MAA 4212 Improper Integrals


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1 Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which we intuitively feel ought to be integrble, but which re not Riemnn integrble ccording to the definition. For exmple, the expression t dt mkes no sense s Riemnn integrl, since the integrnd is not defined t t = 0. Even if we fix tht problem, by defining function tht s t 1/2 for t > 0 nd (sy) 0 for t = 0, this new function is still not Riemnnintegrble over [0, 1] becuse it isn t bounded. However, if formlly mke the chnge of vribles t = u 2 ( formlly mens shoot first, sk questions bout vlidity lter ), the integrl bove gets trnsformed into du, which is s nice n integrl s they come. Furthermore, if we go bck to our originl integrl nd think of it s representing re under curve, there is useful sense in which this re is finite: tke the re below the curve between t = ɛ nd t = 1, nd let ɛ 0. Either wy of looking t the originl integrl, the nswer we formlly clculte is 2. These considertions suggest tht we ought to enlrge the clss of functions we re willing to cll integrble, nd modify our definition of integrl. The types of integrls we ll del with in this hndout re often clled improper integrls, but we ll simply cll them integrls here. Terminology. In this hndout, the words integrl nd integrble will not be synonymous with Riemnn integrl nd Riemnnintegrble. (In Rosenlicht, they re synonymous, but we will need to be clerer here on wht notion of integrtion we re tlking bout.) We will use nottion Riemnn f(t)dt (with the word Riemnn in front of the integrl sign) to denote the Riemnn integrl. Whenever we write hypotheses such s Let f : [, b] R, we understnd this s shorthnd for Let, b R with < b nd let f : [, b] R; nlogous interprettions pply if [, b] is replced by (, b], [, b), or (, b). Also, we write lim x nd lim x in plce of lim x, lim x + respectively. 1 Integrls over bounded intervls. Definition 1. We will sy tht relvlued function f is GRintegrble (for generlized Riemnn integrble ) on the intervl [, b] if either (i) f is defined on (, b] nd is Riemnnintegrble over [y, b] for ll y (, b], nd lim y Riemnn y f(t)dt exists; or 1
2 (ii) f is defined on [, b) nd is Riemnnintegrble over [, y] for ll y [, b), nd lim y b Riemnn y f(t)dt exists. (This is temporry definition tht will be generlized nd finlized in Definition 3.) Note tht if f stisfies both conditions (i) nd (ii), then it is Riemnnintegrble over [, b]. In prticulr, every Riemnnintegrble function is GRintegrble. Note. The terms generlized Riemnn integrl nd GRintegrble re specific to these notes; they re not stndrd terminology. Exercise. 1. Suppose f : [, b] R is Riemnnintegrble on [, b]. Prove tht for ll c [, b], the functions g, h defined by g(x) = Riemnn x c f(t)dt, h(x) = Riemnn c x f(t) dt re continuous. In prticulr, Exercise 1 implies tht if f is Riemnnintegrble on [, b], then lim [Riemnn y y f(t)dt] = Riemnn f(t)dt (1) nd lim [Riemnn y b y f(t)dt] = Riemnn f(t)dt. (2) This suggests using equtions (1) nd (2) to define the integrl in certin nonriemnnintegrble cses. Definition 2. Let f : [, b] R. If f stisfies condition (i) in Definition 1, we define the generlized Riemnn integrl f(t)dt = lim y [Riemnn Similrly if f stisfies condition (ii) in Definition 1, we define f(t)dt = lim y b [Riemnn In both cses we will sy tht f is GRintegrble on [, b]. y y f(t)dt]. (3) f(t)dt]. (4) Note. In plce of sying f is GRintegrble, we often sy the integrl exists or the integrl converges. Equtions (1) (2) show tht there is no mbiguity in Definition 2; if f stisfies both (i) nd (ii) in Definition 1, then the limits in equtions (3) nd (4) re equl. Moreover, if f is Riemnnintegrble on [c, d] then d c f(t)dt = Riemnn d c f(t)dt. Hence if f is GRintegrble on [, b] we my write equtions (3) nd (4) more simply s 2
