Math 360: A primitive integral and elementary functions


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1 Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth C: Integrl/functions 1 / 32
2 Setup for the integrl prtitions Definition: (prtitions of n intervl) Let < b, then sequence of numbers = x 0 < x 1 < x 2 < < x n = b is clled prtition of the intervl [, b]. There is no requirement tht the x i s be evenly spced, only tht they be strictly incresing nd tht x 0 = nd x n = b. Sometimes we refer to prtition by P. For instnce, we sy P is refinement of the prtition P (nd write P > P) if ll of the x i s in P re lso in P. The norm of the prtition P (denoted P ) is the mximum of x i x i 1 for i = 1,..., n. (The textbook clls this the gp of P). Given two prtitions P 1 nd P 2 of [, b] it is esy to see tht they hve common refinement just use ll the x s from both prtitions in new P 3. D. DeTurck Mth C: Integrl/functions 2 / 32
3 Setup for the integrl monotonic functions At first, we re only going to consider the integrls of monotonic functions. Let f : [, b] R be monotonic function. For the moment, we ll ssume tht f is incresing (i.e., f (x) f (y) if x y), but everything will work for decresing functions with the obvious djustments. f does not hve to be continuous (nd in fct it cn be llowed to be discontinuous t [countbly ] infinitely mny plces), s long s it is defined for ll x [, b] nd is monotonic. D. DeTurck Mth C: Integrl/functions 3 / 32
4 Setup for the integrl upper nd lower sums For f monotoniclly incresing on [, b], nd P prtition of [, b]. Definition: (upper nd lower sums) The upper sum of f corresponding to the prtition P is U(f, P) = n f (x i )(x i x i 1 ) i=1 nd the lower sum of f corresponding to the prtition P is L(f, P) = n f (x i 1 )(x i x i 1 ) i=1 Since f is incresing, we hve f (x i ) is the mximum vlue of f on [x i 1, x i ] nd f (x i 1 ) is the minimum. Therefore U(f, P) L(f, P) D. DeTurck Mth C: Integrl/functions 4 / 32
5 Setup for the integrl definition Since there s lwys common refinement P for ny prtitions P 1 nd P 2, we hve L(f, P 1 ) L(f, P) U(f, P) U(f, P 2 ) where P is common refinement for P 1 nd P 2. So for ny pir of prtitions P 1 nd P 2 we hve L(f, P 1 ) U(f, P 2 ). Therefore sup L(f, P) over ll prtitions P is less thn or equl to inf U(f, P). If these re equl then their common vlue is clled the integrl: f (x) dx D. DeTurck Mth C: Integrl/functions 5 / 32
6 Proving existence regulr prtitions To show tht the integrl exists, it is sufficient to find, for ny ε > 0, prtition P such tht U(f, P) < L(f, P) + ε. For monotonic functions, we cn do this by using sufficiently fine regulr prtitions these re prtitions hving the x i s evenly spced (so x i x i 1 = b for ll i = 1,..., n). n Theorem If f : [, b] R is monotonic, then f (x) dx exists. Proof: Given ε > 0, choose n so lrge tht the regulr prtition P f (b) f () with n steps, will hve U(f, P) L(f, P) = < ε. n D. DeTurck Mth C: Integrl/functions 6 / 32
7 Functions of bounded vrition Proposition (Greter generlity) If f (x) cn be written s the sum of two monotonic functions p(x) nd q(x) with p incresing nd q decresing on [, b] (such function is clled function of bounded vrition), then f (x) dx exists nd is equl to p(x) dx + q(x) dx. The set of functions of bounded vrition on closed bounded intervl is ctully quite generl, so even though we strted out with somewht restrictive definition we hve creted pretty powerful form of the integrl. D. DeTurck Mth C: Integrl/functions 7 / 32
8 Bsic properties Becuse the upper nd lower sums hve these properties, it follows tht the integrl does: 1 Linerity: αf (x) + βg(x) dx = α f (x) dx + β g(x) dx for constnts α, β nd f nd g re functions of bounded vrition on [, b]. 2 Monotonicity: If f (x) g(x) for ll x [, b] then b f (x) dx b g(x) dx. 3 If f (x) is of bounded vrition on [, b], then so is f (x) nd f (x) dx f (x) dx. D. DeTurck Mth C: Integrl/functions 8 / 32
