The Riemann Integral. Chapter 1

Chpter The Riemnn Integrl now of some universities in Englnd where the Lebesgue integrl is tught in the first yer of mthemtics degree insted of the Riemnn integrl, but now of no universities in Englnd where students lern the Lebesgue integrl in the first yer of mthemtics degree. (Approximte quottion ttributed to T. W. Körner) Let f : [,b] R be bounded (not necessrily continuous) function on compct (closed, bounded) intervl. We will define wht it mens for f to be Riemnn integrble on [,b] nd, in tht cse, define its Riemnn integrl f. The integrl of f on [,b] is rel number whose geometricl interprettion is the signed re under the grph y = f(x) for x b. This number is lso clled the definite integrl of f. By integrting f over n intervl [,x] with vrying right end-point, we get function of x, clled the indefinite integrl of f. The most importnt result bout integrtion is the fundmentl theorem of clculus, which sttes tht integrtion nd differentition re inverse opertions in n ppropritely understood sense. Among other things, this connection enbles us to compute mny integrls explicitly. Integrbility is less restrictive condition on function thn differentibility. Roughly speking, integrtion mkes functions smoother, while differentition mkes functions rougher. For exmple, the indefinite integrl of every continuous function exists nd is differentible, wheres the derivtive of continuous function need not exist (nd generlly doesn t). The Riemnn integrl is the simplest integrl to define, nd it llows one to integrte every continuous function s well s some not-too-bdly discontinuous functions. There re, however, mny other types of integrls, the most importnt of which is the Lebesgue integrl. The Lebesgue integrl llows one to integrte unbounded or highly discontinuous functions whose Riemnn integrls do not exist, nd it hs better mthemticl properties thn the Riemnn integrl. The definition of the Lebesgue integrl requires the use of mesure theory, which we will not

2. The Riemnn Integrl describe here. In ny event, the Riemnn integrl is dequte for mny purposes, nd even if one needs the Lebesgue integrl, it s better to understnd the Riemnn integrl first... Definition of the Riemnn integrl We sy tht two intervls re lmost disjoint if they re disjoint or intersect only t common endpoint. For exmple, the intervls [,] nd [,3] re lmost disjoint, wheres the intervls [,2] nd [,3] re not. Definition.. Let I be nonempty, compct intervl. A prtition of I is finite collection {I,I 2,...,I n } of lmost disjoint, nonempty, compct subintervls whose union is I. A prtition of [,b] with subintervls = [x k,x k ] is determined by the set of endpoints of the intervls = x < x < x 2 < < x n < x n = b. Abusing nottion, we will denote prtition P either by its intervls P = {I,I 2,...,I n } or by the set of endpoints of the intervls P = {x,x,x 2,...,x n,x n }. We ll dopt either nottion s convenient; the context should mke it cler which one is being used. There is lwys one more endpoint thn intervl. Exmple.2. The set of intervls {[,/5],[/5,/4],[/4,/3],[/3,/2],[/2,]} is prtition of [,]. The corresponding set of endpoints is {,/5,/4,/3,/2,}. We denote the length of n intervl I = [,b] by I = b. Note tht the sum of the lengths = x k x k of the lmost disjoint subintervls in prtition {I,I 2,...,I n } ofn intervl I is equl to length of the whole intervl. This is obvious geometriclly; lgebriclly, it follows from the telescoping series n n = (x k x k ) k= k= = x n x n +x n x n 2 + +x 2 x +x x = x n x = I. Suppose tht f : [,b] R is bounded function on the compct intervl I = [,b] with M = supf, m = inf f. I I

.. Definition of the Riemnn integrl 3 If P = {I,I 2,...,I n } is prtition of I, let M k = sup f, m k = inf f. These suprem nd infim re well-defined, finite rel numbers since f is bounded. Moreover, m m k M k M. If f is continuous on the intervl I, then it is bounded nd ttins its mximum nd minimum vlues on ech subintervl, but bounded discontinuous function need not ttin its supremum or infimum. We define the upper Riemnn sum of f with respect to the prtition P by n n U(f;P) = M k = M k (x k x k ), k= nd the lower Riemnn sum of f with respect to the prtition P by n n L(f;P) = m k = m k (x k x k ). k= Geometriclly, U(f;P) is the sum of the res of rectngles bsed on the intervls tht lie bove the grph of f, nd L(f;P) is the sum of the res of rectngles tht lie below the grph of f. Note tht k= k= m(b ) L(f;P) U(f;P) M(b ). Let Π(,b), or Π for short, denote the collection of ll prtitions of [,b]. We define the upper Riemnn integrl of f on [,b] by U(f) = inf P Π U(f;P). The set {U(f;P) : P Π} of ll upper Riemnn sums of f is bounded from below by m(b ), so this infimum is well-defined nd finite. Similrly, the set {L(f;P) : P Π} of ll lower Riemnn sums is bounded from bove by M(b ), nd we define the lower Riemnn integrl of f on [,b] by L(f) = sup L(f;P). P Π Theseupper nd lowersums nd integrlsdepend onthe intervl [,b]s well sthe function f, but to simplify the nottion we won t show this explicitly. A commonly used lterntive nottion for the upper nd lower integrls is U(f) = f, L(f) = Note the use of lower-upper nd upper-lower pproximtions for the integrls: we tke the infimum of the upper sums nd the supremum of the lower sums. As we show in Proposition.3 below, we lwys hve L(f) U(f), but in generl the upper nd lower integrls need not be equl. We define Riemnn integrbility by their equlity. f.

4. The Riemnn Integrl Definition.3. A bounded function f : [,b] R is Riemnn integrble on [,b] if its upper integrl U(f) nd lower integrl L(f) re equl. In tht cse, the Riemnn integrl of f on [,b], denoted by f(x)dx, f, f [,b] or similr nottions, is the common vlue of U(f) nd L(f). An unbounded function is not Riemnn integrble. In the following, integrble will men Riemnn integrble, nd integrl will men Riemnn integrl unless stted explicitly otherwise..2. Exmples of the Riemnn integrl Let us illustrte the definition of Riemnn integrbility with number of exmples. Exmple.4. Define f : [,] R by { /x if < x, f(x) = if x =. Then x dx isn t defined s Riemnn integrl becuse f is unbounded. In fct, if < x < x 2 < < x n < is prtition of [,], then sup f =, [,x ] so the upper Riemnn sums of f re not well-defined. An integrl with n unbounded intervl of integrtion, such s x dx, lso isn t defined s Riemnn integrl. In this cse, prtition of [, ) into finitely mny intervls contins t lest one unbounded intervl, so the corresponding Riemnn sum is not well-defined. A prtition of [, ) into bounded intervls (for exmple, = [k,k+] with k N) gives n infinite series rther thn finite Riemnn sum, leding to questions of convergence. One cn interpret the integrls in this exmple s limits of Riemnn integrls, or improper Riemnn integrls, dx = lim x ǫ + ǫ x dx, r dx = lim x r x dx, but these re not proper Riemnn integrls in the sense of Definition.3. Such improper Riemnn integrls involve two limits limit of Riemnn sums to define the Riemnn integrls, followed by limit of Riemnn integrls. Both of the improper integrls in this exmple diverge to infinity. (See Section..)

.2. Exmples of the Riemnn integrl 5 Next, we consider some exmples of bounded functions on compct intervls. Exmple.5. The constnt function f(x) = on [,] is Riemnn integrble, nd dx =. To show this, let P = {I,I 2,...,I n } be ny prtition of [,] with endpoints Since f is constnt, nd therefore {,x,x 2,...,x n,}. M k = sup f =, m k = inf f = U(f;P) = L(f;P) = for k =,...,n, n (x k x k ) = x n x =. k= Geometriclly, this eqution is the obvious fct tht the sum of the res of the rectngles over (or, equivlently, under) the grph of constnt function is exctly equl to the re under the grph. Thus, every upper nd lower sum of f on [,] is equl to, which implies tht the upper nd lower integrls U(f) = inf U(f;P) = inf{} =, P Π re equl, nd the integrl of f is. L(f) = sup L(f;P) = sup{} = P Π Moregenerlly, thesmergumentshowsthteveryconstntfunction f(x) = c is integrble nd cdx = c(b ). The following is n exmple of discontinuous function tht is Riemnn integrble. Exmple.6. The function is Riemnn integrble, nd f(x) = { if < x if x = f dx =. To show this, let P = {I,I 2,...,I n } be prtition of [,]. Then, since f(x) = for x >, M k = sup f =, m k = inf f = for k = 2,...,n. The first intervl in the prtition is I = [,x ], where < x, nd M =, m =, since f() = nd f(x) = for < x x. It follows tht Thus, L(f) = nd U(f;P) = x, L(f;P) =. U(f) = inf{x : < x } =,

6. The Riemnn Integrl so U(f) = L(f) = re equl, nd the integrl of f is. In this exmple, the infimum of the upper Riemnn sums is not ttined nd U(f;P) > U(f) for every prtition P. A similr rgument shows tht function f : [,b] R tht is zero except t finitely mny points in [, b] is Riemnn integrble with integrl. The next exmple is bounded function on compct intervl whose Riemnn integrl doesn t exist. Exmple.7. The Dirichlet function f : [,] R is defined by { if x [,] Q, f(x) = if x [,]\Q. Tht is, f is one t every rtionl number nd zero t every irrtionl number. This function is not Riemnn integrble. If P = {I,I 2,...,I n } is prtition of [,], then M k = supf =, m k = inf =, since every intervl of non-zero length contins both rtionl nd irrtionl numbers. It follows tht U(f;P) =, L(f;P) = for every prtition P of [,], so U(f) = nd L(f) = re not equl. The Dirichlet function is discontinuous t every point of [, ], nd the morl of the lst exmple is tht the Riemnn integrl of highly discontinuous function need not exist..3. Refinements of prtitions As the previous exmples illustrte, direct verifiction of integrbility from Definition.3 is unwieldy even for the simplest functions becuse we hve to consider ll possible prtitions of the intervl of integrtion. To give n effective nlysis of Riemnn integrbility, we need to study how upper nd lower sums behve under the refinement of prtitions. Definition.8. A prtition Q = {J,J 2,...,J m } is refinement of prtition P = {I,I 2,...,I n } if every intervl in P is n lmost disjoint union of one or more intervls J l in Q. Equivlently, if we represent prtitions by their endpoints, then Q is refinement of P if Q P, mening tht every endpoint of P is n endpoint of Q. We don t require tht every intervl or even ny intervl in prtition hs to be split into smller intervls to obtin refinement; for exmple, every prtition is refinement of itself. Exmple.9. Consider the prtitions of [, ] with endpoints P = {,/2,}, Q = {,/3,2/3,}, R = {,/4,/2,3/4,}. Thus, P, Q, nd R prtition [,] into intervls of equl length /2, /3, nd /4, respectively. Then Q is not refinement of P but R is refinement of P.