3 f(t)dt = lim y y y f(t)dt = lim y b f(t)dt. (5) In the exercises below we will develop useful comprison test for telling whether certin functions re GRintegrble. To prove the comprison test vlid it is helpful to hve the following simple lemm, which is n nlog of (nd is equivlent to) the Cuchy criterion for sequences. (This is essentilly the Proposition on p. 74 of Rosenlicht, but it ws phrsed there in too nrrow wy for our purposes.) Lemm 1. Let (E, d), (E, d ) be metric spces. Assume tht p 0 is cluster point of E, tht E is complete, nd tht f is function from E to E. Then lim p p0 f(p) exists iff for ll ɛ > 0 there exists δ > 0 such tht d (f(p), f(q)) < ɛ whenever d(p, p 0 ) nd d(q, p 0 ) re both < δ. Proof. ( ) Assume the limit exists nd hs vlue L E. Let ɛ > 0. Then there exists δ such tht d(p, p 0 ) < δ implies d (f(p), L) < ɛ/2. Hence if p, q B δ (p 0 ), the tringle inequlity implies d (f(p), f(q)) < ɛ. ( ) Let ɛ > 0, nd choose δ > 0 such tht p, q B δ (p 0 ) implies d (f(p), f(q)) < ɛ/2. Since p 0 is cluster point of E there exists sequence {p n } converging to p 0. Let N be such tht n N implies p n B δ (p 0 ). Then for n, m N we hve d (f(p n ), f(p m )) < ɛ/2, so the sequence {f(p n )} is Cuchy. Since E is complete, this sequence converges, sy to L. Let n N be such tht d (f(p n ), L) < ɛ/2 nd let q B δ (p 0 ). Then d (f(q), L) d (f(q), f(p n )) + d (f(p n ), L) < ɛ, so lim q p0 f(q) exists (nd equls L). An equivlent form of this lemm is the first sentence of: Lemm 1. Let the hypotheses be s in Lemm 1. Then lim p p0 f(p) exists iff for every sequence {p n } tht converges to p 0, the sequence f(p n ) converges. If lim p p0 f(p) exists, then it equls lim n f(p n ) for every sequence {p n } tht converges to p 0. Remrk. We hve used weker version of this lemm before: lim p p0 f(p) exists iff there exists L such tht for every sequence {p n } tht converges to p 0, the sequence f(p n ) converges to L. Lemm 1 strengthens the impliction by removing the ssumption tht the sequences {f(p n )} hve the sme limit. Proof of Lemm 1. ( direction of first sentence of conclusion): Turn the crnk. ( direction of first sentence of conclusion, nd second sentence of conclusion) Suppose tht whenever p n p 0, {f(p n )} converges. Choose two such sequences {p n }, {q n } nd suppose f(p n ) L 1 while f(q n ) L 2. The spliced sequence p 1, q 1, p 2, q 2,... lso converges to p 0, but if L 1 L 2, the sequence f(p 1 ), f(q 1 ), f(p 2 ), f(q 2 ),... cnnot converge. Hence L 1 = L 2 ; i.e for ll sequences {p n } converging to p 0, the limiting vlue of f(p n ) is the sme, sy L. Now suppose tht lim p p0 f(p) L or doesn t exist. Then there exists 3
4 ɛ > 0 such tht for ll n, there exists p n B 1/n (p 0 ) such tht d (f(p n ), L) ɛ. Since p n p 0, this is contrdiction. Exercises. 2. Let b >. Prove tht f : x 1 is GRintegrble on [, b] iff p < 1. (Here p cn (x ) p be ny rel number. Thus for 0 < p < 1, f is GRintegrble but not Riemnnintegrble on [, b].) 3. Let b >. Assume f is Riemnnintegrble on [y, b] whenever < y b. (Note tht ny function continuous on (, b] stisfies this criterion, even if not defined or continuous t.) Assume there exists function g : (, b] R, GRintegrble on [, b], such tht f(x) g(x), x (, b]. Prove tht f is GRintegrble on [, b] nd tht f(x)dx g(x)dx. 4. Let b >. Assume f is continuous on (, b] nd tht, for some p < 1, the function x (x ) p f(x) is bounded on (, b]. Prove tht f is GRintegrble on [, b]. 5. Stte the nlogs of exercises 13 with the roles of nd b reversed (i.e. with (, b] replced by [, b)). Essentilly the sme proofs work, of course. 6. Without using ny reference to trigonometric functions or their inverses, prove tht t 2 dt exists (i.e. tht (1 t 2 ) 1/2 is GRintegrble on [0, 1]). Remrk: The vlue of this integrl cn be tken s the definition of π/2. In this hndout, we will refer to point x 0 R s singulrity of f if x 0 is in the closure of the domin of f, but there exists no closed intervl contining x 0 over which f, extended rbitrrily to x 0 if x 0 / domin(f), is Riemnnintegrble. To mke the concept more concrete, it s useful to picture function which is defined nd continuous ner x 0 but not t x 0, nd for which lim x x0 f(x) does not exist (e.g. 0 is singulrity of x 1/x). Grdenvriety functions to which Definitions 1 nd 2 pply re functions tht re continuous on the interior of n intervl but hve singulrity t one endpoint or the other, but not both. We next wnt to extend our definition of GRintegrble on [, b] to include functions tht hve singulrities t more thn one point (e.g. both endpoints) nd/or t n interior point of the intervl of integrtion. Our extension will be bsed on the next exercise, which you should think of s generliztion of the Proposition t the bottom of p. 123 in Rosenlicht. Exercise. 7. Prove tht f is GRintegrble over [, b], in the sense of Definition 2, if nd only if t lest one of the following two conditions holds: (i) for ll c (, b), f is GRintegrble over (, c] nd Riemnnintegrble over [c, b], or (ii) for ll c (, b), f is Riemnnintegrble over [, c] nd GRintegrble over [c, b). When the integrls exist, prove tht 4
