Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004

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1 Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October Definite Integrls In this section we revisit the definite integrl tht you were introduced to when you first studied clculus. You undoubtedly lerned tht given positive function f over n intervl [, b] the definite integrl f(x) dx provided it ws defined, ws number equl to the re under the grph of f over [, b]. You lso likely lerned tht the definite integrl ws defined s limit of Riemnn sums. The Riemnn sums you most likely used involved prtitioning [, b] into n uniform subintervls of length (b )/n nd evluting f t either the right-hnd endpoint, the left-hnd endpoint, or the midpoint of ech subintervl. At the time your understnding of the notion of limit ws likely more intuitive thn rigorous. In this section we present rigorous development of the definite integrl built upon the rigorous understnding of limit tht you hve studied erlier in this course Prtitions nd Drboux Sums. In the pproch tken here, we will consider very generl prtitions of the intervl [, b], not just those with uniform subintervls. Definition 1.1. Let [, b] R. A prtition of the intervl [, b] is specified by n N, nd {x i } n [, b] such tht = x 0 < x 1 < < x n 1 < x n = b. This prtition is denoted P = [x 0, x 1,, x n 1, x n ]. Ech x i for i = 0,, n is clled prtition point of P, nd for ech i = 1,, n the intervl [x i 1, x i ] is clled i th subintervl induced by P. The prtition thickness, denoted P, is defined by P = mx { x i x i 1 : i = 1,, n }. The pproch tken here is not bsed on Riemnn sums, but rther on Drboux sums. This is becuse Drboux sums re well-suited for nlysis by the tools we hve developed to estblish the existence of limits. We will be ble to recover results bout Riemnn sums becuse, s we will show, every Riemnn sum is bounded by two Drboux sums. 1

2 2 Let f : [, b] R be bounded. Set (1) m = inf { f(x) : x [, b] }, M = sup { f(x) : x [, b] }. Becuse f is bounded, one knows tht < m M <. Let P = [x 0,, x n ] be prtition of [, b]. For ech i = 1,, n set m i = inf { f(x) : x [x i 1, x i ] }, Clerly m m i M i M. M i = sup { f(x) : x [x i 1, x i ] }. Definition 1.2. The lower nd upper Drboux sums ssocited with the function f nd prtition P re respectively defined by (2) L(f, P) = U(f, P) = m i (x i x i 1 ), M i (x i x i 1 ). Clerly, the Drboux sums stisfy the bounds (3) m (b ) L(f, P) U(f, P) M (b ). These inequlities will ll be equlities when f is constnt. Remrk. A Riemnn sum ssocited with the prtition P is specified by selecting qudrture point q i [x i 1, x i ] for ech i = 1,, n. Let Q = (q 1,, q n ) be the n-tuple of qudrture points. The ssocited Riemnn sum is then (4) R(f, P, Q) = f(q i ) (x i x i 1 ). It is esy to see tht for ny choice of qudrture points Q one hs the bounds (5) L(f, P) R(f, P, Q) U(f, P). Moreover, one cn show tht L(f, P) = inf { R(f, P, Q) : Q re qudrture points for P }, U(f, P) = sup { R(f, P, Q) : Q re qudrture points for P }. The bounds (5) re thereby shrp.

3 1.2. Refinements. We now introduce the notion of refinement of prtition. Definition 1.3. Given prtition P of n intervl [, b], prtition P of [, b] is clled refinement of P provided every prtition point of P is prtition point of P. If P = [x 0, x 1,, x n 1, x n ] nd P is refinement of P then P induces prtition of ech [x i 1, x i ], which we denote by Pi. For exmple, if P = [x 0, x 1,, x n 1, x n ] with x j i = x i for ech i = 0,, n then Pi = [x j i 1,, x j i ]. Observe tht (6) L(f, P ) = L(f, Pi ), U(f, P ) = U(f, Pi ). Moreover, upon pplying the bounds (3) to P i for ech i = 1,, n, we obtin the bounds (7) m i (x i x i 1 ) L(f, P i ) U(f, P i ) M i (x i x i 1 ). This observtion is key to the proof of the following. Lemm 1.1. Refinement Lemm. Let f : [, b] R be bounded. Let P be prtition of [, b] nd P be refinement of P. Then (8) L(f, P) L(f, P ) U(f, P ) U(f, P). 3 Proof. It follows from (2), (7), nd (6) tht L(f, P) = m i (x i x i 1 ) L(f, P i ) = L(f, P ) U(f, P ) = U(f, P i ) M i (x i x i 1 ) = U(f, P) Comprisons. A key step in our development will be to develop comprisons of L(f, P 1 ) nd U(f, P 2 ) for ny two prtitions P 1 nd P 2, of [, b].

