LECTURE. INTEGRATION AND ANTIDERIVATIVE.


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1 ANALYSIS FOR HIGH SCHOOL TEACHERS LECTURE. INTEGRATION AND ANTIDERIVATIVE. ROTHSCHILD CAESARIA COURSE, 2015/6 1. Integrtion Historiclly, it ws the problem of computing res nd volumes, tht triggered development of the modern clculus Definite integrl nd the re under the grph. Computing the re of the shpe bounded, sy, by the grph of f(x) = x 2, the horizontl xis y = 0 nd the verticl line x = 1, cn be done by elementry methods: this mounts to computtion of the sums n 1 ( ) k 2 Sn 1 n ( ) k 2 = n n, 1 S+ n = n n k=0 for lrge n N. One cn prove by induction n explicit formul for the sum n 1 k2 = 1 n3 6n(n+1)(2n+1) = 3 + n2 2 + n 6 nd conclude tht S± n = 1 3 +o(1) s n. However, if the function f(x) = x 2 is replced by more generl polynomil of higher degree, then the summtion becomes more nd more sophisticted. Alterntive is required. But first of ll, wht is the re? more specificlly, wht is the re of the trpeze between the grph of positive function f : [, b] R +, the xis y = 0 nd two lines x = nd x = b? The definition is bsed on the ide of the inequlity vlid for polygons on R 2 : if P, Q re two polygons with P Q, then re(p ) re(q). Let P = (x 0,..., x n ) be n points on [, b], = x 0 < x 1 < < x n = b (it will be clled n nprtition), nd f : [, b] R function on [, b]. Define the two uxiliry functions, f P (x) = f + P (x) = k=1 inf f(x), if x [x k 1, x k ], x [x k 1,x k ] sup f(x), if x [x k 1, x k ]. x [x k 1,x k ] The grph of ech of these functions is polygon on the (x, y)plne, hence the re S ± P (f) cn be esily computed for ny prtition P. Since f P (x) f(x) f + P (x), we lwys hve the inequlity S P (f) S+ P (f). If the re Dte: December 22,
2 2 ROTHSCHILD CAESARIA COURSE, 2015/6 S(f) under the grph of f exists in ny sense, it must seprte the two sets, S P (f) S(f) S+ P (f) for ny two prtitions P, P. If there is more thn one such number, we re in trouble, but if this number is unique, there is no other choice but to cll it the re. Definition 1. A function f : [, b] R is clled (Riemnn) integrble (on this segment), if sup S P (f) = inf P P S+ P (f), the supremum/infimum being tken over ll nprtitions P for ll n. The (unique) common vlue bove is clled the Riemnn integrl of f nd denoted f(x) dx. The corresponding sums re clled lower (resp., upper) integrl sums for f on [, b]. Remrk 1. An unbounded (on finite segment) function f does not dmit either lower or upper pproximtion f ± so one of integrl sums if ±. Thus by definition it cnnot be Riemnn integrble. Remrk 2. The nottion is both historiclly nd mthemticlly justified: it is for good reson tht insted of the function f we integrte the differentil f(x) dx. Lter we will lern tht this nottion is well dpted mnemoniclly to the chnge of the independent vrible x. By the wy, the nottion of the derivtive f () lso concels the choice of the independent vrible, so in sense the oldfshioned nottion df dx () is better! Obviously, the constnt function f(x) = c is integrble on ny (finite) intervl, nd c dx = c(b ). The functions f(x) = x2 nd f(x) = x 3 re lso integrble, s the bove computtions show (one hs to ensure tht there is no gp between the lower nd the upper sums). Theorem 1. Any function f : [, b] R continuous on [, b], is integrble. Some functions re nonintegrble, however. Exmple 1. The Dirichlet function f(x) = 1 or 0 depending on whether x is rtionl or not, is nonintegrble on ny segment. Indeed, since both rtionl nd irrtionl numbers re dense, ll upper sums will be b, while ll lower sums re zero. Exmple 2. If f is integrble on [, b] nd g differs from f only in finitely mny points, then g is lso integrble on [, b]. Exmple 3. The function f(x) = 1/x is nonintegrble on the segment [0, 1] regrdless of the wy how one defines f(0). Remrk 3. Integrbility on infinite segments is out of question currently: ny finite prtition will necessrily involve the infinite difference x n x n 1, lthough sometimes the res of the infinite shpes re finite. Remrk 4. If f is integrble on [, b], then it is integrble on [, c] for ny c (, b). Prove it!