5 (End of exercise 7.) f(t)dt = c f(t)dt + c f(t)dt. (6) Now suppose we hve function f defined on [, b], continuous except for n interior singulrity t single point c (, b). If the GRintegrls over [, c] nd [c, b] exist, we cn define the GR integrl of f over [, b] by tking eqution (6) s definition. Similrly, if f is continuous on the interior but singulr t both endpoints, nd if for some interior point c the GR integrls over [, c] nd [c, b] exist, then we cn gin tke (6) s definition of the lefthnd side. Finlly, if we hve function which is continuous [, b] except for singulr points s 1,..., s n, we cn chop up [, b] into finite number of subintervls on which f hs only one singulrity (intersperse nonsingulr points y i with the s i s), nd use the nlog of eqution (6) with one term for ech subintervl to define the lefthnd side. Looking over wht we ve just sid, we see tht we never relly needed f to be continuous off the set of singulr points (though tht s most commonly wht we see in prctice). Our forml definition becomes: Definition 3. Suppose the relvlued function f hs the following property: there exist points s 0, s 1,..., s n+1, with = s 0 nd b = s n+1, such tht f is defined t every point of [, b] except possibly the s i s, nd such tht for 0 i n, f is GRintegrble over [s i, s i+1 ] in the sense of Definition 1. Then we sy tht f is GRintegrble over [, b], nd define n si+1 f(t)dt = f(t)dt. (7) s i i=0 (Note tht we don t require f to be singulr t the s i s; we simply llow it. In generl, to pply Definition 1 we ll hve to intersperse nonsingulr points between the singulr points.) There is potentil problem with this definition. In generl, if f is GRintegrble over [, b], there will be infinitely mny choices for the s i. For exmple, if f(x) = [x(1 x)] 1/2 on [0,1], we could choose s 1 to be ny number strictly between 0 nd 1. For the integrl over [0, 1] to be welldefined, we need to know tht the righthnd side of (6) does not depend on where we put the nonsingulr s i s. Exercise. 8. Prove tht for functions GRintegrble ccording to Definition 3, f(t)dt is welldefined (i.e. does not depend on the choice of the points s i ). 2 Integrls over unbounded intervls. Next we wnt to llow for the possibility of integrting functions over infinite intervls (e.g. [0, )). The most intuitive wy to do this is the following. Definition 4. Let R. We sy f is GRintegrble on [, ) (or f(t)dt 5
6 exists, or f(t)dt converges ) iff (i) for every y >, f is GRintegrble over [, y] (in the sense of Definition 3), nd (ii) lim y y f(t)dt exists. In the GRintegrble cse, we define f(t)dt to be the vlue of this limit. We define GRintegrble over (, ] nd f(t)dt similrly. We sy f is GRintegrble on (, ) if it is GRintegrble on (, 0] nd [0, ), in which cse we define f(t)dt = 0 f(t)dt + 0 f(t)dt. Exercises. 9. Prove tht, in the definition of integrbility over (, ), the number 0 could hve been replced by ny rel number without chnging the set of functions being clled GRintegrble or (in the GRintegrble cse) the vlue of the integrl. 10. Let > 0. Prove tht 11. Determine ll vlues of p for which 0 1 dx exists iff p > 1. (Here p cn be ny rel number.) x p 1 dx exists. x p 12. Let R. Assume f is GRintegrble on [, y] for ll y >. Assume there exists function g : (, ) R, GRintegrble on [, ), such tht f(x) g(x), x >. Prove tht f is GRintegrble on [, ) nd tht f(x)dx g(x)dx. 13. Let R. Assume f is continuous on (, ) nd tht, for some p > 1 nd some q < 1, the function x ((x ) p + (x ) q ) f(x) is bounded on (, ). Prove tht f(t)dt exists. 14. Let ɛ > 0 nd let r > 1. Prove tht, no mtter how smll ɛ is or how lrge r is, 0 x r e ɛx dx converges. 