cn be etended to include the closed intervl [, ]. Then one cn write sin d c = sin +, [, ]. Smple Problem Prove tht for indenite integrl I n () =R (+ ) n d on R the following recurrent reltion holds I n+ () = n ( + ) + n n n I n(), R, (0.) 7

18 0.3 Indenite Integrl 8 where n N. Solution: Choosing f :, R nd G : (+ ) n, R one nds F :, R nd g : n (+ ) n+, R. Then ccording to theorem.. I n () = = = This esily simplies to formul (.0) ( + ) + n d = n ( + ) n+ ( + ) + n + n ( + ) d n d = n+ ( + ) n+ ( + ) n I n+() + n I n n (), R. Note Note tht the N-integrl formul (.9) cn be written in the form (N) 0 f(t) G(t) dt + (N) for I nd 0 I (compre with theorem 5.0.). Substitution Method 0 F (t) g(t) dt = [F ()G()] 0 The bsic theorems describing the substitution method re obtined from the theorem on the derivtive of composite function (the chin rule): THEOREM Let I nd I be intervls in R nd f : I R. Let function φ : I I be dierentible on I. Then for f P (I ), nd F being strongly primitive function to f on I the composite function F f is strongly primitive function to function (f φ)φ on I. Proof. Since F (t) = f(t) for t I then ccording to theorem.8. on the derivtive of composite function the following equlity holds: for I. (F φ) () = (F φ)(). φ () = (f φ)(). φ () Note Using the indenite integrl the sttement in theorem 3 cn be epressed in the form (f φ)()φ () d = f(t) dt (0.3) for t = φ(), I. Alterntively, the N-integrl cn be used in the form for I, 0 I. (N) (f φ)(s)φ (s) ds = (N)φ() f(t) dt 0 φ( 0 ) 8

19 0.3 Indenite Integrl 9. Theorem 3 cn be pplied in the following sense: when clculting the integrl on the left side of eqution (.) in the ctul clcultion one introduces φ() = t nd symbol φ ()d is then replced by dt. Smple Problem Clculte I := ln d on the miml set. Solution: Let φ : ln, R + \ {}. Restrictions φ (0,) conditions of theorem 3. Therefore Š dt = nd φ (, ) stisfy the nd hence I = t dt c = ln t I c = ln ln, for t (, 0) (0, + ) (0, ) (, + ). If the primitive function to function (f φ). φ is known then the primitive function to function f cn be clculted using the following theorem: THEOREM Let I nd I be intervls in R nd f : I R. Let function φ : I I be bijection nd hs proper derivtive φ (t) 0 on intervl I. If (f φ). φ P (I ) nd G is its primitive function on I, then function G φ (φ is the inverse function to φ) is primitive to f on I. Proof. A nite set M I eists such tht t I \ M : G (t) = (f φ)(t)φ (t). Since φ is bijection set M := φ(m ) is nite. Then ccording to the theorem on the derivtive of composite (theorem.8.) nd inverse (theorem 4.8.) functions the following reltion holds for ll I \ M (G φ ) () = (G φ )(). (φ ) () = [(f φ) φ ](). The lst equlity completes the proof. f() d = (f φ)(t)φ (t) dt φ φ )() = f() Note Using the indenite integrl the sttement in theorem 4 cn be epressed by the following formul for t = φ (), I. Alterntively, the N-integrl cn be used in the form (N) f(s) ds = (N)φ () (f φ)(t)φ (t) dt 0 φ ( 0 ) 9

20 0.3 Indenite Integrl 0 for I, 0 I.. In order to clculte integrl R f()d theorem 4 cn be used with substitutions = φ(t) nd d = φ (t)dt. Smple Problem ClculteR + d on R. Solution:. Consider φ : t sinh t, t R. φ is bijection nd for t R we hve φ (t) = cosh t > 0. The ssumptions of theorem 4 re therefore stised. Hence + d = È + sinh t cosh t dt = cosh t dt for t = sinh = ln(+ + ), R. (We hve pplied the identity cosh sinh =, for R.) Furthermore we hve cosh t dt = e t dt + dt + e t dt = c t sinh t, t R. 4 Tht implies + d = c ln( + + ) + 4 sinh[ ln( + + )] for R.. Using integrtion by prts with f : +, R nd G :, R one gets + d c = ln( + + ) + +, R. Formlly, the two methods give dierent results. However, one cn work out the derivtives of the two results nd esily see tht both results re correct. As n independent check, one cn recll theorem.0., prt (ii): R : + = sinh[ ln( + + )] + c. Choosing = 0 in this formul we get c = 0 nd hence the equlity of the two results. Note Note tht theorem 3 cn be used with much weker ssumptions bout function φ thn theorem 4. In theorem 3, the bijection is not required nd neither is the ssumption φ (t) 0. Problems. Find recurrent reltions for the following indenite intergls: ) I n () =R sin n d, b) J n () =R cos n d.. Use both integrtion by prts nd substitution method for the clcultion of indenite integrlr d where (, ) (, + ), nd compre the two results. 0