.3. Refinements of prtitions 7 Given two prtitions, neither one need be refinement of the other. However, two prtitions P, Q lwys hve common refinement; the smllest one is R = P Q, mening tht the endpoints of R re exctly the endpoints of P or Q (or both). Exmple.. Let P = {,/2,} nd Q = {,/3,2/3,}, s in Exmple.9. Then Q isn t refinement of P nd P isn t refinement of Q. The prtition S = P Q, or S = {,/3,/2,2/3,}, is refinement of both P nd Q. The prtition S is not refinement of R, but T = R S, or T = {,/4,/3,/2,2/3,3/4,}, is common refinement of ll of the prtitions {P,Q,R,S}. As we show next, refining prtitions decreses upper sums nd increses lower sums. (The proof is esier to understnd thn it is to write out drw picture!) Theorem.. Suppose tht f : [,b] R is bounded, P is prtitions of [,b], nd Q is refinement of P. Then Proof. Let U(f;Q) U(f;P), L(f;P) L(f;Q). P = {I,I 2,...,I n }, Q = {J,J 2,...,J m } be prtitions of [,b], where Q is refinement of P, so m n. We list the intervls in incresing order of their endpoints. Define M k = sup f, m k = inf f, M l = sup J l f, m l = inf J l f. Since Q is refinement of P, ech intervl in P is n lmost disjoint union of intervls in Q, which we cn write s = q k l=p k J l for some indices p k q k. If p k < q k, then is split into two or more smller intervls in Q, nd if p k = q k, then belongs to both P nd Q. Since the intervls re listed in order, we hve If p k l q k, then J l, so p =, p k+ = q k +, q n = m. M l M k, m k m l for p k l q k. Using the fct tht the sum of the lengths of the J-intervls is the length of the corresponding I-intervl, we get tht q k l=p k M l J l q k q k M k J l = M k J l = M k. l=p k l=p k

8. The Riemnn Integrl It follows tht m n q k n U(f;Q) = M l J l = M l J l M k = U(f;P) l= k=l=p k k= Similrly, q k q k m l J l m k J l = m k, l=p k l=p k nd n q k n L(f;Q) = m l J l m k = L(f;P), k= l=p k k= which proves the result. It follows from this theorem tht ll lower sums re less thn or equl to ll upper sums, not just the lower nd upper sums ssocited with the sme prtition. Proposition.2. If f : [,b] R is bounded nd P, Q re prtitions of [,b], then L(f;P) U(f;Q). Proof. Let R be common refinement of P nd Q. Then, by Theorem., L(f;P) L(f;R), U(f;R) U(f;Q). It follows tht L(f;P) L(f;R) U(f;R) U(f;Q). An immedite consequence of this result is tht the lower integrl is lwys less thn or equl to the upper integrl. Proposition.3. If f : [,b] R is bounded, then Proof. Let A = {L(f;P) : P Π}, L(f) U(f). B = {U(f;P) : P Π}. From Proposition.2, b for every A nd b B, so Proposition 2.9 implies tht supa infb, or L(f) U(f)..4. The Cuchy criterion for integrbility The following theorem gives criterion for integrbility tht is nlogous to the Cuchy condition for the convergence of sequence. Theorem.4. A bounded function f : [,b] R is Riemnn integrble if nd only if for every ǫ > there exists prtition P of [,b], which my depend on ǫ, such tht U(f;P) L(f;P) < ǫ.

.4. The Cuchy criterion for integrbility 9 Proof. First, suppose tht the condition holds. Let ǫ > nd choose prtition P tht stisfies the condition. Then, since U(f) U(f;P) nd L(f;P) L(f), we hve U(f) L(f) U(f;P) L(f;P) < ǫ. Since this inequlity holds for every ǫ >, we must hve U(f) L(f) =, nd f is integrble. Conversely, suppose tht f is integrble. Given ny ǫ >, there re prtitions Q, R such tht U(f;Q) < U(f)+ ǫ 2, L(f;R) > L(f) ǫ 2. Let P be common refinement of Q nd R. Then, by Theorem., U(f;P) L(f;P) U(f;Q) L(f;R) < U(f) L(f)+ǫ. Since U(f) = L(f), the condition follows. If U(f;P) L(f;P) < ǫ, then U(f;Q) L(f;Q) < ǫ for every refinement Q of P, so the Cuchy condition mens tht function is integrble if nd only if its upper nd lower sums get rbitrrily close together for ll sufficiently refined prtitions. It is worth considering in more detil wht the Cuchy condition in Theorem.4 implies bout the behvior of Riemnn integrble function. Definition.5. The oscilltion of bounded function f on set A is k= osc f = sup f inf f. A A A If f : [,b] R is bounded nd P = {I,I 2,...,I n } is prtition of [,b], then n n n U(f;P) L(f;P) = supf inf f = oscf. A function f is Riemnn integrble if we cn mke U(f;P) L(f;P) s smll s we wish. This is the cse if we cn find sufficiently refined prtition P such tht the oscilltion of f on most intervls is rbitrrily smll, nd the sum of the lengths of the remining intervls (where the oscilltion of f is lrge) is rbitrrily smll. For exmple, the discontinuous function in Exmple.6 hs zero oscilltion on every intervl except the first one, where the function hs oscilltion one, but the length of tht intervl cn be mde s smll s we wish. Thus, roughly speking, function is Riemnn integrble if it oscilltes by n rbitrry smll mount except on finite collection of intervls whose totl length is rbitrrily smll. Theorem.87 gives precise sttement. One direct consequence of the Cuchy criterion is tht function is integrble if we cn estimte its oscilltion by the oscilltion of n integrble function. Proposition.6. Suppose tht f,g : [,b] R re bounded functions nd g is integrble on [,b]. If there exists constnt C such tht k= osc I f Cosc I g on every intervl I [,b], then f is integrble. k=

. The Riemnn Integrl Proof. If P = {I,I 2,...,I n } is prtition of [,b], then n [ ] U (f;p) L(f;P) = supf inff = C k= n k= osc f n k= osc g C[U(g;P) L(g;P)]. Thus, f stisfies the Cuchy criterion in Theorem.4 if g does, which proves tht f is integrble if g is integrble. We cn lso give sequentil chrcteriztion of integrbility. Theorem.7. A bounded function f : [,b] R is Riemnn integrble if nd only if there is sequence (P n ) of prtitions such tht In tht cse, lim [U(f;P n) L(f;P n )] =. n f = lim n U(f;P n) = lim n L(f;P n). Proof. First, suppose tht the condition holds. Then, given ǫ >, there is n n N such tht U(f;P n ) L(f;P n ) < ǫ, so Theorem.4 implies tht f is integrble nd U(f) = L(f). Furthermore, since U(f) U(f;P n ) nd L(f;P n ) L(f), we hve U(f;P n ) U(f) = U(f;P n ) L(f) U(f;P n ) L(f;P n ). Since the limit of the right-hnd side is zero, the squeeze theorem implies tht It lso follows tht lim U(f;P n) = U(f) = n lim L(f;P n) = lim U(f;P n) lim [U(f;P n) L(f;P n )] = n n n Conversely, if f is integrble then, by Theorem.4, for every n N there exists prtition P n such tht U(f;P n ) L(f;P n ) < n, f f. nd U(f;P n ) L(f;P n ) s n. Note tht if the limits of U(f;P n ) nd L(f;P n ) both exist nd re equl, then lim [U(f;P n) L(f;P n )] = lim U(f;P n) lim L(f;P n), n n n so the conditions of the theorem re stisfied. Conversely, the proof of the theorem shows tht if the limit of U(f;P n ) L(f;P n ) is zero, then the limits of U(f;P n )

.5. Integrbility of continuous nd monotonic functions nd L(f;P n ) both exist nd re equl. This isn t true for generl sequences, where one my hve lim( n b n ) = even though lim n nd limb n don t exist. Theorem.7 provides one wy to prove the existence of n integrl nd, in some cses, evlute it. Exmple.8. Let P n be the prtition of [,] into n-intervls of equl length /n with endpoints x k = k/n for k =,,2,...,n. If = [(k )/n,k/n] is the kth intervl, then sup f = x 2 k, inf = x 2 k since f is incresing. Using the formul for the sum of squres n k 2 = 6 n(n+)(2n+), we get nd U(f;P n ) = L(f;P n ) = n k= n k= k= (See Figure.8.) It follows tht x 2 k n = n 3 n k 2 = 6 k= x 2 k n = n n 3 k 2 = 6 k= lim U(f;P n) = lim L(f;P n) = n n 3, nd Theorem.7 implies tht x 2 is integrble on [,] with x 2 dx = 3. ( + )( 2+ ) n n ( )( 2 ). n n The fundmentl theorem of clculus, Theorem.45 below, provides much esier wy to evlute this integrl..5. Integrbility of continuous nd monotonic functions The Cuchy criterion leds to the following fundmentl result tht every continuous function is Riemnn integrble. To provethis, we use the fct tht continuous function oscilltes by n rbitrrily smll mount on every intervl of sufficiently refined prtition. Theorem.9. A continuous function f : [,b] R on compct intervl is Riemnn integrble. Proof. A continuous function on compct set is bounded, so we just need to verify the Cuchy condition in Theorem.4. Let ǫ >. A continuous function on compct set is uniformly continuous, so there exists δ > such tht f(x) f(y) < ǫ for ll x,y [,b] such tht x y < δ. b

2. The Riemnn Integrl Upper Riemnn Sum =.44.8.6 y.4.2.2.4.6.8 x Lower Riemnn Sum =.24.8.6 y.4.2.2.4.6.8 x Upper Riemnn Sum =.385.8.6 y.4.2.2.4.6.8 x Lower Riemnn Sum =.285.8.6 y.4.2.2.4.6.8 x Upper Riemnn Sum =.3434.8.6 y.4.2.2.4.6.8 x Lower Riemnn Sum =.3234.8.6 y.4.2.2.4.6.8 x Figure. Upper nd lower Riemnn sumsfor Exmple.8 with n = 5,,5 subintervls of equl length.

.5. Integrbility of continuous nd monotonic functions 3 Choose prtition P = {I,I 2,...,I n } of [,b] such tht < δ for every k; for exmple, we cn tke n intervls of equl length (b )/n with n > (b )/δ. Since f is continuous, it ttins its mximum nd minimum vlues M k nd m k on the compct intervl t points x k nd y k in. These points stisfy x k y k < δ, so M k m k = f(x k ) f(y k ) < ǫ b. The upper nd lower sums of f therefore stisfy n n U(f;P) L(f;P) = M k m k = k= k= n (M k m k ) k= < ǫ b < ǫ, nd Theorem.4 implies tht f is integrble. n Exmple.2. The function f(x) = x 2 on [,] considered in Exmple.8 is integrble since it is continuous. Another clss of integrble functions consists of monotonic (incresing or decresing) functions. Theorem.2. A monotonic function f : [,b] R on compct intervl is Riemnn integrble. Proof. Suppose tht f is monotonic incresing, mening tht f(x) f(y) for x y. Let P n = {I,I 2,...,I n } be prtition of [,b] into n intervls = [x k,x k ], of equl length (b )/n, with endpoints k= x k = +(b ) k n, k =,,...,n,n. Since f is incresing, M k = sup f = f(x k ), m k = inf f = f(x k ). Hence, summing telescoping series, we get n U(f;P n ) L(U;P n ) = (M k m k )(x k x k ) k= = b n n [f(x k ) f(x k )] = b n [f(b) f()]. It follows tht U(f;P n ) L(U;P n ) s n, nd Theorem.7 implies tht f is integrble. k=

4. The Riemnn Integrl.2.8 y.6.4.2.2.4.6.8 x Figure 2. The grph of the monotonic function in Exmple.22 with countbly infinite, dense set of jump discontinuities. The proof for monotonic decresing function f is similr, with sup f = f(x k ), inf f = f(x k ), or we cn pply the result for incresing functions to f nd use Theorem.23 below. Monotonic functions needn t be continuous, nd they my be discontinuous t countbly infinite number of points. Exmple.22. Let {q k : k N} be n enumertion of the rtionl numbers in [,) nd let ( k ) be sequence of strictly positive rel numbers such tht k =. Define f : [,] R by f(x) = k Q(x) k, k= Q(x) = {k N : q k [,x)}. for x >, nd f() =. Tht is, f(x) is obtined by summing the terms in the series whose indices k correspond to the rtionl numbers q k < x. For x =, this sum includes ll the terms in the series, so f() =. For every < x <, there re infinitely mny terms in the sum, since the rtionls re dense in [,x), nd f is incresing, since the number of terms increses with x. By Theorem.2, f is Riemnn integrble on [, ]. Although f is integrble, it hs countbly infinite number of jump discontinuities t every rtionl number in [,), which re dense in [,], The function is continuous elsewhere (the proof is left s n exercise).