15. Suppose f is defined on [, ) nd f(x) 0 x. Prove tht if f(x)dx exists, then lim inf x f(x) = 0. (Note: previously we defined lim inf only for sequences, but you should be ble to figure out how to extend the definition to the current sitution.) Remrk. The sttement in exercise 15 would be flse if lim inf were replced by lim. First, lim x f(x) doesn t hve to exist for f(x)dx to exist. Second, it is even possible for the integrl to converge if there is sequence x n for which f(x n ). As n exmple, consider function f which is zero most plces, except for tringulr spikes centered t the positive integers. For the spike centered t n, let the bse of the spike hve width 2 2n nd height 2 2 n, so tht the tringle hs re 2 n. Then it s not hrd to show tht f is GRintegrble on [0, ) nd tht the integrl equls the (convergent) geometric series 1 2 n, even though f(n). If we drop the restriction tht f be nonnegtive, it is esy to come up with other exmples of counterintuitive phenomen; see exercises ChngeofVribles Formul. Often forml chnge of vribles mde to simplify the computtion of n integrl turns proper integrl into n improper one, or vicevers. Sometimes chnge of vribles turns one improper integrl into nother. One needs to know whether such chnges of vrible re vlid. For simplicity, we will stte the theorem only for functions 6
7 defined on n intervl of the form (, b] or [b, ) nd which stisfy the corresponding prts of Definitions 1 or 4. A more generl sttement isn t hrd to prove, but is messier to stte. ChngeofVribles Theorem. Suppose f is continuous on n intervl I, where either (i) I = (, b] or (ii) I = [b, ). Let φ : I R be continuous on I nd continuously differentible on the interior of I. In cse (i), suppose tht lim t φ(t) = c, where we llow the symbol c to stnd for rel number or for. (Thus we ssume tht either the limit exists, or tht lim t φ(t) =.) Similrly, in cse (ii), suppose tht lim φ(t) = c, t where gin c my stnd for. Then in ech cse ((i) nd (ii)) either both of the integrls f(φ(t))φ (t)dt, c f(x)dx (8) exist, or neither does. (If c = or c φ(b), see the lst remrk t the end of these notes.) When the integrls exist, they re equl. Proof. We will write out the proof only for the cse in which I = (, b] nd c < φ(b) is rel number; the other cses re similr. Assume the second integrl in (8) exists. Note tht for u >, the hypotheses of Corollry 3 on p. 128 of Rosenlicht (the chngeofvribles formul for proper integrls) re stisfied with U = (u, b). By hypothesis lim φ(b) y c y f(x)dx exists nd lim u φ(u) = c. Hence c f(x)dx = lim y c y f(x)dx = lim u φ(u) f(x)dx = lim f(φ(t))φ (t)dt. u u Thus the limit on the extreme right, which is the definition of f(φ(t))φ (t)dt, exists nd equls the integrl on the extreme left. Conversely, suppose tht the first integrl in (8) exists. Then, s bove, we hve f(φ(t))φ (t)dt = lim u u f(φ(t))φ (t)dt = lim f(x)dx. u φ(u) Let y n c. Since lim t φ(t) = c, the Intermedite Vlue Theorem implies tht φ mps the intervl (, + δ] onto (c, φ( + δ)] for ll δ < b. Thus there exists sequence u n such tht y n = φ(u n ) for ll n. Continuing the chin of equlities bove, we then hve lim f(x)dx = lim f(x)dx = lim f(x)dx. u φ(u) n φ(u n) n y n 7
8 Applying Lemm 1, lim y c y f(x)dx exists nd equls the first integrl in (8). Exercises. 16. Redo exercise 10 by using exercise 1 nd chnge of vribles. 17. Prove tht if f : [0, ) R is nonnegtive, decresing function, with lim x f(x) = 0, then 0 f(x) sin(x)dx converges. 18. Prove tht 0 sin(x)dx does not converge, but tht 0 sin(x 2 ) dx nd 0 x sin(x 2 )dx do converge. (To understnd wht is enbling the lst two integrls to converge, even though the integrnd is not pproching 0 nd, in the lst integrl, is even unbounded it is instructive to grph the integrnds.) 19. Prove tht 1 0 sin( 1 )dx exists. x Finl remrks. 1. From Definition 3 nd eqution (6) it is cler tht f GRintegrble on [, b] does not require the function f even to be defined t nd b. Even if f() nd f(b) re defined, the vlues f(), f(b) ffect neither GRintegrbility (or Riemnn integrbility for tht mtter) nor the vlue of the integrl. Therefore we define GRintegrble on (, b], GRintegrble on [, b), nd GRintegrble on (, b) ll to men the sme thing s GRintegrble on [, b]. The terminology GRintegrble on (, b) is the most flexible, since it llows for the cses = nd b =. 2. So fr we hve defined integrls f(x) dx only when < b. We extend our definition to llow for b in the sme wy s for Riemnn integrls: In the generlized cse if > b we sy tht f(x) dx exists iff f is GRintegrble on (b, ), in which cse we define f(x) dx = b f(x) dx. (We hve used openintervl nottion here in order include the cses in which or b is infinite.) For R we declre every function to be GRintegrble over the intervl [, ], with f(x) dx = 0. 8
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