9 More bsic properties 4 If < b < c then ˆ c f (x) dx = ˆ c f (x) dx + f (x) dx. b 5 If m inf{f (x) x [, b]} nd M sup{f (x) x [, b]} then (b )m f (x) dx (b )M. 6 Men vlue theorem for integrls: If f is continuous nd of bounded vrition on [, b] then there is c with < c < b such tht f (c) = 1 f (x) dx. b D. DeTurck Mth C: Integrl/functions 9 / 32
10 Fundmentl Theorem integrls of derivtives Fundmentl theorem of clculus I: Integrls of derivtives Let F (x) be differentible function on [, b] with derivtive F (x), nd suppose F is function of bounded vrition on [, b]. Then F (x) dx = F (b) F (). We cn prove this for functions with monotonic derivtives nd then use linerity. If F (x) is monotoniclly incresing, then for ny intervl (x i 1, x i ) prtition P we hve F (x i 1 ) F (x i) F (x i 1 ) x i x i 1 F (x i ) by the men vlue theorem (for derivtives). D. DeTurck Mth C: Integrl/functions 10 / 32
11 Fundmentl Theorem proof conclusion But then L(f, P) = n F (x i 1 )(x i x i 1 ) i=1 n F (x i ) F (x i 1 ) i=1 n F (x i )(x i x i 1 ) = U(f, P). i=1 But the middle sum telescopes to F (x n ) F (x 0 ) = F (b) F (). Now F (b) F () is trpped between sup L(f, P) nd inf U(f, P), both of which re equl to the integrl P f (x) dx. P D. DeTurck Mth C: Integrl/functions 11 / 32
12 Second fundmentl theorem derivtives of integrls Fundmentl theorem of clculus II: Derivtives of integrls Let f (x) be continuous function of bounded vrition on [, b]. Define the function F (x) vi F (x) = ˆ x f (t) dt. Then F is differentible nd F (x) = f (x). First, if h > 0 we hve F (x + h) F (x) = ˆ x+h x f (t) dt = hf (c) for some c between x nd x + h by properties 4 nd 6 (men vlue theorem for integrls). D. DeTurck Mth C: Integrl/functions 12 / 32
13 Second fundmentl theorem proof conclusion Therefore the difference quotient F (x + h) F (x) h = f (c) where x < c < x + h But since f is continuous, f (c) f (x) s h 0 +. ˆ x For h < 0 we use tht F (x + h) F (x) = x+h f (t) dt = hf (c) for some c between x + h nd x, nd the proof goes through s for the h > 0 cse. D. DeTurck Mth C: Integrl/functions 13 / 32
14 Using FTC I to compute integrls Of course, we cn use the first FTC to clculte integrls, once we hve ntiderivtive formuls obtined by turning round derivtive formuls. Right now, though, we don t hve much more thn derivtives of rtionl powers of x. So we hve tht since if r is rtionl number, then d dx (x r ) = rx r 1 x r 1 dx = 1 r (br r ) provided r 0. Of course we usully replce r by r + 1 nd sy x r dx = 1 r + 1 (br+1 r+1 ) provided r 1. D. DeTurck Mth C: Integrl/functions 14 / 32
15 Wht bout r = 1? 1 So now we hve question: wht is dx? Becuse x f (x) = 1/x is monotoniclly decresing on ny intervl [, b] where > 0 the integrl should exist. So, provisionlly, let s define the function L(x) vi: L(x) = ˆ x 1 1 t dt nd see if we cn uncover some of its properties. To begin, we know only tht L(1) = 0 nd L (x) = 1 x. D. DeTurck Mth C: Integrl/functions 15 / 32
16 Multiplictive property of L(x) Consider the function L(x) for positive constnt. By wht we know bout L nd the chin rule we hve d dl L(x) = dx dx = x x = 1 x. So the derivtive of L(x) is the sme s the derivtive of L(x), therefore the two differ by constnt. Wht constnt? Well, putting x = 1 gives tht L(x) t 1 is L(). So L(x) = L() + L(x). Look fmilir? D. DeTurck Mth C: Integrl/functions 16 / 32