.6. Properties of the Riemnn integrl 5 Figure 2 shows the grph of f corresponding to the enumertion {,/2,/3,2/3,/4,3/4,/5,2/5,3/5,4/5,/6,5/6,/7,...} of the rtionl numbers in [,) nd k = 6 π 2 k 2..6. Properties of the Riemnn integrl The integrl hs the following three bsic properties. () Linerity: cf = c f, (2) Monotonicity: if f g, then (3) Additivity: if < c < b, then c f + f c (f +g) = In this section, we prove these properties nd derive few of their consequences. These properties re nlogous to the corresponding properties of sums (or convergent series): n n n n n c k = c k, ( k +b k ) = k + b k ; k= n k k= m k + k= k= k= f = n b k if k b k ; k= n k=m+ k = n k. k=.6.. Linerity. We begin by proving the linerity. First we prove linerity with respect to sclr multipliction nd then linerity with respect to sums. Theorem.23. If f : [,b] R is integrble nd c R, then cf is integrble nd cf = c g. f. k= f + Proof. Suppose tht c. Then for ny set A [,b], we hve sup A cf = csup A f. f, inf cf = cinf f, A A so U(cf;P) = cu(f;p) for every prtition P. Tking the infimum over the set Π of ll prtitions of [,b], we get g. k= U(cf) = inf U(cf;P) = inf cu(f;p) = c inf U(f;P) = cu(f). P Π P Π P Π

6. The Riemnn Integrl Similrly, L(cf;P) = cl(f;p) nd L(cf) = cl(f). If f is integrble, then which shows tht cf is integrble nd we hve Now consider f. Since Therefore U(cf) = cu(f) = cl(f) = L(cf), sup( f) = inf f, A A U( f;p) = L(f;P), cf = c inf A f. ( f) = supf, A L( f;p) = U(f;P). U( f) = inf U( f;p) = inf [ L(f;P)] = sup L(f;P) = L(f), P Π P Π P Π U(f;P) = U(f). L( f) = sup L( f;p) = sup P Π P Π Hence, f is integrble if f is integrble nd [ U(f;P)] = inf P Π ( f) = Finlly, if c <, then c = c, nd successive ppliction of the previous results shows tht cf is integrble with cf = c f. Next, we prove the linerity of the integrl with respect to sums. If f, g re bounded, then f +g is bounded nd It follows tht sup I (f +g) sup I f +sup I f. g, inf(f +g) inf f +infg. I I I osc(f +g) osc f +oscg, I I I so f+g is integrble if f, g re integrble. In generl, however, the upper (or lower) sum of f +g needn t be the sum of the corresponding upper (or lower) sums of f nd g. As result, we don t get (f +g) = simply by dding upper nd lower sums. Insted, we prove this equlity by estimting the upper nd lower integrls of f +g from bove nd below by those of f nd g. Theorem.24. If f,g : [,b] R re integrblefunctions, then f+g is integrble, nd (f +g) = f + f + g g.

.6. Properties of the Riemnn integrl 7 Proof. We first prove tht if f,g : [,b] R re bounded, but not necessrily integrble, then U(f +g) U(f)+U(g), L(f +g) L(f)+L(g). Suppose tht P = {I,I 2,...,I n } is prtition of [,b]. Then n U(f +g;p) = sup(f +g) k= n k= sup f + n k= U(f;P)+U(g;P). sup g Let ǫ >. Since the upper integrl is the infimum of the upper sums, there re prtitions Q, R such tht U(f;Q) < U(f)+ ǫ 2, U(g;R) < U(g)+ ǫ 2, nd if P is common refinement of Q nd R, then It follows tht U(f;P) < U(f)+ ǫ 2, U(g;P) < U(g)+ ǫ 2. U(f +g) U(f +g;p) U(f;P)+U(g;P) < U(f)+U(g)+ǫ. Sincethisinequlityholdsforrbitrryǫ >, wemusthveu(f+g) U(f)+U(g). Similrly, we hve L(f +g;p) L(f;P)+L(g;P) for ll prtitions P, nd for every ǫ >, we get L(f +g) > L(f)+L(g) ǫ, so L(f +g) L(f)+L(g). For integrble functions f nd g, it follows tht U(f +g) U(f)+U(g) = L(f)+L(g) L(f +g). Since U(f +g) L(f +g), we hve U(f +g) = L(f +g) nd f +g is integrble. Moreover, there is equlity throughout the previous inequlity, which proves the result. Although the integrl is liner, the upper nd lower integrls of non-integrble functions re not, in generl, liner. Exmple.25. Define f,g : [,] R by { if x [,] Q, f(x) = g(x) = if x [,]\Q, Tht is, f is the Dirichlet function nd g = f. Then so { if x [,] Q, if x [,]\Q. U(f) = U(g) =, L(f) = L(g) =, U(f +g) = L(f +g) =, U(f +g) < U(f)+U(g), L(f +g) > L(f)+L(g). The product of integrble functions is lso integrble, s is the quotient provided it remins bounded. Unlike the integrl of the sum, however, there is no wy to express the integrl of the product fg in terms of f nd g.

8. The Riemnn Integrl Theorem.26. If f,g : [,b] R re integrble, then fg : [,b] R is integrble. If, in ddition, g nd /g is bounded, then f/g : [,b] R is integrble. Proof. First, we show tht the squre of n integrble function is integrble. If f is integrble, then f is bounded, with f M for some M. For ll x,y [,b], we hve f 2 (x) f 2 (y) = f(x)+f(y) f(x) f(y) 2M f(x) f(y). Tking the supremum of this inequlity over x,y I [,b] nd using Proposition 2.9, we get tht [ ]. sup(f 2 ) inf I I (f2 ) 2M supf inf f I I mening tht osc(f 2 ) 2M osc f. I I If follows from Proposition.6 tht f 2 is integrble if f is integrble. Since the integrl is liner, we then see from the identity fg = [ (f +g) 2 (f g) 2] 4 tht fg is integrble if f, g re integrble. In similr wy, if g nd /g M, then g(x) g(y) = g(x) g(y) M 2 g(x) g(y). g(x)g(y) Tking the supremum of this eqution over x,y I [,b], we get ( ) ( ) [ ] sup inf M 2 supg inf I g I g g, I I mening tht osc I (/g) M 2 osc I g, nd Proposition.6 implies tht /g is integrble if g is integrble. Therefore f/g = f (/g) is integrble..6.2. Monotonicity. Next, we prove the monotonicity of the integrl. Theorem.27. Suppose tht f,g : [,b] R re integrble nd f g. Then f Proof. First suppose tht f is integrble. Let P be the prtition consisting of the single intervl [, b]. Then so g. L(f;P) = inf f (b ), [,b] f L(f;P). If f g, then h = f g, nd the linerity of the integrl implies tht f g = h,

.6. Properties of the Riemnn integrl 9 which proves the theorem. One immedite consequence of this theorem is the following simple, but useful, estimte for integrls. Theorem.28. Suppose tht f : [,b] R is integrble nd Then M = supf, m = inf f. [,b] [,b] m(b ) f M(b ). Proof. Since m f M on [,b], Theorem.27 implies tht which gives the result. m f M, This estimte lso follows from the definition of the integrl in terms of upper nd lower sums, but once we ve estblished the monotonicity of the integrl, we don t need to go bck to the definition. A further consequence is the intermedite vlue theorem for integrls, which sttes tht continuous function on n intervl is equl to its vergevlue t some point. Theorem.29. If f : [,b] R is continuous, then there exists x [,b] such tht f(x) = f. b Proof. Since f is continuous function on compct intervl, it ttins its mximum vlue M nd its minimum vlue m. From Theorem.28, m b f M. By the intermedite vlue theorem, f tkes on every vlue between m nd M, nd the result follows. As shown in the proof of Theorem.27, given linerity, monotonicity is equivlent to positivity, f if f. We remrk tht even though the upper nd lower integrls ren t liner, they re monotone. Proposition.3. If f,g : [,b] R re bounded functions nd f g, then U(f) U(g), L(f) L(g).

2. The Riemnn Integrl Proof. From Proposition 2.2, we hve for every intervl I [,b] tht sup I f supg, inf f inf g. I I I It follows tht for every prtition P of [,b], we hve U(f;P) U(g;P), L(f;P) L(g;P). Tking the infimum of the upper inequlity nd the supremum of the lower inequlity over P, we get U(f) U(g) nd L(f) L(g). We cn estimte the bsolute vlue of n integrl by tking the bsolute vlue under the integrl sign. This is nlogous to the corresponding property of sums: n n n k. k= k= Theorem.3. If f is integrble, then f is integrble nd f f. Proof. First, suppose tht f is integrble. Since f f f, we get from Theorem.27 tht f f f, or f f. To complete the proof, we need to show tht f is integrble if f is integrble. For x,y [,b], the reverse tringle inequlity gives Using Proposition 2.9, we get tht f(x) f(y) f(x) f(y). sup I f inf I f supf inf f, I I mening tht osc I f osc I f. Proposition.6 then implies tht f is integrble if f is integrble. In prticulr, we immeditely get the following bsic estimte for n integrl. Corollry.32. If f : [,b] R is integrble then M = sup f, [,b] f M(b ).

.6. Properties of the Riemnn integrl 2.6.3. Additivity. Finlly, we prove dditivity. This property refers to dditivity with respect to the intervl of integrtion, rther thn linerity with respect to the function being integrted. Theorem.33. Suppose tht f : [,b] R nd < c < b. Then f is Riemnn integrble on [,b] if nd only if it is Riemnn integrble on [,c] nd [c,b]. In tht cse, f = c Proof. Suppose tht f is integrbleon [,b]. Then, givenǫ >, there is prtition P of [,b] such tht U(f;P) L(f;P) < ǫ. Let P = P {c} be the refinement of P obtined by dding c to the endpoints of P. (If c P, then P = P.) Then P = Q R where Q = P [,c] nd R = P [c,b] re prtitions of [,c] nd [c,b] respectively. Moreover, It follows tht U(f;P ) = U(f;Q)+U(f;R), f + c f. L(f;P ) = L(f;Q)+L(f;R). U(f;Q) L(f;Q) = U(f;P ) L(f;P ) [U(f;R) L(f;R)] U(f;P) L(f;P) < ǫ, which proves tht f is integrble on [,c]. Exchnging Q nd R, we get the proof for [c,b]. Conversely, if f is integrble on [,c] nd [c,b], then there re prtitions Q of [,c] nd R of [c,b] such tht U(f;Q) L(f;Q) < ǫ 2, U(f;R) L(f;R) < ǫ 2. Let P = Q R. Then U(f;P) L(f;P) = U(f;Q) L(f;Q)+U(f;R) L(f;R) < ǫ, which proves tht f is integrble on [,b]. Finlly, with the prtitions P, Q, R s bove, we hve Similrly, f U(f;P) = U(f;Q)+U(f;R) < L(f;Q)+L(f;R)+ǫ < c f + c f +ǫ. f L(f;P) = L(f;Q)+L(f;R) > U(f;Q)+U(f;R) ǫ > c f + c f ǫ. Since ǫ > is rbitrry, we see tht f = c f + c f.