21 0.4 Integrtion of Specil Types of Functions 3. Clculte (on R) ) f() d, where f :,, > b) d. Answers. ) I n () = n sinn cos + n I n n () for n N, n ; b) J n () = n cosn sin + n I n n () for n N, n.. ln ) 3 for ; nd + sgn for > ; 3 6 b) F : tn, R \ {0}; F (0) = 0. (Use the following chnge of vribles: = t.) 0.4 Integrtion of Specil Types of Functions The gol of this section is to show the wys how to nd indenite integrls of some importnt nd frequent clsses of rel functions. For prcticl purposes we limit ourselves to show only the description of severl useful lgorithms without the mbition to provide the rigorous formultions. Hence in ech prticulr cse the vlidity conditions must be considered seprtely Integrtion of Rtionl Functions Recll tht in theorem we hve introduced the decomposition of purely rtionl single vrible function into prtil frctions. According to this theorem nd theorem.0.3 the problem of integrtion of such rtionl functions is reduced to the problem of integrtion of individul prtil frctions. Since for k N nd k ( ) k d c = ( ) k, R \ {} k nd d = c ln, R \ {} then the only remining problem in the integrtion of rtionl single vrible functions is the clcultion of the primitive function to function A+B, R, where A, B, p ( +p+q) k

22 0.4 Integrtion of Specil Types of Functions nd q re rel numbers, k N, nd the discriminnt p 4q of the qudrtic epression in the denomintor is negtive. Tking the decomposition A + B ( + p + q) k = A + p ( + p + q) + B Ap k ( + p + q) k into ccount the indenite integrl of the rst term cn esily be found bsed on theorem (the rst theorem on substitution) using the chnge of vribles + p + q = t while A + p ( + p + q) k d c = A A ( k) ( + p + q) k, R, k + p + p + q d c = A ln( + p + q), R. The denomintor of the second term cn be epnded ccording to + p + q = + p + q p Š q p 4 = q p 4Œ p qq p = Then the R primitive function to the second term cn be obtined using the chnge of vrible t + R p (theorem 3.0.3). This substitution trnsforms the originl 4 / d dt integrl into integrl. The ltter integrl cn be obtined using the ( +p+q) k (t +) k recurrent formul (.0.3) found in problem The procedure described bove cn now be summrised in THEOREM The indenite integrl of n rbitrry rtionl single vrible function on its denition domin belongs to the set of the elementry functions Integrtion of Trigonometric Functions Net, we shll eplicitly show the substitutions used to trnsform the integrls of type R(sin, cos ) d, M R (0.4) into integrls of rtionl functions. Function R : (u, v) R(u, v) from R to R is rtionl function of two vribles. Integrl (.) cn be clculted using the substitutions shown below (nd theorem 3.0.3). Ech of the substitutions trnsforms the integrnd into rtionl single vrible function.. If R is n odd function in the rst of its two vribles, R( sin, cos ) = R(sin, cos ) then substitution cos = t, R, cn be used.

23 0.4 Integrtion of Specil Types of Functions 3. If R(sin, cos ) = R(sin, cos ) then substitution sin = t, R, cn be used. 3. If R( sin, cos ) = R(sin, cos ) then substitution tn = t, Sk Š π + kπ, π + kπ, cn be used. 4. For n rbitrry rtionl function R of two vribles substitution tn = t cn be used. (This substitution cn lso be used in the previous three cses lthough the clcultion is most of the times techniclly more involved.) The lst substitution, tn = t, for ( π, π), t R, leds to the formuls sin = sin cos = tn tn + = cos = cos sin = t + t t t + = tn t, d = + t dt. It is now cler tht integrl () trnsforms into rtionl function in terms of vrible t. Smple Problem ClculteR d on R. +cos Solution: Bsed on the ides bove we cn use either the chnge of vrible tn = t or tn = t. We shll use both substitutions: ) With the chnge of vrible tn = t the conditions of theorem re stised on every intervl Š π + kπ, π + kπ, k. Since sin = cos t one gets cos = on ny intervl Š π + kπ, π + kπ, k, the following reltions hold: + cos d = cos d + cos cos dt = dt + t = Thus one nds + cos d c = +t tn tn c = tn t, t R. Œ +t nd on ny intervl Š π + kπ, π + kπ, k. The clcultion is, however, not completely nished, since the function found t the end is not primitive to function f : ( + cos ), R everywhere on set R. Using the "gluing" construction of emple 3.0. the primitive function (indenite integrl) obtined bove cn now be continuously etended t points π + kπ, k in such wy tht new etension F will be the primitive function to function f : ( + cos ), R on R. 3