3 ANALYSIS FOR HIGH SCHOOL TEACHERS 3 2. The Newton Leibniz formul: the mgic key Theorem 2. Assume tht f : [, b] R is continuous function, nd F (c) = c f(x) dx is the integrl of f over [, c] s function of the right endpoint c [, b]. Then F is differentible function on [, b], F () = 0 nd F (c) = f(c) t ny point c [, b]. Definition 2. A function F is clled n ntiderivtive of f, if F is differentible nd DF = f. Corollry 3 (inversion). If F is n ntiderivtive of f, then f(x) dx = F (b) F (). Computtion of derivtives is n explicit process, in prticulr, derivtive of n elementry function is gin n elementry function. Thus there is Google wy to compute integrls: sk computers to differentite ll possible functions nd serch mong the results to find out whether your specific function is somewhere in the dtbse. There re esy rules which follow from the corresponding rules of the differentition: linerity (ntiderivtive of liner combintion is liner combintion of ntiderivtives); ntileibniz rule: if f = uv nd ntiderivtive of v is known, V = v, then one cn express the nswer in terms of uv nd the ntiderivtive of the function g = u V. There re no priori resons why the ltter ntiderivtive is esier to compute, but sometimes it indeed is. This trick is clled integrtion by prts. However, the thumb rule is tht you should be quite lucky to find n explicit ( elementry ) ntiderivtive for your function. Exmple 4. If u = x, v = e x, then V = e x, so the ntiderivtive for xe x cn be expressed through the function uv = xe x nd the ntiderivtive of u V = e x which is gin the exponent e x. Repeting the sme trick, one cn reduce computtion of the ntiderivtive of x 2 e x to tht of xe x, lredy known by the previous trick. It looks like trechery, but surprisingly it works nd, moreover, llows to define the vlue of the fctoril F (n) = n! from nturl to ll rel (nd even complex) vlues of the rgument (Euler). Mysteriously, ( 1 2 )! = 1 2 π... Exmple 5. All monomils x n, n Z, re ntiderivtives of the monomils x n+1 n+1, except for n = 1: there is no lgebric function f stisfying the eqution f = 1 x. Antiderivtive of x 1 is the logrithm ln x. This is how the logrithmic nd exponentil functions enter the lgebric world. Exmple 6. Antiderivtives of the elementry trigonometric functions (when they cn be computed) re usully trigonometric.
4 4 ROTHSCHILD CAESARIA COURSE, 2015/6 The bridge between the trigonometric functions nd lgebric functions is the identity (rctn x) = x 2. This hs mzing consequences! Trigonometric constnts (e.g., π, the most mysterious number in mth) cn be expressed through the rtionl series. Upon the second thought, π/2 is the re between the curve x 2 +y 2 = 1 (the unit circle) nd the upper hlfplne {y 0}... Integrtion is source of mny mircles Chnge of vribles. Assume tht f is integrble on [, b], x = ϕ(t) is monotonously growing differentible function, ϕ(a) =, ϕ(b) = b. Theorem 4. f(x) dx = B A f(ϕ(t)) dϕ (t) dt = dt First proof. Use the chin rule for derivtion B A (f ϕ) dϕ. d df (ϕ(t)) (F (ϕ(t)) = dϕ(t) dt dx dt pply it to the cse where F = f nd integrte both sides from A to B. Second proof. Consider prtition P nd its ϕpreimge Q = {A = t 0 < t 1 < < t n = B}, ϕ(t i ) = x i. Then the vlues of the functions f ± (ϕ(t)) on [t k 1, t k ] coincide with the vlues of f ± on [x k 1, x k ], but x k x k 1 by the Lgrnge theorem is ϕ (ξ k )(t k t k 1 ) for some intermedite points ξ k (t k 1, t k ). If the derivtive ϕ is continuous, the upper nd lower sums S ± P (f ϕ) converge to the integrl of (f ϕ) ϕ The Stieltjes integrl. One cn slightly generlize the notion of the Riemnn integrl. Let f, g be two functions on [, b], with f continuous nd g monotonous (growing). Insted of the Riemnn sums S ± P (f) = k f ± (x)(x k x k 1 ) one cn consider the Stieltjes sums S ± P (f, g) = f ± (x)(g(x k ) g(x k 1 )). k These sums converge in the sense tht sup S P (f, g) = inf P P S+ P (f, g). The common vlue is the Stieltjes integrl denoted by f(x) dg(x). Exmple 7. If g(x) hs continuous derivtive g, then the Stieltjes integrl is equl to the Riemnn integrl, f dg = f(x)g (x) dx.