22. The Riemnn Integrl We cn extend the dditivity property of the integrl by defining n oriented Riemnn integrl. Definition.34. If f : [,b] R is integrble, where < b, nd c b, then f = b f, c c f =. With this definition, the dditivity property in Theorem.33 holds for ll,b,c R for which the oriented integrls exist. Moreover, if f M, then the estimte in Corollry.32 becomes f M b for ll,b R (even if b). The oriented Riemnn integrl is specil cse of the integrl of differentil form. It ssigns vlue to the integrl of one-form f dx on n oriented intervl..7. Further existence results for the Riemnn integrl In this section, we prove severl further useful conditions for the existences of the Riemnn integrl. First, we show tht chnging the vlues of function t finitely mny points doesn t chnge its integrbility of the vlue of its integrl. Proposition.35. Suppose tht f,g : [,b] R nd f(x) = g(x) except t finitely mny points x [,b]. Then f is integrble if nd only if g is integrble, nd in tht cse f = Proof. It is sufficient to prove the result for functions whose vlues differ t single point, sy c [,b]. The generl result then follows by induction. Since f, g differ t single point, f is bounded if nd only if g is bounded. If f, g re unbounded, then neither one is integrble. If f, g re bounded, we will show tht f, g hve the sme upper nd lower integrls becuse their upper nd lower sums differ by n rbitrrily smll mount with respect to prtition tht is sufficiently refined ner the point where the functions differ. Suppose tht f, g re bounded with f, g M on [,b] for some M >. Let ǫ >. Choose prtition P of [,b] such tht g. U(f;P) < U(f)+ ǫ 2. Let Q = {I,...,I n } be refinement of P such tht < δ for k =,...,n, where δ = ǫ 8M.

.7. Further existence results for the Riemnn integrl 23 Then g differs from f on t most two intervls in Q. (There could be two intervls if c is n endpoint of the prtition.) On such n intervl we hve sup g supf sup g +sup f 2M, nd on the remining intervls, sup Ik g sup Ik f =. It follows tht U(g;Q) U(f;Q) < 2M 2δ < ǫ 2. Using the properties of upper integrls nd refinements, we obtin U(g) U(g;Q) < U(f;Q)+ ǫ 2 U(f;P)+ ǫ 2 < U(f)+ǫ. Since this inequlity holds for rbitrry ǫ >, we get tht U(g) U(f). Exchnging f nd g, we see similrly tht U(f) U(g), so U(f) = U(g). An nlogous rgument for lower sums (or n ppliction of the result for upper sums to f, g) shows tht L(f) = L(g). Thus U(f) = L(f) if nd only if U(g) = L(g), in which cse f = g. Exmple.36. The function f in Exmple.6 differs from the -function t one point. It is integrble nd its integrl is equl to. The conclusion of Proposition.35 cn fil if the functions differ t countbly infinite number of points. One reson is tht we cn turn bounded function into n unbounded function by chnging its vlues t n infinite number of points. Exmple.37. Define f : [,] R by { n if x = /n for n N, f(x) = otherwise Then f is equl to the -function except on the countbly infinite set {/n : n N}, but f is unbounded nd therefore it s not Riemnn integrble. The result is still flse, however, for bounded functions tht differ t countbly infinite number of points. Exmple.38. The Dirichlet function in Exmple.7 is bounded nd differs from the -function on the countbly infinite set of rtionls, but it isn t Riemnn integrble. The Lebesgue integrl is better behved thn the Riemnn intgerl in this respect: two functions tht re equl lmost everywhere, mening tht they differ on set of Lebesgue mesure zero, hve the sme Lebesgue integrls. In prticulr, two functions tht differ on countble set re equl lmost everywhere (see Section.2). The next proposition llows us to deduce the integrbility of bounded function on n intervl from its integrbility on slightly smller intervls.

24. The Riemnn Integrl Proposition.39. Suppose tht f : [,b] R is bounded nd integrble on [,r] for every < r < b. Then f is integrble on [,b] nd f = lim r b Proof. Since f is bounded, f M on [,b] for some M >. Given ǫ >, let r = b ǫ 4M (where we ssume ǫ is sufficiently smll tht r > ). Since f is integrble on [,r], there is prtition Q of [,r] such tht r f. U(f;Q) L(f;Q) < ǫ 2. ThenP = Q {b}isprtitionof[,b]whoselstintervlis[r,b]. The boundedness of f implies tht supf inf f 2M. [r,b] [r,b] Therefore ( U(f;P) L(f;P) = U(f;Q) L(f;Q)+ sup [r,b] ) f inf f (b r) [r,b] < ǫ +2M (b r) = ǫ, 2 so f is integrble on [,b] by Theorem.4. Moreover, using the dditivity of the integrl, we get r f f = f M (b r) s r b. r An obvious nlogous result holds for the left endpoint. Exmple.4. Define f : [,] R by { sin(/x) if < x, f(x) = if x =. Then f is bounded on [, ]. Furthemore, f is continuous nd therefore integrble on [r,] for every < r <. It follows from Proposition.39 tht f is integrble on [,]. The ssumption in Proposition.39 tht f is bounded on [,b] is essentil. Exmple.4. The function f : [,] R defined by { /x for < x, f(x) = for x =, is continuous nd therefore integrble on [r,] for every < r <, but it s unbounded nd therefore not integrble on [, ].

.7. Further existence results for the Riemnn integrl 25.5 y.5 2 3 4 5 6 x Figure 3. Grph of the Riemnn integrble function y = sin(/sinx) in Exmple.43. As corollry of this result nd the dditivity of the integrl, we prove generliztion of the integrbility of continuous functions to piecewise continuous functions. Theorem.42. If f : [,b] R is bounded function with finitely mny discontinuities, then f is Riemnn integrble. Proof. By splitting the intervl into subintervls with the discontinuities of f t n endpoint nd using Theorem.33, we see tht it is sufficient to prove the result if f is discontinuous only t one endpoint of [,b], sy t b. In tht cse, f is continuous nd therefore integrble on ny smller intervl [,r] with < r < b, nd Proposition.39 implies tht f is integrble on [, b]. Exmple.43. Define f : [,2π] R by { sin(/sinx) if x,π,2π, f(x) = if x =,π,2π. Thenf isboundedndcontinuousexcepttx =,π,2π,soitisintegrbleon[,2π] (see Figure 3). This function doesn t hve jump discontinuities, but Theorem.42 still pplies. Exmple.44. Define f : [,/π] R by { sgn[sin(/x)] if x /nπ for n N, f(x) = if x = or x /nπ for n N,

26. The Riemnn Integrl.8.6.4.2.2.4.6.8.5..5.2.25.3 Figure 4. Grph of the Riemnn integrble function y = sgn(sin(/x)) in Exmple.44. where sgn is the sign function, if x >, sgnx = if x =, if x <. Then f oscilltes between nd countbly infinite number of times s x + (see Figure 4). It hs jump discontinuities t x = /(nπ) nd n essentil discontinuity t x =. Nevertheless, it is Riemnn integrble. To see this, note tht f is bounded on [, ] nd piecewise continuous with finitely mny discontinuities on [r,] for every < r <. Theorem.42 implies tht f is Riemnn integrble on [r,], nd then Theorem.39 implies tht f is integrble on [,]..8. The fundmentl theorem of clculus In the integrl clculus I find much less interesting the prts tht involve only substitutions, trnsformtions, nd the like, in short, the prts tht involve the known skillfully pplied mechnics of reducing integrls to lgebric, logrithmic, nd circulr functions, thn I find the creful nd profound study of trnscendentl functions tht cnnot be reduced to these functions. (Guss, 88) The fundmentl theorem of clculus sttes tht differentition nd integrtion re inverse opertions in n ppropritely understood sense. The theorem hs two prts: in one direction, it sys roughly tht the integrl of the derivtive is the originl function; in the other direction, it sys tht the derivtive of the integrl is the originl function.

.8. The fundmentl theorem of clculus 27 In more detil, the first prt sttes tht if F : [,b] R is differentible with integrble derivtive, then F (x)dx = F(b) F(). This result cn be thought of s continuous nlog of the corresponding identity for sums of differences, n (A k A k ) = A n A. k= The second prt sttes tht if f : [,b] R is continuous, then d dx x f(t)dt = f(x). This is continuous nlog of the corresponding identity for differences of sums, k k j j = k. j= j= The proof of the fundmentl theorem consists essentilly of pplying the identities for sums or differences to the pproprite Riemnn sums or difference quotients nd proving, under pproprite hypotheses, tht they converge to the corresponding integrls or derivtives. We ll split the sttement nd proof of the fundmentl theorem into two prts. (The numbering of the prts s I nd II is rbitrry.).8.. Fundmentl theorem I. First we prove the sttement bout the integrl of derivtive. Theorem.45 (FundmentltheoremofclculusI). IfF : [,b] Riscontinuous on [,b] nd differentible in (,b) with F = f where f : [,b] R is Riemnn integrble, then Proof. Let be prtition of [,b]. Then f(x)dx = F(b) F(). P = { = x,x,x 2,...,x n,x n = b} F(b) F() = n [F(x k ) F(x k )]. k= The function F is continuous on the closed intervl [x k,x k ] nd differentible in the open intervl (x k,x k ) with F = f. By the men vlue theorem, there exists x k < c k < x k such tht F(x k ) F(x k ) = f(c k )(x k x k ). Since f is Riemnn integrble, it is bounded, nd it follows tht m k (x k x k ) F(x k ) F(x k ) M k (x k x k ),

28. The Riemnn Integrl where M k = sup f, m k = inf f. [x k,x k ] [x k,x k ] Hence, L(f;P) F(b) F() U(f;P) for every prtition P of [,b], which implies tht L(f) F(b) F() U(f). Since f is integrble, L(f) = U(f) nd F(b) F() = f. In Theorem.45, we ssume tht F is continuous on the closed intervl [,b] nd differentible in the open intervl (, b) where its usul two-sided derivtive is defined nd is equl to f. It isn t necessry to ssume the existence of the right derivtive of F t or the left derivtive t b, so the vlues of f t the endpoints re rbitrry. By Proposition.35, however, the integrbility of f on [, b] nd the vlue of its integrl do not depend on these vlues, so the sttement of the theorem mkes sense. As result, we ll sometimes buse terminology, nd sy tht F is integrble on [,b] even if it s only defined on (,b). Theorem.45 imposes the integrbility of F s hypothesis. Every function F tht is continuously differentible on the closed intervl [, b] stisfies this condition, but the theorem remins true even if F is discontinuous, Riemnn integrble function. Exmple.46. Define F : [,] R by { x 2 sin(/x) if < x, F(x) = if x =. Then F is continuous on [, ] nd, by the product nd chin rules, differentible in (,]. It is lso differentible but not continuously differentible t, with F ( + ) =. Thus, { F cos(/x)+2xsin(/x) if < x, (x) = if x =. The derivtive F is bounded on [,] nd discontinuous only t one point (x = ), so Theorem.42 implies tht F is integrble on [,]. This verifies ll of the hypotheses in Theorem.45, nd we conclude tht F (x)dx = sin. There re, however, differentible functions whose derivtives re unbounded or so discontinuous tht they ren t Riemnn integrble. Exmple.47. Define F : [,] R by F(x) = x. Then F is continuous on [,] nd differentible in (,], with F (x) = 2 x for < x. This function is unbounded, so F is not Riemnn integrble on [,], however we define its vlue t, nd Theorem.45 does not pply.