24 0.4 Integrtion of Specil Types of Functions 4 tn For k let us denote F k : tn tn tn F k+ : nd C k := 8><>: Œ, π Œ + kπ, π + kπ, π + (k + )π, π + (k + )π lim F k+() ( π +(k+)π ) + Then the primitive function F is tn tn F : tn tn, π π 4, = π lim ( F k() = π π +kπ ). Š, Š π + kπ, π + kπ π + (k + )π, π + (k + )π + kπ, k. Besides tht F lso + is the strongly primitive function to f on R. Reclling theorem this cn be seen from π F + kπ = lim F () = lim ( π +kπ ) + ( π +kπ ) + + cos = π + kπ nd hence F π + kπ = = + cos π. = = F b) Substitution φ : tn lso stises the conditions of theorem on n intervl ( π + kπ, π + kπ), k. Then cos = ( t ), cos =, t R. For ny (+t = ) +t ( π + kπ, π + kπ) + cos d = cos d t + + cos cos dt, t R. t 4 + The rtionl function tht hs been obtined here is more complicted thn in cse ). The only benet of this substitution is tht we nd primitive functions on lrger intervls. However, since we re seeking the primitive function on R the obtined result must gin be etended by "gluing" Integrtion of Irrtionl Functions ) An indenite integrl of type R", + b c + dœp/q# d where, b, c nd d re rel numbers nd p nd q re integers, q 0, cn be trnsformed into n integrl of rtionl functions using chnge of vrible + b Œ/q = t. c + d 4

25 0.4 Integrtion of Specil Types of Functions Smple Problem Trnsform 3 d on the denition domin of the + integrnd into n integrl of rtionl functions. Solution: The denition domin of the integrted function is (, 0) (0, + ). Here we shll use the substitution ( + ) /6 = t tht stises the conditions of theorem on these intervls (t (0, ) (, + )). According to the theorem, on ech of the intervls d = 3 + 6( + d )5/6 c t 5 ( + t 3 ) = dt. 6( + ) 5/6 t b) The integrtion of n irrtionl function of type R(, + b + c), where R is n irrtionl function of two vribles, depends on the eistence of the rel roots of qudrtic form + b + c (, b nd c re rel numbers).. If < re the rel roots such tht + b + c = ( )( ) for R then (ssuming tht < 0) one cn write )s + b + c = ( for (, ) nd the clcultion of the remining integrl cn be ccomplished using the method of prt ).. If the qudrtic term hs no rel roots then function + b + c hs non-empty denition domin only for > 0 nd c > 0. The integrtion of this function cn be reformulted in terms of the integrtion of rtionl function by one of the following Euler substitutions: + b + c = + t or + b + c = t + c. These substitutions cn lso be used if + b + c hs rel roots. In such cse, the rst substitution is useful for > 0 nd the second one for c > 0. When used in ctul clcultions it is necessry to check the intervls where the substitution theorems re stised in terms of the new vribles Integrtion of Trnscendent Functions ) The integrl of trnscendent function R(e ), A (, + ) ( R is rtionl function of rel single vrible) cn be turned into the integrl of rtionl function using the chnge of vrible t = e. b) The integrl of function R(ln ), A (, + ) is trnsformed into n integrl of rtionl function using the chnge of vrible ln = t. Finlly we note tht mstering the methods of integrtion requires independent nd honest clcultion of sucient number of problems nd eercices. 5

26 0.4 Integrtion of Specil Types of Functions 6 Problems. Prove in full detil tht ll substitutions suggested in this section trnsform the respective integrls into integrls of rtionl functions.. Clculte d ) R for R. 4 b) (+ 3 d for 0. ) d c) + for (, ) (, + ) ++ sin d) I() :=R cos d for R e)rqe Answers. ) b) c) e + sin 8 +cos 8 d for [0, + ). 4 ln [tn ( + ) + tn ( )] ; tn 6 ; 3 (z + ) + ln z 4 z + 3 where z = ; d) I() =È + 4 tn tn È È4 + tn tn 4 È4, π 4 + kπ, k ; + π 4 È + È 4 + π π I( π 4 + kπ ) = lim I() ; π 4 + kπ e) ln(e + e + ) + sin (e ). 6