4 4 Definition 1.4. Given two prtitions, P 1 nd P 2, of [, b] define P 1 P 2 to be the prtition whose set of prtition points is the union of the prtition points of P 1 nd the prtition points of P 2. We cll P 1 P 2 the supremum of P 1 nd P 2. It is esy to rgue tht P 1 P 2 is the smllest prtition of [, b] tht is refinement of both P 1 nd P 2. Lemm 1.2. Comprison Lemm. Let f : [, b] R be bounded. Let P 1 nd P 2 be prtitions of [, b]. Then (9) L(f, P 1 ) U(f, P 2 ). Proof. Becuse P 1 P 2 is refinement of both P 1 nd P 2, it follows from the Refinement Lemm tht L(f, P 1 ) L(f, P 1 P 2 ) U(f, P 1 P 2 ) U(f, P 2 ). Becuse the prtitions P 1 nd P 2 on either side of inequlity (9) re independent, we my obtin shrper bounds by tking the supremum over P 1 on the right-hnd side, or the infimum over P 2 on the left-hnd side. Indeed, we prove the following. Lemm 1.3. Shrp Comprison Lemm. Let f : [, b] R be bounded. Let (10) L(f) = sup { L(f, P) : P is prtition of [, b] }, U(f) = inf { U(f, P) : P is prtition of [, b] }. Let P 1 nd P 2 be prtitions of [, b]. Then (11) L(f, P 1 ) L(f) U(f) U(f, P 2 ). Remrk. Becuse it is cler from (10) tht L(f) nd U(f) depend on [, b], strictly speking these quntities should be denoted L(f, [, b]) nd U(f, [, b]). This would be necessry if more thn one intervl ws involved in the discussion. However, tht is not the cse here. We therefore embrce the less cluttered nottion. Proof. If we tke the infimum of the right-hnd side of (9) over P 2, we obtin L(f, P 1 ) U(f). If we then tke the supremum of the left-hnd side bove over P 1, we obtin L(f) U(f). The bound (11) then follows.

5 1.4. Definition of the Definite Integrl. We re now redy to define the definite integrl. You will find different definitions of the definite integrl in different books. Here we will used the definition found in Fitzptrick s book. We will then give theorem tht shows this definition is equivlent to nother one commonly found in other books. Definition 1.5. Let f : [, b] R be bounded. Then f is sid to be integrble over [, b] whenever there exists unique A R such tht (12) L(f, P) A U(f, P) for every prtition P of [, b]. In this cse we cll A the definite integrl of f over [, b] nd denote it by f. Remrk. The Shrp Comprison Lemm shows tht (12) holds for every A [L(f), U(f)]. The key thing to be estblished when using the bove definition is therefore the uniqueness of such n A. We now give the following chrcteriztions of integrbility. Theorem 1.1. Integrbility Theorem. Let f : [, b] R be bounded. Then the following re equivlent: (1) f is integrble over [, b]; (2) L(f) = U(f); (3) for every ǫ > 0 there exists prtition P of [, b] such tht 0 U(f, P) L(f, P) < ǫ. Proof. We first show tht (1) = (2). Suppose tht (2) is flse. Then L(f) < U(f). Observe tht the Shrp Comprison Lemm shows tht for every A [L(f), U(f)] one hs L(f, P 1 ) A U(f, P 2 ) for ny prtitions P 1 nd P 2 of [, b]. Hence, there re mny vlues of A tht stisfy (12), whereby f is not integrble. It follows tht (1) = (2). Next we show tht (2) = (3). Let ǫ > 0. By the definition (10) of L(f) nd U(f), we cn find prtitions P 1 nd P 2 of [, b] such tht L(f) ǫ 2 < L(f, P 1 ) L(f), U(f) U(f, P 2 ) < U(f) + ǫ 2. Now let P = P 1 P 2. Becuse the Comprison Lemm implies tht L(f, P 1 ) L(f, P) nd U(f, P) U(f, P 2 ), it follows from the bove 5