.8. The fundmentl theorem of clculus 29 We cn, however, interpret the integrl of F on [,] s n improper Riemnn integrl. The function F is continuously differentible on [ǫ,] for every < ǫ <, so 2 x dx = ǫ. Thus, we get the improper integrl ǫ lim ǫ + ǫ 2 dx =. x The construction of function with bounded, non-integrble derivtive is more involved. It s not sufficient to give function with bounded derivtive tht is discontinuous t finitely mny points, s in Exmple.46, becuse such function is Riemnn integrble. Rther, one hs to construct differentible function whose derivtive is discontinuous on set of nonzero Lebesgue mesure; we won t give n exmple here. Finlly, we remrk tht Theorem.45 remins vlid for the oriented Riemnn integrl, since exchnging nd b reverses the sign of both sides..8.2. Fundmentl theorem of clculus II. Next, we prove the other direction of the fundmentl theorem. We will use the following result, of independent interest, which sttes tht the verge of continuous function on n intervl pproches the vlue of the function s the length of the intervl shrinks to zero. The proof uses common trick of tking constnt inside n verge. Theorem.48. Suppose tht f : [,b] R is integrble on [,b] nd continuous t. Then +h lim f(x)dx = f(). h + h Proof. If k is constnt, we hve k = h +h kdx. (Tht is, the vergeof constnt is equl to the constnt.) We cn therefore write h +h f(x)dx f() = h +h [f(x) f()] dx. Let ǫ >. Since f is continuous t, there exists δ > such tht f(x) f() < ǫ for x < +δ. It follows tht if < h < δ, then +h f(x)dx f() h h sup +h f(x) f() h ǫ, which proves the result.

3. The Riemnn Integrl A similr proof shows tht if f is continuous t b, then b lim f = f(b), h + h b h nd if f is continuous t < c < b, then c+h lim f = f(c). h + 2h c h More generlly, if {I h : h > } is ny collection of intervls with c I h nd I h s h +, then lim f = f(c). h + I h I h The ssumption in Theorem.48 tht f is continuous t the point bout which we tke the verges is essentil. Exmple.49. Let f : R R be the sign function if x >, f(x) = if x =, if x <. Then h lim f(x)dx =, lim f(x)dx =, h + h h + h h nd neither limit is equl to f(). In this exmple, the limit of the symmetric verges h lim f(x)dx = h + 2h h is equl to f(), but this equlity doesn t hold if we chnge f() to nonzero vlue, since the limit of the symmetric verges is still. The second prt of the fundmentl theorem follows from this result nd the fct tht the difference quotients of F re verges of f. Theorem.5 (Fundmentl theorem of clculus II). Suppose tht f : [,b] R is integrble nd F : [,b] R is defined by F(x) = x f(t)dt. Then F is continuous on [,b]. Moreover, if f is continuous t c b, then F is differentible t c nd F (c) = f(c). Proof. First, note tht Theorem.33 implies tht f is integrble on [,x] for every x b, so F is well-defined. Since f is Riemnn integrble, it is bounded, nd f M for some M. It follows tht x+h F(x+h) F(x) = f(t)dt M h, which shows tht F is continuous on [, b] (in fct, Lipschitz continuous). x

.8. The fundmentl theorem of clculus 3 Moreover, we hve F(c+h) F(c) h = h c+h c f(t)dt. It follows from Theorem.48 tht if f is continuous t c, then F is differentible t c with [ ] F(c+h) F(c) F c+h (c) = lim = lim f(t)dt = f(c), h h h h where we use the pproprite right or left limit t n endpoint. The ssumption tht f is continuous is needed to ensure tht F is differentible. Exmple.5. If then F(x) = f(x) = x { for x, for x <, f(t)dt = c { x for x, for x <. The function F is continuous but not differentible t x =, where f is discontinuous, since the left nd right derivtives of F t, given by F ( ) = nd F ( + ) =, re different..8.3. Consequences of the fundmentl theorem. The first prt of the fundmentl theorem, Theorem.45, is the bsic computtionl tool in integrtion. It llows us to compute the integrl of of function f if we cn find n ntiderivtive; tht is, function F such tht F = f. There is no systemtic procedure for finding ntiderivtives. Moreover, even if one exists, n ntiderivtive of n elementry function (constructed from power, trigonometric, nd exponentil functions nd their inverses) my not be nd often isn t expressible in terms of elementry functions. Exmple.52. For p =,,2,..., we hve [ ] d dx p+ xp+ = x p, nd it follows tht x p dx = p+. We remrk tht once we hve the fundmentl theorem, we cn use the definition of the integrl bckwrds to evlute limit such s [ ] n lim n n p+ k p = p+, k= since the sum is the upper sum of x p on prtition of [,] into n intervls of equl length. Exmple.8 illustrtes this result explicitly for p = 2.

32. The Riemnn Integrl Two importnt generl consequences of the first prt of the fundmentl theorem re integrtion by prts nd substitution (or chnge of vrible), which come from inverting the product rule nd chin rule for derivtives, respectively. Theorem.53 (Integrtion by prts). Suppose tht f,g : [,b] R re continuous on [,b] nd differentible in (,b), nd f, g re integrble on [,b]. Then fg dx = f(b)g(b) f()g() f gdx. Proof. The function fg is continuous on [,b] nd, by the product rule, differentible in (, b) with derivtive (fg) = fg +f g. Since f, g, f, g re integrble on [,b], Theorem.26 implies tht fg, f g, nd (fg), re integrble. From Theorem.45, we get tht fg dx+ which proves the result. f gdx = f gdx = f(b)g(b) f()g(), Integrtion by prts sys tht we cn move derivtive from one fctor in n integrl onto the other fctor, with chnge of sign nd the ppernce of boundry term. The product rule for derivtives expresses the derivtive of product in terms of the derivtives of the fctors. By contrst, integrtion by prts doesn t give n explicit expression for the integrl of product, it simply replces one integrl by nother. This cn sometimes be used to simplify n integrl nd evlute it, but the importnce of integrtion by prts goes fr beyond its use s n integrtion technique. Exmple.54. For n =,,2,3,..., let I n (x) = x t n e t dt. If n, integrtion by prts with f(t) = t n nd g (t) = e t gives I n (x) = x n e x +n x Also, by the fundmentl theorem, I (x) = It then follows by induction tht where, s usul,! =. x I n (x) = n! t n e t dt = x n e x +ni n (x). [ e t dt = e x. e x n k= ] x k, k! Since x k e x s x for every k =,,2,..., we get the improper integrl t n e t dt = lim r r t n e t dt = n!.

.8. The fundmentl theorem of clculus 33 This formul suggests n extension of the fctoril function to complex numbers z C, clled the Gmm function, which is defined for Rz > by the improper, complex-vlued integrl Γ(z) = t z e t dt. In prticulr, Γ(n) = (n )! for n N. The Gm function is n importnt specil function, which is studied further in complex nlysis. Next we consider the chnge of vrible formul for integrls. Theorem.55 (Chnge of vrible). Suppose tht g : I R differentible on n open intervl I nd g is integrble on I. Let J = g(i). If f : J R continuous, then for every,b I, Proof. Let f (g(x))g (x)dx = F(x) = x g(b) g() f(u)du. f(u)du. Since f is continuous, Theorem.5 implies tht F is differentible in J with F = f. The chin rule implies tht the composition F g : I R is differentible in I, with (F g) (x) = f (g(x))g (x). This derivtive is integrble on [,b] since f g is continuous nd g is integrble. Theorem.45, the definition of F, nd the dditivity of the integrl then imply tht which proves the result. f (g(x))g (x)dx = (F g) dx = F (g(b)) F (g()) = g(b) g() F (u)du, A continuous function mps n intervl to n intervl, nd it is one-to-one if nd only if it is strictly monotone. An incresing function preserves the orienttion of the intervl, while decresing function reverses it, in which cse the integrls re understood s pproprite oriented integrls. There is no ssumption in this theorem tht g is invertible, nd the result remins vlid if g is not monotone. Exmple.56. For every >, the incresing, differentible function g : R R defined by g(x) = x 3 mps (,) one-to-one nd onto ( 3, 3 ) nd preserves orienttion. Thus, if f : [,] R is continuous, f(x 3 ) 3x 2 dx = 3 3 f(u)du.

34. The Riemnn Integrl.5.5 y.5.5 2.5.5.5.5 2 x Figure 5. Grphs of the error function y = F(x) (blue) nd its derivtive, the Gussin function y = f(x) (green), from Exmple.58. The decresing, differentible function g : R R defined by g(x) = x 3 mps (,) one-to-one nd onto ( 3, 3 ) nd reverses orienttion. Thus, f( x 3 ) ( 3x 2 )dx = 3 3 f(u)du = 3 3 f(u)du. The non-monotone, differentible function g : R R defined by g(x) = x 2 mps (,) onto [, 2 ). It is two-to-one, except t x =. The chnge of vribles formul gives f(x 2 ) 2xdx = f(u)du =. 2 The contributions to the originl integrl from [,] nd [,] cncel since the integrnd is n odd function of x. One consequence of the second prt of the fundmentl theorem, Theorem.5, is tht every continuous function hs n ntiderivtive, even if it cn t be expressed explicitly in terms of elementry functions. This provides wy to define trnscendentl functions s integrls of elementry functions. Exmple.57. One wy to define the logrithm ln : (, ) R in terms of lgebric functions is s the integrl lnx = x 2 t dt. The integrl is well-defined for every < x < since /t is continuous on the intervl [,x] (or [x,] if < x < ). The usul properties of the logrithm follow from this representtion. We hve (lnx) = /x by definition, nd, for exmple, mking the substitution s = xt in the second integrl in the following eqution,

.8. The fundmentl theorem of clculus 35.8.6.4.2 y.2.4.6.8 2 4 6 8 x Figure 6. Grphs of the Fresnel integrl y = S(x) (blue) nd its derivtive y = sin(πx 2 /2) (green) from Exmple.59. when dt/t = ds/s, we get lnx+lny = x t dt+ y x t dt = xy t dt+ xy x s ds = dt = ln(xy). t We cn lso define mny non-elementry functions s integrls. Exmple.58. The error function erf(x) = 2 x e t2 dt π is n nti-derivtive on R of the Gussin function f(x) = 2 e x2. π The error function isn t expressible in terms of elementry functions. Nevertheless, it is defined s limit of Riemnn sums for the integrl. Figure 5 shows the grphs of f nd F. The nme error function comes from the fct tht the probbility of Gussin rndom vrible deviting by more thn given mount from its men cn be expressed in terms of F. Error functions lso rise in other pplictions; for exmple, in modeling diffusion processes such s het flow. Exmple.59. The Fresnel sine function S is defined by x ( ) πt 2 S(x) = sin dt. 2 The function S is n ntiderivtive of sin(πt 2 /2) on R (see Figure 6), but it cn t be expressed in terms of elementry functions. Fresnel integrls rise, mong other plces, in nlysing the diffrction of wves, such s light wves. From the perspective of complex nlysis, they re closely relted to the error function through the Euler formul e iθ = cosθ +isinθ.