17 Power property of L How bout L(x r ) if r Q? we hve d dx L(x r ) = 1 d x r dx x r rx r 1 = x r = r x = r dl(x) dx. So the derivtive of L(x r ) is r times the derivtive of L(x). Moreover, t x = 1, we hve L(x r ) x=1 = 0. This shows tht L(x r ) = rl(x). Agin, look fmilir? So fr, we hve defined L(x) only for x 1. But we could define it for 0 < x < 1 either by sying tht L(x) = L(1/x) or else by ˆ 1 1 sying tht L(x) = dx. It s n esy exercise to show tht x x these definitions gree. Likewise, if we define L(x) = L( x) for x < 0 we will hve L (x) = 1/x there s well. D. DeTurck Mth C: Integrl/functions 17 / 32
18 The rnge of ln(x) We hve defined the function L(x) s the integrl of 1/x with L(1) = 0 (nd L( 1) = 0). So L(x) grees with the nturl logrithm function ln(x). Since 1 1 x 1 2 for 1 < x < 2, we hve tht 1 > ln(2) > 1 2. Therefore ln(2 n ) > n 2 which shows tht ln(x) s x. Likewise, ln(1/2) < 1/2, so ln(1/2 n ) < n/2, showing tht ln(x) s x 0 +. Therefore the rnge of ln(x) is ll of R, nd ln(x) is strictly monotoniclly incresing function, since its derivtive is positive. Therefore it hs n inverse function tht will mp R to (0, ). Let s cll the inverse function E(x) nd study its properties. D. DeTurck Mth C: Integrl/functions 18 / 32
19 The inverse of ln x nd rbitrry powers The function E(x) is defined by E(ln(x)) = x for x > 0 nd ln(e(x)) = x for ll x R. Becuse ln(1) = 0 we hve E(0) = 1. And since ln(e()e(b)) = ln(e()) + ln(e(b)) = + b, we hve E( + b) = E()E(b). Likewise, since for R nd r Q, we hve E(r) = E() r. ln(e() r ) = r ln(e()) = r, Therefore E(x) behves like rising some number to the xth power, nd it gives us wy to define x y for ll rel y, nmely x y = E(y ln x). D. DeTurck Mth C: Integrl/functions 19 / 32
20 The number e We hve E(x) is like rising some number to the xth power. Wht number? Well, tht would be E(1), which we ll cll e. So e is the number such tht ln e = 1. Becuse ln 2 = ˆ 2 1 ˆ dt < 1 nd ln 4 = t 1 t dt > > 1 we know tht 2 < e < 4. We ll get better estimtes lter. But for now, we ll write e x for E(x). Derivtive: If y = e x then x = ln y so 1 = 1 y in other words d dx ex = e x. dy dy, nd so dx dx = y, D. DeTurck Mth C: Integrl/functions 20 / 32
21 An importnt differentil eqution So e x is solution of the differentil eqution y = y with initil vlue y(0) = 1. More generlly Importnt! The unique solution of the initilvlue problem y = ky, y(0) = C is y = Ce kx. It is esy to check tht Ce kx stisfies the differentil eqution nd the initil condition. If f (x) were nother solution, then f = kf nd f (0) = C nd we could clculte: d dx ( ) f (x) Ce kx = 1C (e kx f (x) ke kx f (x)) = e kx (kf kf ) = 0 C so e kx f (x) is constnt. Wht constnt? Evlute t zero to get tht it s 1, so f (x) = Ce kx fter ll. D. DeTurck Mth C: Integrl/functions 21 / 32
22 Better estimtes of e Now tht we know tht e x stisfies y = y, we know tht ll the derivtives of e x re e x. In prticulr, ll of them re equl to 1 t x = 0. Therefore, by the Tylor estimte: e = e 1 = ! + 1 2! + 1 3! n! + M (n + 1)! where M is the vlue of the (n + 1)st derivtive of e x evluted somewhere between 0 nd 1. From our erlier primitive estimte we know tht 1 < M < 4. So for ny n > 0 we cn write 1 (n + 1)! < e [ ! + 1 2! + 1 3! n! ] < 4 (n + 1)! D. DeTurck Mth C: Integrl/functions 22 / 32
23 e is irrtionl Use the lst inequlity to show tht e is not rtionl number. If it were, then e = p/q for some positive integers p nd q nd we would hve 1 (n + 1)! < p q Multiply by n! nd get 1 (n + 1) < n!p q n! [ ! + 1 2! + 1 3! ] < n! [ ! + 1 2! + 1 3! ] < n! 4 (n + 1)! 4 (n + 1) If n q then the number in the middle is n integer, but if n > 4 then we ve trpped n integer between two positive rtionl numbers both of which re less thn 1. A contrdiction! D. DeTurck Mth C: Integrl/functions 23 / 32
24 Deciml pproximtion of e Strt gin from the Tylor estimte e = e 1 = ! + 1 2! + 1 3! n! + where we know tht 1 < M < 4. If we evlute this for n = 19 we get tht M (n + 1)! < e < Probbly close enough for most purposes. D. DeTurck Mth C: Integrl/functions 24 / 32