6 6 inequlities tht L(f) ǫ 2 < L(f, P) L(f), U(f) U(f, P) < U(f) + ǫ 2. Hence, if L(f) = U(f) one thereby concludes tht ( 0 U(f, P) L(f, P) < U(f) + ǫ ) ( L(f) ǫ ) = ǫ. 2 2 This shows tht (2) = (3). Finlly, we show tht (3) = (1). Suppose tht (1) is flse. Then by the Shrp Comprison Lemm there exists A 1 nd A 2 such tht L(f, P) A 1 < A 2 U(f, P) for every prtition P of [, b]. One thereby hs tht U(f, P) L(f, P) A 2 A 1 for every prtition P of [, b]. Hence, (3) must be flse. It follows tht (1) = (2). Remrk. Property (3) of the Integrbility Theorem provides very useful criterion for estblishing the integrbility of function f. We will exploit this in the next section. Remrk. It follows from the Integrbility Theorem tht one could well dopt the following lterntive definition of the definite integrl. Definition 1.5. Let f : [, b] R be bounded. Then f is sid to be integrble over [, b] whenever L(f) = U(f). In this cse we cll this common vlue the definite integrl of f over [, b] nd denote it by f. This is definition of the definite integrl tht is commonly found in textbooks. 2. Integrble Functions nd Integrls In this section we use the Integrbility Theorem to develop criteri to identify integrble functions. We lso begin the tsk of evluting definite integrls Evluting Integrls vi Riemnn Sums. We now mke the connection with the notion of definite integrl s the limit of sequence of Riemnn sums.

7 Theorem 2.1. Riemnn Sums Convergence Theorem. Let f : [, b] R be bounded. Let {P n } n=1 be sequence of prtitions of [, b] such tht ( (13) lim U(f, P n ) L(f, P n ) ) = 0. n Let Q n be ny qudrture set ssocited with P n. Then f is integrble over [, b] nd (14) f = lim n R(f, P n, Q n ), where the Riemnn sums R(f, P, Q) re defined by (4). Remrk. The content of this theorem is tht once one hs found sequence of prtitions P n such tht (13) holds, then the integrl f exists nd my be evluted s the limit of ny ssocited sequence of Riemnn sums (14). This theorem thereby splits the tsk of evluting definite integrls into two steps. The first step is by fr the esier. It is rre integrnd f for which one cn find sequence of Riemnn tht llow one to evlute the limit in (14). Proof. Given (13), the fct tht f is integrble over [, b] follows directly from criterion (3) of the Integrbility Theorem. The bounds on Riemnn sums given by (5) yield the inequlities L(f, P n ) R(f, P n, Q n ) U(f, P n ), while, becuse f is integrble, we lso hve the inequlities L(f, P n ) It follows from these inequlities tht L(f, P n ) U(f, P n ) L(f, P n ) which implies tht R(f, P n, Q n ) f U(f, P n ). R(f, P n, Q n ) U(f, P n ) f f f U(f, P n ) L(f, P n ), f U(f, P n ) L(f, P n ). Becuse (13) sttes tht the right-hnd side bove vnishes s n tends to, the limit (14) follows. 7