36. The Riemnn Integrl 8 6 4 2 2 4 2 2 3 Figure 7. Grphs of the exponentil integrl y = Ei(x) (blue) nd its derivtive y = e x /x (green) from Exmple.6. Exmple.6. The exponentil integrl Ei is non-elementry function defined by x e t Ei(x) = t dt. Its grph is shown in Figure 7. This integrl hs to be understood, in generl, s n improper, principl vlue integrl, nd the function hs logrithmic singulrity t x = (see Exmple.83 below for further explntion). The exponentil integrl rises in physicl pplictions such s het flow nd rditive trnsfer. It is lso relted to the logrithmic integrl li(x) = x by li(x) = Ei(lnx). The logrithmic integrl is importnt in number theory, nd it gives n symptotic pproximtion for the number of primes less thn x s x. Roughlyspeking, thedensityofthe primesnerlrgenumberxiscloseto/lnx. Discontinuous functions my or my not hve n ntiderivtive, nd typiclly they don t. Drboux provedtht everyfunction f : (,b) R tht is the derivtive of function F : (,b) R, where F = f t every point of (,b), hs the intermedite vlue property. Tht is, if < c < d < b, then for every y between f(c) nd f(d) there exists n x between c nd d such tht f(x) = y. A continuous derivtive hs this property by the intermedite vlue theorem, but discontinuous derivtive lso hs it. Thus, functions without the intermedite vlue property, such s ones with jump discontinuity or the Dirichlet function, don t hve n ntiderivtive. For exmple, the function F in Exmple.5 is not n ntiderivtive of the step function f on R since it isn t differentible t. In deling with functions tht re not continuously differentible, it turns out to be more useful to bndon the ide of derivtive tht is defined pointwise dt lnt

.9. Integrls nd sequences of functions 37 everywhere (pointwise vlues of discontinuous functions re somewht rbitrry) nd introduce the notion of wek derivtive. We won t define or study wek derivtives here..9. Integrls nd sequences of functions A fundmentl question tht rises throughout nlysis is the vlidity of n exchnge in the order of limits. Some sort of condition is lwys required. In this section, we consider the question of when the convergence of sequence of functions f n f implies the convergence of their integrls f n f. There re mny inequivlent notions of the convergence of functions. The two we ll discuss here re pointwise nd uniform convergence. Recll tht if f n,f : A R, then f n f pointwise on A s n if f n (x) f(x) for every x A. On the other hnd, f n f uniformly on A if for every ǫ > there exists N N such tht n > N implies tht f n (x) f(x) < ǫ for every x A. Equivlently, f n f uniformly on A if f n f s n, where f = sup{ f(x) : x A} denotes the sup-norm of function f : A R. Uniform convergence implies pointwise convergence, but not conversely. As we show first, the Riemnn integrl is well-behved with respect to uniform convergence. The drwbck to uniform convergence is tht it s strong form of convergence, nd we often wnt to use weker form, such s pointwise convergence, in which cse the Riemnn integrl my not be suitble..9.. Uniform convergence. The uniform limit of continuous functions is continuous nd therefore integrble. The next result shows, more generlly, tht the uniform limit of integrble functions is integrble. Furthermore, the limit of the integrls is the integrl of the limit. Theorem.6. Suppose tht f n : [,b] R is Riemnn integrble for ech n N ndf n f uniformlyon[,b]sn. Thenf : [,b] RisRiemnnintegrble on [,b] nd f = lim n Proof. The uniform limit of bounded functions is bounded, so f is bounded. The min sttement we need to prove is tht f is integrble. Let ǫ >. Since f n f uniformly, there is n N N such tht if n > N then f n (x) ǫ b < f(x) < f n(x)+ ǫ for ll x b. b It follows from Proposition.3 tht ( L f n ǫ ) L(f), b f n. U(f) U ( f n + ǫ ). b

38. The Riemnn Integrl Since f n is integrble nd upper integrls re greter thn lower integrls, we get tht which implies tht f n ǫ L(f) U(f) U(f) L(f) 2ǫ. f n +ǫ, Sinceǫ > isrbitrry,weconcludethtl(f) = U(f),sof isintegrble. Moreover, it follows tht for ll n > N we hve f n f ǫ, which shows tht f n f s n. Once we know tht the uniform limit of integrble functions is integrble, the convergence of the integrls lso follows directly from the estimte f n f = (f n f) f n f (b ) s n. Exmple.62. The function f n : [,] R defined by f n (x) = n+cosx ne x +sinx converges uniformly on [, ] to f(x) = e x, since for x n+cosx ne x +sinx e x = cosx e x sinx ne x +sinx 2 n. It follows tht n+cosx lim n ne x +sinx dx = Exmple.63. Every power series e x dx = e. f(x) = + x+ 2 x 2 + + n x n +... with rdius of convergence R > converges uniformly on compct intervls inside the intervl x < R, so we cn integrte it term-by-term to get x f(t)dt = x+ 2 x 2 + 3 2x 3 + + n+ nx n+ +... for x < R. As one exmple, if we integrte the geometric series x = +x+x2 + +x n +... for x <, we get power series for ln, ( ) ln = x+ x 2 x2 + 3 x3 + n xn +... for x <.

.9. Integrls nd sequences of functions 39 For instnce, tking x = /2, we get the rpidly convergent series ln2 = n2 n n= for the irrtionl number ln2.693. This series ws known nd used by Euler. Although we cn integrte uniformly convergent sequences, we cnnot in generl differentite them. In fct, it s often esier to prove results bout the convergence of derivtives by using results bout the convergence of integrls, together with the fundmentl theorem of clculus. The following theorem provides sufficient conditions for f n f to imply tht f n f. Theorem.64. Let f n : (,b) R be sequence of differentible functions whose derivtives f n : (,b) R re integrble on (,b). Suppose tht f n f pointwise nd f n g uniformly on (,b) s n, where g : (,b) R is continuous. Then f : (,b) R is continuously differentible on (,b) nd f = g. Proof. Choose some point < c < b. Since f n is integrble, the fundmentl theorem of clculus, Theorem.45, implies tht f n (x) = f n (c)+ x c f n for < x < b. Since f n f pointwise nd f n g uniformly on [,x], we find tht f(x) = f(c)+ Since g is continuous, the other direction of the fundmentl theorem, Theorem.5, implies tht f is differentible in (,b) nd f = g. In prticulr, this theorem shows tht the limit of uniformly convergent sequence of continuously differentible functions whose derivtives converge uniformly is lso continuously differentible. The key ssumption in Theorem.64 is tht the derivtives f n converge uniformly, not just pointwise; the result is flse if we only ssume pointwise convergence of the f n. In the proof of the theorem, we only use the ssumption tht f n (x) converges t single point x = c. This ssumption together with the ssumption tht f n g uniformly implies tht f n f pointwise (nd, in fct, uniformly) where x f(x) = lim n f n(c)+ Thus, the theorem reminstrue if we replcethe ssumption tht f n f pointwise on (,b) by the weker ssumption tht lim n f n (c) exists for some c (,b). This isn t n importnt chnge, however, becuse the restrictive ssumption in the theorem is the uniform convergence of the derivtives f n, not the pointwise (or uniform) convergence of the functions f n. The ssumption tht g = limf n is continuous is needed to show the differentibility of f by the fundmentl theorem, but the result result true even if g isn t continuous. In tht cse, however, different (nd more complicted) proof is required. c g. x c g.

4. The Riemnn Integrl.9.2. Pointwise convergence. On its own, the pointwise convergence of functions is never sufficient to imply convergence of their integrls. Exmple.65. For n N, define f n : [,] R by { n if < x < /n, f n (x) = if x = or /n x. Then f n pointwise on [,] but f n = for every n N. By slightly modifying these functions to { n 2 if < x < /n, f n (x) = if x = or /n x, we get sequence tht converges pointwise to but whose integrls diverge to. The fct tht the f n re discontinuous is not importnt; we could replce the step functions by continuous tent functions or smooth bump functions. The behvior of the integrl under pointwise convergence in the previous exmple is unvoidble. A much worse feture of the Riemnn integrl is tht the pointwise limit of integrble functions needn t be integrble t ll, even if it is bounded. Exmple.66. Let {q k : k N} be n enumertion of the rtionls in [,] nd define f n : [,] R by { if x = q k for k n, f n (x) = otherwise. The ech f n is Riemnn integrble since it differs from the zero function t finitely mnypoints. However, f n f pointwise on [,] to the Dirichlet function f, which is not Riemnn integrble. This is nother plce where the Lebesgue integrl hs better properties thn the Riemnn integrl. The pointwise (or pointwise lmost everywhere) limit of Lebesgue integrble functions is Lebesgue integrble. As Exmple.65 shows, we still need conditions to ensure the convergence of the integrls, but there re quite simple nd generl conditions for the Lebesgue integrl (such s the monotone convergence nd dominted convergence theorems)... Improper Riemnn integrls The Riemnn integrl is only defined for bounded function on compct intervl (or finite union of such intervls). Nevertheless, we frequently wnt to integrte n unbounded function or function on n infinite intervl. One wy to interpret such n integrl is s limit of Riemnn integrls; this limit is clled n improper Riemnn integrl.

.. Improper Riemnn integrls 4... Improper integrls. First, we define the improper integrl of function tht fils to be integrble t one endpoint of bounded intervl. Definition.67. Suppose tht f : (, b] R is integrble on [c, b] for every < c < b. Then the improper integrl of f on [,b] is f = lim ǫ + The improper integrl converges if this limit exists (s finite rel number), otherwise it diverges. Similrly, if f : [,b) R is integrble on [,c] for every < c < b, then f = lim ǫ + +ǫ ǫ We use the sme nottion to denote proper nd improper integrls; it should be cler from the context which integrls re proper Riemnn integrls (i.e., ones given by Definition.3) nd which re improper. If f is Riemnn integrble on [, b], then Proposition.39 shows tht its improper nd proper integrls gree, but n improper integrl my exist even if f isn t integrble. Exmple.68. If p >, the integrl x p dx isn t defined s Riemnn integrl since /x p is unbounded on (,]. The corresponding improper integrl is dx = lim xp ǫ + ǫ f. f. x p dx. For p, we hve ǫ p dx = ǫ xp p, so the improper integrl converges if < p <, with x p dx = p, nd diverges to if p >. The integrl lso diverges (more slowly) to if p = since ǫ x dx = ln ǫ. Thus, we get convergent improper integrl if the integrnd /x p does not grow too rpidly s x + (slower thn /x). We define improper integrls on n unbounded intervl s limits of integrls on bounded intervls. Definition.69. Suppose tht f : [, ) R is integrble on [,r] for every r >. Then the improper integrl of f on [, ) is f = lim r r f.