25 Trig? How should we pproch trigonometric functions? Let s strt with the function ˆ x 1 A(x) = 1 + t 2 dt Clerly A(x) is n incresing function nd since A(x) < ˆ x 1 1 t 2 dt = 2 1 x, we hve tht A(x) is defined for ll x R (we hve A(x) is n odd function of x) nd A(x) < 2 for ll x. Since A(x) is incresing nd bounded bove, it must pproch limit s x. Let s cll this limit π/2: Definition of π π = 2 lim x A(x). D. DeTurck Mth C: Integrl/functions 25 / 32
26 Inverse functions Since A(x) is monotoniclly incresing function from R to the intervl ( 1 2 π, 1 2π), it hs n inverse function tht we will cll T (x) from ( 1 2 π, 1 2π) to R. Since d dx A(x) = 1, if x = A(y) (or y = T (x)) we hve 1 + x 2 1 = y 2 dy dx Therefore d dx T (x) = 1 + T (x)2. From our definition of π/2, we see tht lim T (x) = x π/2 D. DeTurck Mth C: Integrl/functions 26 / 32
27 S(x) nd C(x) Next. let S(x) nd C(x) be the functions stisfying S(x) = T (x) 1 + T (x) 2 nd C(x) = T (x) 2. It s immeditely pprent tht S(x) 2 + C(x) 2 = 1. Moreover d dx S(x) = = 1 + T 2 T T TT 1+T T 2 = (1 + T 2 ) 2 T 2 (1 + T 2 ) (1 + T 2 ) 3/2 1 (1 + T 2 = C(x) ) 1/2 nd likewise d C(x) = S(x). dx D. DeTurck Mth C: Integrl/functions 27 / 32
28 The differentil eqution for S(x) nd C(x) Both S(x) nd C(x) stisfy the secondorder DE: y + y = 0. Additionlly S(0) = 0 nd S (0) = 1 nd C(0) = 1 nd C (0) = 0. A new kind of initilvlue problem! Proposition There is t most one solution of the initilvlue problem y + y = 0 y(0) = y (0) = b. If there were two, cll them y nd z, then their difference u = y z would stisfy u + u = 0 with u(0) = u (0) = 0. But then d dx (u2 + u 2 ) = 2uu + 2u u = 2u (u + u ) = 0, so this quntity is constnt, nd in fct is zero becuse it is zero when x = 0. But tht mens u 2 = 0, so u = 0 nd so y = z. D. DeTurck Mth C: Integrl/functions 28 / 32
29 Extensions, periodicity At this point, becuse T (x) is defined only for 1 2 π < x < 1 2π, the functions S(x) nd C(x) hve these sme restrictions. However, since T (x) s x 1 2 π, you cn see tht S(x) 1 nd C(x) 0 s x 1 2 π. It follows tht C (x) 1 nd S (x) 0 there, nd becuse we hve tht S(π x) nd C(π x) re solutions of y + y = 0, we see we cn extend S nd C to be defined up to π. Keep repeting this trick to get tht S nd C re defined for ll x nd re periodic with period 2π. D. DeTurck Mth C: Integrl/functions 29 / 32
30 More trig identities Addition formuls: Wht is sin( + b)? Well, the function y = sin( + x) stisfies y + y = 0 nd y(0) = sin nd y (0) = cos. Therefore we hve y = sin( + x) = sin cos x + cos sin x Get ddition formuls for other trig functions in similr wy. D. DeTurck Mth C: Integrl/functions 30 / 32
31 Integrtion by substitution We puse to do the proof of fmilir integrtion technique, which is the chin rule in reverse: Integrtion by subsitution Suppose f : [, b] R nd g : [c, d] R re of bounded vrition, nd tht the derivtive of g exists nd is of bounded vrition on (c, d). Further, ssume tht the imge of g : (c, d) R is contined in the intervl (, b). Then ˆ d c f (g(x))g (x) dx = Proof: Apply FTC to the function H(x) = ˆ x c f (g(t)) dt ˆ g(d) g(c) ˆ g(x) g(c) f (x) dx. f (t) dt. D. DeTurck Mth C: Integrl/functions 31 / 32
32 Wht is π? We ve defined π/2 to be the first positive zero of cos x, but now we cn give it geometric significnce. To do so, we ll use the substitution g(x) = sin x on the following integrl: ˆ x 2 dx = ˆ π/2 = = 0 ˆ π/2 0 1 sin 2 θ cos θ dθ cos 2 θ dθ = ( ) θ sin 2θ π/ ˆ π/2 0 = π 4 1 (1 + cos 2θ) dθ 2 This shows tht π is the re of the unit circle. At lst. D. DeTurck Mth C: Integrl/functions 32 / 32
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