8 Monotone nd Piecewise Monotone Functions. We now use the Riemnn Sums Convergence Theorem to show tht the clss of integrble functions includes the clss of monotone functions. Recll tht this clss is defined s follows. Definition 2.1. A function f : [, b] R is sid to be monotoniclly incresing provided tht x < y = f(x) f(y) for every x, y [, b]. A function f : [, b] R is sid to be monotoniclly decresing provided tht x < y = f(x) f(y) for every x, y [, b]. If function is either monotoniclly incresing or monotoniclly decresing then it is sid to be monotone. It is clssicl fct tht monotone function over [, b] is continuous t ll but t most countble number of points where it hs jump discontinuity. Theorem 2.2. Let f : [, b] R be monotone. Then f is integrble over [, b]. Moreover, for ny sequences P n of prtitions of [, b] nd Q n of ssocited qudrture points such tht P n 0 s n, one hs tht f = lim R(f, P n, Q n ), n where the Riemnn sums R(f, P, Q) re defined by (4). Proof. For ny prtition P = [x 0,, x n ] we hve the following bsic estimte. Becuse f is monotone, over ech subintervl [x i 1, x i ] one hs tht M i m i = f(x i ) f(x i 1 ). We thereby obtin the bsic estimte 0 U(f, P) L(f, P) = (M i m i ) (x i x i 1 ) P (M i m i ) = P = P f(b) f(), f(xi ) f(x i 1 ) where P = mx{x i x i 1 : i = 1,, n} is the thickness of P. Here we hve used the fct tht, becuse f is monotone, the terms f(x i ) f(x i 1 ) re either ll nonnegtive, or ll nonpositive. This

9 fct llows us to pss the bsolute vlue outside the sum, which then telescopes. We now pply the bove bsic estimte to our sequence P n of prtitions, which shows tht 0 U(f, P n ) L(f, P n ) P n f(b) f() 0 s n. The result then follows from the Riemnn Sums Convergence Theorem. Exmple. One cn use this theorem to show tht for every k N the function x x k is integrble over [0, b] nd tht [ b (15) x k k+1 ] dx = lim i k = bk+1 n n k+1 k The detils of this clcultion re presented in the book for the cses k = 0, 1, 2 with b = 1. Here we present the generl cse. Define S k (n) = i k. In order to prove (15) we must estblish the limit 1 (16) lim n n k+1 Sk (n) = 1 k + 1. We do this below by induction on k. Proof. Clerly S 0 (n) = n, so tht limit (16) holds for k = 0. Now ssume tht for some l 1 limit (16) holds for every k < l. By telescoping sum, binomil expnsion, nd the definition of S k (n), one obtins the identity (n + 1) l+1 1 = = = [ (i + 1) l+1 i l+1] l j=0 l j=0 (l + 1)! j!(l j + 1)! ij (l + 1)! j!(l j + 1)! Sj (n) = (l + 1) S l (n) + l 1 j=0 (l + 1)! j!(l j + 1)! Sj (n). 9

10 10 Upon solving for S l (n) nd dividing by n l+1, we obtin the reltion (17) [ 1 n l+1 Sl (n) = 1 (n + 1) l+1 1 l + 1 n l+1 n l+1 Becuse we know l 1 j=0 (l + 1)! j!(l j + 1)! (n + 1) l+1 1 lim = 1, lim n n l+1 n n = 0, l+1 nd becuse, by the induction hypothesis, we know lim n 1 n l+1 Sj (n) = 0 for every j < l, ] 1 n l+1 Sj (n). we cn pss to the n limit in reltion (17). We thereby estblish tht limit (16) holds for k = l. We now use the Riemnn Sums Convergence Theorem to show tht the clss of integrble functions includes the clss of piecewise monotone functions. Recll tht this clss is defined s follows. Definition 2.2. A function f : [, b] R is sid to be piecewise monotone over [, b] provided there exists prtition [p 0,, p m ] of [, b] such tht f is monotone over [p j 1, p j ] for every j = 1,, m. Theorem 2.3. Piecewise Monotone Integrbility Theorem. Let f : [, b] R be piecewise monotone over [, b]. Then f is integrble over [, b]. Moreover, for ny sequences P n of prtitions of [, b] nd Q n of ssocited qudrture points such tht P n 0 s n, one hs tht f = lim n R(f, P n, Q n ), where the Riemnn sums R(f, P, Q) re defined by (4). Proof. Let [p 0,, p m ] be prtition of [, b] such tht f is monotone over [p j 1, p j ] for ech j = 1,, m. Let P = [x 0,, x n ] be ny prtition of [, b]. The key step will be to estblish the bound (18) (M i m i ) m f(pj ) f(p j 1 ). j=1