5 ANALYSIS FOR HIGH SCHOOL TEACHERS 5 Exmple 8. Assume tht g(x) is step function, g(x) = c i if x [x i 1, x i ) for some prtition P, c 1 < c 2 < < c n. Then the Stieltjes integrl of continuous function f reduces to finite sum, n 1 f dg = f(x i )(c i+1 c i ). i=1 Thus the Stieltjes integrl interpoltes between the genuine integrtion nd the finite sums (if g hs jumps). Theorem 5 (chnge of vribles in the Stieltjes integrl). If ϕ : [A, B] [, b], t ϕ(t) is monotone differentible chnge of vribles, f, g : [, b] R re two functions such tht the Stieltjes integrl f dg exists, then f dg = B A F dg, F = f ϕ, G = g ϕ. Proof. This is tutologicl sttement relted to the upper/lower sums. Contempltion of the Stieltjes integrl nd its trnsformtion clrifies the reson, why we integrte not function f, but rther differentil df = f(x) dx. The Newton Leibnitz formul then becomes the obvious identity df = F (b) F () Integrbility of discontinuous functions. Integrtion improves the regulrity of functions. Exmple 9. If f(x) = sign(x) = ±1 for ±x > 0, then (regrdless of the choice for f(0)) this function is integrble on ny segment, nd 0 signx dx = x. Note tht the function F (x) = x is not exctly ntiderivtive of f: F is nondifferentible t = 0. Yet F is continuous nd F () = f() for ll 0. This exmple cn simplified even more: let f be continuous nd F = f. Consider the function f which differs from f t one point only. Then f is discontinuous t this point, yet integrble, nd F is n lmost ntiderivtive of f s well. Proposition 6. Any bounded function hving only finitely mny points of discontinuity, is integrble. In fct, stronger sttement cn be proved. Definition 3. We sy tht subset C R is εsmll for some ε > 0, if it cn be covered by finitely mny open intervls of totl length less thn ε. The set C hs zero length, if it is εsmll for ny ε > 0. Of course, ll finite sets hve zero length. One cn construct infinite sets of zero length, e.g., C = { 1 n : n N} (prove it!).
6 6 ROTHSCHILD CAESARIA COURSE, 2015/6 Theorem 7. A bounded function continuous on the complement to set of length zero, is integrble. This is lredy very close to the necessry nd sufficient condition of integrbility (Lebesgue, erly 20th century). Definition 4. A subset C R is sid to hve zero mesure, if for ny ε > 0 it cn be covered by countbly mny open intervls of totl length less thn ε > 0. Exmple 10. Any countble subset of R, e.g., C = Q, hs mesure zero. Theorem 8. A bounded function is integrble if nd only if it is continuous on the complement to set of mesure zero. 3. Conclusions The problem of computing res, volumes etc., cn be solved by computing definite (Riemnn) integrl. By the Newton Leibniz fundmentl theorem, insted of clculting the limits of integrl sums, one cn serch for n ntiderivtive. Unlike derivtion which is lwys explicitly computble (hence cn be trusted to computers mnging symbolic computtions), ntiderivtions cn be relly new functions not dmitting explicit expression. Integrtion regulrizes functions (mkes them behving better). The Riemnn pproch to integrtion is not the only one possible: there exist other constructions (most notbly the Lebesgue integrl) which cn be pplied to broder clss of functions. One cn work out some wys to define (nd compute) integrls of unbounded functions or integrls over infinite intervls. For instnce, the re under the grph of the function f(x) = 1 on the whole rel line R, is 1+x 2 equl to π. In fct, integrls over the entire rel line R re sometimes esier to compute thn integrls over finite segments (the resons for this lie in the complex domin).
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