42. The Riemnn Integrl Similrly, if f : (,b] R is integrble on [r,b] for every r < b, then f = lim r Let s consider the convergence of the integrl of the power function in Exmple.68 t infinity rther thn t zero. Exmple.7. Suppose p >. The improper integrl r ( r p ) dx = lim dx = lim xp r xp r p converges to /(p ) if p > nd diverges to if < p <. It lso diverges (more slowly) if p = since dx = lim x r r r f. dx = lim lnr =. x r Thus, wegetconvergentimproperintegrlifthe integrnd/x p decyssufficiently rpidly s x (fster thn /x). A divergent improper integrl my diverge to (or ) s in the previous exmples, or if the integrnd chnges sign it my oscillte. Exmple.7. Define f : [, ) R by f(x) = ( ) n for n x < n+ where n =,,2,... Then r f nd n f = { if n is n odd integer, if n is n even integer. Thus, the improper integrl f doesn t converge. More generl improper integrls my be defined s finite sums of improper integrls of the previous forms. For exmple, if f : [,b]\{c} R is integrble on closed intervls not including < c < b, then f = lim δ + c δ f + lim ǫ + nd if f : R R is integrble on every compct intervl, then f = lim s c s f + lim r where we split the integrl t n rbitrry point c R. Note tht ech limit is required to exist seprtely. Exmple.72. If f : [,] R is continuous nd < c <, then we define s n improper integrl f(x) c δ f(x) f(x) dx = lim dx+ lim dx. x c /2 δ + x c /2 ǫ + c+ǫ x c /2 Integrls like this one pper in the theory of integrl equtions. c+ǫ r c f, f;

.. Improper Riemnn integrls 43 Exmple.73. Consider the following integrl, clled Frullni integrl, I = f(x) f(bx) x We ssume tht,b > nd f : [, ) R is continuous function whose limit s x exists; we write this limit s f( ) = lim x f(x). We interpret the integrl s n improper integrl I = I +I 2 where I = lim ǫ + ǫ f(x) f(bx) x dx, dx. I 2 = lim r r f(x) f(bx) x Consider I. After mking the substitutions s = x nd t = bx nd using the dditivity property of the integrl, we get tht ( ) f(s) b f(t) ǫb f(t) b f(t) I = lim ds dt = lim dt dt. ǫ + s ǫb t ǫ + t t ǫ To evlute the limit, we write ǫb ǫ f(t) t dt = = ǫb ǫ ǫb ǫ f(t) f() t f(t) f() t ǫ ǫb dt+f() dt+f()ln ǫ t dt ). Assuming for definiteness tht < < b, we hve ǫb f(t) f() ( ) b dt t mx{ f(t) f() : ǫ t ǫb} ǫ s ǫ +, since f is continuous t. It follows tht ( ) b f(t) I = f()ln t A similr rgument gives I 2 = f( )ln Adding these results, we conclude tht f(x) f(bx) x dt. ( ) b f(t) + dt. t dx = {f() f( )}ln ( b ( ) b...2. Absolutely convergent improper integrls. The convergence of improper integrls is nlogous to the convergence of series. A series n converges bsolutely if n converges, nd conditionlly if n converges but n diverges. We introduce similr definition for improper integrls nd provide test for the bsolute convergence of n improper integrl tht is nlogous to the comprison test for series. dx.

44. The Riemnn Integrl Definition.74. An improper integrl f is bsolutely convergent if the improper integrl f converges, nd conditionlly convergent if f converges but f diverges. As prt of the next theorem, we prove tht n bsolutely convergent improper integrl converges (similrly, n bsolutely convergent series converges). Theorem.75. Suppose tht f,g : I R re defined on some finite or infinite intervl I. If f g nd the improper integrl I g converges, then the improper integrl f converges bsolutely. Moreover, n bsolutely convergent improper I integrl converges. Proof. To be specific, we suppose tht f,g : [, ) R re integrble on [,r] for r > nd consider the improper integrl f = lim r A similr rgument pplies to other types of improper integrls. First, suppose tht f. Then r f r so r f is monotonic incresing function of r tht is bounded from bove. Therefore it converges s r. g r f. In generl, we decompose f into its positive nd negtive prts, f = f + f, f = f + +f, f + = mx{f,}, g, f = mx{ f,}. We hve f ± g, so the improper integrls of f ± converge by the previous rgument, nd therefore so does the improper integrl of f: ( r r ) f = lim f + f r r = lim = r f + r f r f + lim Moreover, since f ± f, we see tht f + nd f converge if f converges, nd therefore so does f. Exmple.76. Consider the limiting behvior of the error function erf(x) in Exmple.58 s x, which is given by 2 π f. e x2 dx = 2 π lim r r e x2 dx. The convergence of this improper integrl follows by comprison with e x, for exmple, since e x2 e x for x,

.. Improper Riemnn integrls 45.4.3 y.2.. 5 5 x Figure 8. Grph of y = (sinx)/( + x 2 ) from Exmple.77. The dshed green lines re the grphs of y = ±/x 2. nd e x dx = lim r r e x ( dx = lim e e r) = r e. This rgument proves tht the error function pproches finite limit s x, but it doesn t give the exct vlue, only n upper bound 2 e x2 dx M, M = 2 e x2 dx+ π π e. Numericlly, M.26. In fct, one cn show tht 2 e x2 dx =. π The stndrd trick (pprently introduced by Lplce) uses double integrtion, polr coordintes, nd the substitution u = r 2 : ( 2 e dx) x2 = e x2 y 2 dxdy = π/2 = π 4 ( ) e r2 rdr dθ e u du = π 4. This forml computtion cn be justified rigorously, but we won t do tht here. Exmple.77. The improper integrl sinx dx = lim +x2 r r sinx +x 2 dx

46. The Riemnn Integrl converges bsolutely, since sinx +x 2 dx = nd (see Figure 8) sinx +x 2 x 2 for x, sinx +x 2 dx+ sinx +x 2 dx dx <. x2 The vlue of this integrl doesn t hve n elementry expression, but by using contour integrtion from complex nlysis one cn show tht sinx +x 2 dx = 2e Ei() e Ei( ).6468, 2 where Ei is the exponentil integrl function defined in Exmple.6. Improper integrls, nd the principl vlue integrls discussed below, rise frequently in complex nlysis, nd mny such integrls cn be evluted by contour integrtion. Exmple.78. The improper integrl sinx x dx = lim r r sinx x dx = π 2 converges conditionlly. We leve the proof s n exercise. Comprison with the function /x doesn t imply bsolute convergence t infinity becuse the improper integrl /xdx diverges. There re mny wys to show tht the exct vlue of the improper integrl is π/2. The stndrd method uses contour integrtion. Exmple.79. Consider the limiting behvior of the Fresnel sine function S(x) in Exmple.59 s x. The improper integrl ( ) πx 2 r ( ) πx 2 sin dx = lim sin dx = 2 r 2 2. converges conditionlly. This exmple my seem surprising since the integrnd sin(πx 2 /2) doesn t converge to s x. The explntion is tht the integrnd oscilltes more rpidly with incresing x, leding to more rpid cnceltion between positive nd negtive vlues in the integrl (see Figure 6). The exct vlue cn be found by contour integrtion, gin, which shows tht ( ) πx 2 sin dx = ) exp ( πx2 dx. 2 2 2 Evlution of the resulting Gussin integrl gives /2...3. Principl vlue integrls. Some integrls hve singulrity tht is too strong for them to converge s improper integrls but, due to cnceltion, they hve finite limit s principl vlue integrl. We begin with n exmple. Exmple.8. Consider f : [,]\{} defined by f(x) = x.

.. Improper Riemnn integrls 47 The definition of the integrl of f on [,] s n improper integrl is dx = lim x δ + δ dx+ lim x ǫ + = lim δ +lnδ lim ǫ +lnǫ. ǫ x dx Neither limit exists, so the improper integrl diverges. (Formlly, we get.) If, however, we tke δ = ǫ nd combine the limits, we get convergent principl vlue integrl, which is defined by p.v. dx = lim x ǫ + ( ǫ x dx+ ǫ ) x dx = lim ǫ +(lnǫ lnǫ) =. The vlue of is wht one might expect from the oddness of the integrnd. A cnceltion in the contributions from either side of the singulrity is essentilly to obtin finite limit. The principl vlue integrl of /x on non-symmetric intervl bout still exists but is non-zero. For exmple, if b >, then ( b ǫ ) b p.v. dx = lim x ǫ + x dx+ x dx = lim (lnǫ+lnb lnǫ) = lnb. ǫ + ǫ The crucil feture if principl vlue integrl is tht we remove symmetric intervl round singulr point, or infinity. The resulting cnceltion in the integrl of non-integrble function tht chnges sign cross the singulrity my led to finite limit. Definition.8. Iff : [,b]\{c} Risintegrbleonclosedintervlsnotincluding < c < b, then the principl vlue integrl of f is ( b c ǫ ) p.v. f = lim f + f. ǫ + If f : R R is integrble on compct intervls, then the principl vlue integrl is p.v. f = lim r If the improper integrl exists, then the principl vlue integrl exists nd is equl to the improper integrl. As Exmple.8 shows, the principl vlue integrl my exist even if the improper integrl does not. Of course, principl vlue integrl my lso diverge. r Exmple.82. Consider the principl vlue integrl p.v. dx = lim x2 ǫ + = lim ǫ + ( ǫ ( 2 ǫ 2 r f. c+ǫ x 2 dx+ ) =. ǫ ) x 2 dx In this cse, the function /x 2 is positive nd pproches on both sides of the singulrity t x =, so there is no cnceltion nd the principl vlue integrl diverges to.

48. The Riemnn Integrl Principl vlue integrls rise frequently in complex nlysis, hrmonic nlysis, nd vriety of pplictions. Exmple.83. Consider the exponentil integrl function Ei given in Exmple.6, x e t Ei(x) = t dt. Ifx <, the integrndis continuous for < t x, nd the integrlis interpreted s n improper integrl, x e t x e t dt = lim t r r t dt. This improper integrl converges bsolutely by comprison with e t, since e t t et for < t, nd e t dt = lim r r e t dt = e. If x >, then the integrnd hs non-integrble singulrity t t =, nd we interpret it s principl vlue integrl. We write x e t t dt = e t t dt+ x e t t dt. The first integrl is interpreted s n improper integrl s before. The second integrl is interpreted s principl vlue integrl x e t ( ǫ e t x p.v. dt = lim t ǫ + t dt+ e t ) t dt. This principl vlue integrl converges, since p.v. x e t t dt = x e t t dt+p.v. x ǫ x t dt = e t dt+lnx. t The first integrl mkes sense s Riemnn integrl since the integrnd hs removble singulrity t t =, with ( e t ) lim =, t t so it extends to continuous function on [,x]. Finlly, if x =, then the integrnd is unbounded t the left endpoint t =. The corresponding improper or principl vlue integrl diverges, nd Ei() is undefined. Exmple.84. Let f : R R nd ssume, for simplicity, tht f hs compct support, mening tht f = outside compct intervl [ r,r]. If f is integrble, we define the Hilbert trnsform Hf : R R of f by the principl vlue integrl Hf(x) = π p.v. f(t) x t dt = π lim ǫ + ( x ǫ f(t) x t dt+ x+ǫ f(t) x t dt ).

.. Riemnn sums 49 Here, x plys the role of prmeter in the integrl with respect to t. We use principl vlue becuse the integrnd my hve non-integrble singulrity t t = x. Since f hs compct support, the intervls of integrtion re bounded nd there is no issue with the convergence of the integrls t infinity. For exmple, suppose tht f is the step function { for x, f(x) = for x < or x >. If x < or x >, then t x for t, nd we get proper Riemnn integrl Hf(x) = π x t dt = π ln x x. If < x <, then we get principl vlue integrl Hf(x) = π lim Thus, for x, we hve ǫ + = π lim ǫ + ( x ǫ x t dt+ π ( ( )] x ǫ +ln ǫ) x ) [ ln = π ln ( x x Hf(x) = π ln x x. x+ǫ ) x t dt The principl vlue integrl with respect to t diverges if x =, becuse f(t) hs jump discontinuity t the point where t = x. Consequently the vlues Hf(), Hf() of the Hilbert trnsform of the step function re undefined... Riemnn sums An lterntive wy to define the Riemnn integrl is in terms of the convergence of Riemnn sums. This ws, in fct, Riemnn s originl definition, which he gve in 854 in his Hbilittionsschrift ( kind of post-doctorl disserttion required of Germn cdemics), building on previous work of Cuchy who defined the integrl for continuous functions. It is interesting to note tht the topic of Riemnn s Hbilittionsschrift ws not integrtion theory, but Fourier series. Riemnn introduced n nlyticl definition of the integrl long the wy so tht he could stte his results more precisely. In fct, lmost ll of the fundmentl developments of rigorous rel nlysis in the nineteenth century were motivted by problems relted to Fourier series nd their convergence. Upper nd lower sums were introduced by Drboux, nd they simplify the theory. We won t use Riemnn sums here, but we will explin the equivlence of the definitions. We ll sy, temporrily, tht function is Drboux integrble if it stisfies Definition.3.