11 11 Once this is done we cn obtin the bsic estimte 0 U(f, P) L(f, P) = P P (M i m i 1 ) (M i m i ) (x i x i 1 ) m f(pj ) f(p j 1 ), where P = mx{x i x i 1 : i = 1,, n} is the thickness of P. By then pplying the bove bsic estimte to our sequence P n of prtitions, we see tht 0 U(f, P n ) L(f, P n ) m P n f(pj ) f(p j 1 ) 0 s n. The result would then follows from the Riemnn Sums Convergence Theorem. All tht remins to be done is estblish the bound (18). This is esy to do when P is refinement of [p 0,, p m ]. When P is not refinement of [p 0,, p m ] one simply replces P with P [p 0,, p m ]. We leve the detils of these rguments s n exercise Piecewise Integrbility. A key tool for building up the clss of integrble functions is the the following lemm. Lemm 2.1. Piecewise Integrbility Lemm. Let f : [, b] R be bounded. Let P = [p 0,, p k ] be prtition of [, b] such tht f is integrble over [p i 1, p i ] for every i = 1,, k. Then f is integrble over [, b]. Moreover, (19) f = k pi p i 1 f. Proof. Let ǫ > 0. Becuse f is integrble over [p i 1, p i ] there exists prtition P i of [p i 1, p i ] such tht 0 U(f, P i ) L(f, P i ) < ǫ k.

12 12 Let P be the refinement of P such tht P i is the induced prtition of [p i 1, p i ]. One then sees tht 0 U(f, P ) L(f, P ) k ( = U(f, P i ) L(f, Pi )) < k ǫ k = ǫ. Hence, by item (3) of the Integrbility Theorem, f is integrble. Let P be prtition of [, b]. Without loss of generlity we my ssume tht P is refinement of P, otherwise pss to the prtition P P. Let Pi be the induced prtition of [p i 1, p i ]. Becuse f is integrble over [p i 1, p i ] we hve tht pi L(f, Pi ) f U(f, Pi ). p i 1 Summing these inequlities over i = 1,, k yields k pi L(f, P ) f U(f, P ). p i 1 Formul (19) then follows Piecewise Continuous Functions. We now show tht ll functions tht re piecewise continuous over [, b] re lso intergrble over [, b]. We first recll the definition of piecewise continuous function. Definition 2.3. A function f : [, b] R is sid to be piecewise continuous if it is bounded nd there exists prtition P = [x 0,, x n ] of [, b] such tht f is continuous over (x i 1, x i ) for every i = 1,, n. We remrk tht piecewise continuous functions re discontinuous t only finite number of points. Still, the clss of piecewise continuous functions includes some firly wild functions. For exmple, it contins the function 1 + sin(1/x) if x > 0, f(x) = 4 if x = 0, 1 + sin(1/x) if x < 0, considered over [ 1, 1]. As wild s this function looks, it is continuous everywhere except t the point x = 0. We will need two lemms. Lemm 2.2. Let f : [, b] R be continuous. Then f is integrble over [, b].

13 Proof. Let ǫ > 0. Becuse f is uniformly continuous over [, b], there exists δ > 0 such tht x y < δ = f(x) f(y) < ǫ for every x, y [, b]. b Pick n N such tht b n < δ. Let P = [x 0,, x n ] be the prtition of [, b] with x i = + i b for every i = 0,, n. n For every i = 1,, n over the subintervl [x i 1, x i ] one hs x i x i 1 = b n, M i m i ǫ b. One thereby sees tht 0 U(f, P) L(f, P) = (M i m i )(x i x i 1 ) = b n < b n (M i m i ) ǫ b = b n n ǫ b = ǫ. Hence, by item (3) of the Integrbility Theorem, f is integrble. Lemm 2.3. Let f : [, b] R be bounded nd f : (, b) R be continuous. Then f is integrble over [, b]. Proof. Let ǫ > 0. Let δ > 0 such tht ǫ δ < 3(M m), δ < b. 2 where, s before, m nd M re defined by (1). Becuse f is continuous over [ + δ, b δ], by the previous lemm it is integrble over [ + δ, b δ]. By the Integrbility Theorem, there exists prtition P δ of [ + δ, b δ] such tht 0 U(f, P δ ) L(f, P δ ) < ǫ 3. Let P be the prtition of [, b] given by P = [x 0, x 1,, x n 1, x n ] where P δ = [x 1,, x n 1 ]. One hs x 1 x 0 = x n x n 1 = δ, M 1 m 1 < M m, M n m n < M m. 13

14 14 One thereby sees tht 0 U(f, P) L(f, P) = (M i m i )(x i x i 1 ) n 1 = (M 1 m 1 )δ + (M n m n )δ + (M i m i )(x i x i 1 ) i=2 = (M 1 m 1 )δ + (M n m n )δ + U(f, P δ ) L(f, P δ ) ǫ < 2 (M m) 3(M m) + ǫ 3 = ǫ. Hence, by item (3) of the Integrbility Theorem, f is integrble. Theorem 2.4. Let f : [, b] R be piecewise continuous. Then f is integrble over [, b]. Proof. The theorem follows from the previous lemm nd the Piecewise Integrbility Lemm. 3. The First Fundmentl Theorem of Clculus The business of evluting integrls by tking limits of Riemnn sums is usully either difficult or impossible. However, s you hve known since you first studied integrtion, for mny integrnds there is must esier wy. Theorem 3.1. First Fundmentl Theorem of Clculus. Let f : [, b] R be integrble. Suppose tht F : [, b] R is continuous, tht F : (, b) R is differentible, nd tht (20) F (x) = f(x) for every x (, b). Then f = F(b) F(). Remrk. This theorem essentilly reduces the problem of evluting definite integrls to tht of finding n explicit solution of the differentil eqution (20). While such n explicit solution cnnot lwys be found, for wide clss of integrnds f. Proof. We must show tht for every prtition P of [, b] one hs (21) L(f, P) F(b) F() U(f, P). Let P = [x 0,, x n ] be n rbitrry prtition of [, b]. For every i = 1,, n one knows tht F : [x i 1, x i ] R is continuous, nd tht

15 F : (x i 1, x i ) R is differentible. Then by the Lgrnge Men Vlue Theorem there exists q i (x i 1, x i ) such tht F(x i ) F(x i 1 ) = F (q i ) (x i x i 1 ) = f(q i ) (x i x i 1 ). Becuse m i f(q i ) M i, we see from the bove tht m i (x i x i 1 ) F(x i ) F(x i 1 ) M i (x i x i 1 ). Finlly, dding these inequlities yields (21). The following is n immedite corollry of the First Fundmentl Theorem of Clculus. Corollry 3.1. Let F : [, b] R be continuous, F : (, b) R be differentible, nd F : (, b) R be continuous nd bounded. Let f be ny extension of F to [, b]. Then f = F(b) F(). Exmple. Let F be defined over [0, 1] by { x cos(log(1/x)) if 0 < x 1, F(x) = 0 if x = 0. Then F is continuous over [0, 1] nd differentible over (0, 1] with F (x) = cos(log(1/x)) + sin(log(1/x)). As this function is bounded, we hve 1 0 [cos(log(1/x)) + sin(log(1/x))] dx = F(1) F(0) = 0. Here the integrnd cn be ssigned ny vlue t x = 0. 15

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