5. The Riemnn Integrl To give Riemnn s definition, we define tgged prtition (P,C) of compct intervl [,b] to be prtition of the intervl together with set P = {I,I 2,...,I n } C = {c,c 2,...,c n } of points such tht c k for k =,...,n. (Think of c k s tg ttched to.) If f : [,b] R, then we define the Riemnn sum of f with respect to the tgged prtition (P, C) by n S(f;P,C) = f(c k ). Tht is, insted of using the supremum or infimum of f on the kth intervl in the sum, we evlute f t n rbitrry point in the intervl. Roughly speking, function is Riemnn integrble if its Riemnn sums pproch the sme vlue s the prtition is refined, independently of how we choose the points c k. As mesure of the refinement of prtition P = {I,I 2,...,I n }, we define the mesh (or norm) of P to be the mximum length of its intervls, k= mesh(p) = mx k n = mx k n x k x k. Definition.85. A bounded function f : [,b] R is Riemnn integrble on [,b] if there exists number R R with the following property: For every ǫ > there is δ > such tht S(f;P,C) R < ǫ for every tgged prtition (P,C) of [,b] with mesh(p) < δ. In tht cse, R = f is the Riemnn integrl of f on [,b]. Note tht L(f;P) S(f;P,C) U(f;P), so the Riemnn sums re squeezed between the upper nd lower sums. The following theorem shows tht the Drboux nd Riemnn definitions led to the sme notion of the integrl, so it s mtter of convenience which definition we dopt s our strting point. Theorem.86. A function is Riemnn integrble (in the sense of Definition.85) if nd only if it is Drboux integrble (in the sense of Definition.3). In tht cse, the Riemnn nd Drboux integrls of the function re equl. Proof. First, suppose tht f : [,b] R is Riemnn integrble with integrl R. Then f must be bounded; otherwise f would be unbounded in some intervl of every prtition P, nd we could mke its Riemnn sums with respect to P rbitrrily lrge by choosing suitble point c k, contrdicting the definition of R. Let ǫ >. There is prtition P = {I,I 2,...,I n } of [,b] such tht S(f;P,C) R < ǫ 2

.. Riemnn sums 5 for every set of points C = {c k : k =,...,n}. If M k = sup Ik f, then there exists c k such tht ǫ M k 2(b ) < f(c k). It follows tht n M k ǫ n 2 < f(c k ), k= mening tht U(f;P) ǫ/2 < S(f;P,C). Since S(f;P,C) < R+ǫ/2, we get tht k= U(f) U(f;P) < R+ǫ. Similrly, if m k = inf Ik f, then there exists c k such tht ǫ n m k + 2(b ) > f(c k), m k + ǫ n 2 > f(c k ), nd L(f;P)+ǫ/2 > S(f;P,C). Since S(f;P,C) > R ǫ/2, we get tht These inequlities imply tht k= L(f) L(f;P) > R ǫ. L(f)+ǫ > R > U(f) ǫ for every ǫ >, nd therefore L(f) R U(f). Since L(f) U(f), we conclude tht L(f) = R = U(f), so f is Drboux integrble with integrl R. Conversely, suppose tht f is Drboux integrble. The min point is to show tht if ǫ >, then U(f;P) L(f;P) < ǫ not just for some prtition but for every prtition whose mesh is sufficiently smll. Let ǫ > be given. Since f is Drboux integrble. there exists prtition Q such tht U(f;Q) L(f;Q) < ǫ 4. Suppose tht Q contins m intervls nd f M on [,b]. We clim tht if δ = ǫ 8mM, then U(f;P) L(f;P) < ǫ for every prtition P with mesh(p) < δ. To prove this clim, suppose tht P = {I,I 2,...,I n } is prtition with mesh(p) < δ. Let P be the smllest common refinement of P nd Q, so tht the endpoints of P consist of the endpoints of P or Q. Since, b re common endpoints of P nd Q, there re t most m endpoints of Q tht re distinct from endpoints of P. Therefore, t most m intervls in P contin dditionl endpoints of Q nd re strictly refined in P, mening tht they re the union of two or more intervls in P. Now consider U(f;P) U(f;P ). The terms tht correspond to the sme, unrefined intervls in P nd P cncel. If is strictly refined intervl in P, then the corresponding terms in ech of the sums U(f;P) nd U(f;P ) cn be estimted by M nd their difference by 2M. There re t most m such intervls nd < δ, so it follows tht k= U(f;P) U(f;P ) < 2(m )Mδ < ǫ 4.

52. The Riemnn Integrl Since P is refinement of Q, we get nd U(f;P) < U(f;P )+ ǫ 4 U(f;Q)+ ǫ 4 < L(f;Q)+ ǫ 2. It follows by similr rgument tht L(f;P ) L(f;P) < ǫ 4, L(f;P) > L(f;P ) ǫ 4 L(f;Q) ǫ 4 > U(f;Q) ǫ 2. Since L(f;Q) U(f;Q), we conclude from these inequlities tht for every prtition P with mesh(p) < δ. U(f;P) L(f;P) < ǫ If D denotes the Drboux integrl of f, then we hve L(f;P) D U(f,P), L(f;P) S(f;P,C) U(f;P). Since U(f;P) L(f;P) < ǫ for every prtition P with mesh(p) < δ, it follows tht S(f;P,C) D < ǫ. Thus, f is Riemnn integrble with Riemnn integrl D. Finlly, we give necessry nd sufficient condition for Riemnn integrbility tht ws proved by Riemnn himself (854). To stte the condition, we introduce some nottion. Let f;[,b] R be bounded function. If P = {I,I 2,...,I n } is prtition of [,b] nd ǫ >, let A ǫ (P) {,...,n} be the set of indices k such tht osc f = sup f inf f ǫ for k A ǫ (P). Similrly, let B ǫ (P) {,...,n} be the set of indices such tht osc f < ǫ for k B ǫ (P). Tht is, the oscilltion of f on is lrge if k A ǫ (P) nd smll if k B ǫ (P). We denote the sum of the lengths of the intervls in P where the oscilltion of f is lrge by s ǫ (P) =. k A ǫ(p) Fixing ǫ >, we sy tht s ǫ (P) s mesh(p) if for every η > there exists δ > such tht mesh(p) < δ implies tht s ǫ (P) < η. Theorem.87. AboundedfunctionisRiemnnintegrbleifndonlyifs ǫ (P) s mesh(p) for every ǫ >. Proof. Let f : [,b] R be bounded with f M on [,b] for some M >. First, suppose tht the condition holds, nd let ǫ >. If P is prtition of [,b], then, using the nottion bove for A ǫ (P), B ǫ (P) nd the inequlity osc f 2M,

.2. The Lebesgue criterion for Riemnn integrbility 53 we get tht U(f;P) L(f;P) = n k= = k A ǫ(p) osc f 2M k A ǫ(p) osc f + +ǫ 2Ms ǫ (P)+ǫ(b ). k B ǫ(p) k B ǫ(p) osc f By ssumption, there exists δ > such tht s ǫ (P) < ǫ if mesh(p) < δ, in which cse U(f;P) L(f;P) < ǫ(2m +b ). The Cuchy criterion in Theorem.4 then implies tht f is integrble. Conversely, suppose tht f is integrble, nd let ǫ > be given. If P is prtition, we cn bound s ǫ (P) from bove by the difference between the upper nd lower sums s follows: U(f;P) L(f;P) k A ǫ(p) osc f ǫ k A ǫ(p) = ǫs ǫ (P). Since f is integrble, for every η > there exists δ > such tht mesh(p) < δ implies tht U(f;P) L(f;P) < ǫη. Therefore, mesh(p) < δ implies tht s ǫ (P) [U(f;P) L(f;P)] < η, ǫ which proves the result. This theorem hs the drwbck tht the necessry nd sufficient condition for Riemnn integrbility is somewht complicted nd, in generl, it isn t esy to verify. In the next section, we stte simpler necessry nd sufficient condition for Riemnn integrbility..2. The Lebesgue criterion for Riemnn integrbility Although the Dirichlet function in Exmple.7 is not Riemnn integrble, it is Lebesgue integrble. Its Lebesgue integrl is given by f = A + B where A = [,] Q is the set of rtionl numbers in [,], B = [,] \ Q is the set of irrtionl numbers, nd E denotes the Lebesgue mesure of set E. The Lebesgue mesure of set is generliztion of the length of n intervl which pplies to more generl sets. It turns out tht A = (s is true for ny countble set of rel numbers see Exmple.89 below) nd B =. Thus, the Lebesgue integrl of the Dirichlet function is.

54. The Riemnn Integrl A necessry nd sufficient condition for Riemnn integrbility cn be given in terms of Lebesgue mesure. We will stte this condition without proof, beginning with criterion for set to hve Lebesgue mesure zero. Theorem.88. A set E R hs Lebesgue mesure zero if nd only if for every ǫ > there is countble collection of open intervls {( k,b k ) : k N} such tht E ( k,b k ), (b k k ) < ǫ. k= The open intervls is this theorem re not required to be disjoint, nd they my overlp. Exmple.89. Every countble set E = {x k R : k N} hs Lebesgue mesure zero. To prove this, let ǫ > nd for ech k N define k = x k ǫ 2 k+2, b k = x k + ǫ 2 k+2. Then E k= ( k,b k ) since x k ( k,b k ) nd (b k k ) = k= k= k= so the Lebesgue mesure of E is equl to zero. ǫ 2 k+ = ǫ 2 < ǫ, If E = [,] Q consists of the rtionl numbers in [,], then the set G = k= ( k,b k ) described bove encloses the dense set of rtionls in collection of open intervls the sum of whose lengths is rbitrrily smll. This isn t so esy to visulize. Roughly speking, if ǫ is smll nd we look t section of [,] t given mgnifiction, then we see few of the longer intervls in G with reltively lrge gps between them. Mgnifying one of these gps, we see few more intervls with lrge gps between them, mgnifying those gps, we see few more intervls, nd so on. Thus, the set G hs frctl structure, mening tht it looks similr t ll scles of mgnifiction. We then hve the following result, due to Lebesgue. Theorem.9. A bounded function on compct intervl is Riemnn integrble if nd only if the set of points t which it is discontinuous hs Lebesgue mesure zero. For exmple, the set of discontinuities of the Riemnn-integrble function in Exmple.6 consists of single point {}, which hs Lebesgue mesure zero. On the other hnd, the set of discontinuities of the non-riemnn-integrble Dirichlet function in Exmple.7 is the entire intervl [,], nd its set of discontinuities hs Lebesgue mesure one. Theorem.9 implies tht every bounded function with countble set of discontinuities is Riemnn integrble, since such set hs Lebesgue mesure zero. A specil cse of this result is Theorem.33 tht every bounded function with finitely mny discontinuities is Riemnn integrble. The monotonic function in Exmple.22 is n explicit exmple of Riemnn integrble function with dense, countbly infinite set of discontinuities. A set doesn t hve to be countble to

.2. The Lebesgue criterion for Riemnn integrbility 55 hve Lebesgue mesure zero, nd there re mny uncountble sets whose Lebesgue mesure is zero. Exmple.9. The stndrd middle-thirds Cntor set K [, ] is n uncountble set with Lebesgue mesure zero. The chrcteristic function f : [,] R of K, defined by { if x K f(x) = if x [,]\K, hs K s its set of discontinuities. Therefore, f is Riemnn integrble on [, ], with integrl zero, even though it is discontinuous t uncountbly mny points.