Transcription of Lecture 18 : Improper integrals - IIT Kanpur
1 1 Lecture 18 : Improper integralsWe defined baf(t)dtunder the conditions thatfis defined and bounded on the bounded interval[a, b]. In this Lecture , we will extend the theory of integration to bounded functions defined onunbounded intervals and also to unbounded functions defined on bounded or unbounded integral of the first kind:Supposefis (Riemann) integrable on [a, x] for allx > a, , xaf(t)dtexists for allx > a. If limx xaf(t)dt=Lfor someL R,then we say thatthe Improper integral (of the first kind) af(t)dtconverges toLand we write af(t)dt= , we say that the Improper integral af(t) that the definition of convergence of Improper integrals is similar to the one given forseries.
2 For example, xaf(t)dt, x > ais analogous to the partial sum of a :1. The Improper integral 11t2dtconverges, because, x11t2dt= 1 1x 1 asx .On the other hand, 11tdtdiverges because limx x11tdt= limx logx. In fact, one can showthat 11tpdtconverges to1p 1forp >1 and diverges forp Consider 0te will use substitution in this example. Note that x0te t2dt=12 x20e sds=12(1 e x2) 12as x .3. The integral 0sintdtdiverges, because, x0sintdt= 1 now derive some convergence tests for Improper integrals . These tests are similar to thoseused for series.
3 We do not present the proofs of the following three results as they are similar tothe proofs of the corresponding results for :Supposefis integrable on[a, x]for allx > awheref(t) 0for allt > a. Ifthere existsM >0such that xaf(t)dt Mfor allx athen af(t) result is similar to the result:Ifan 0for all n and the partial sumSn Mfor alln, then proofs are also similar. One uses the above theorem to prove thefollowing theorem which is analogous to the comparison test of the following two results we assume thatfandgare integrable on [a, x] for allx > : (Comparison test)Suppose0 f(t) g(t)for allt > ag(t)dtconverges, then af(t) :1.
4 The Improper integral 1cos2tt2dtconverges, because 0 cos2tt2 The Improper integral 12+sinttdtdiverges, because2+sintt 1t>0 for allt > : (Limit Comparison Test(LCT))Supposef(t) 0andg(t)>0for allx > f(t)g(t)=cwherec6= 0,then both the integrals af(t)dtand ag(t)dtconverge or bothdiverge. In casec= 0, then convergence of ag(t)dtimplies convergence of af(t) :1. The integral 1sin1tdtdiverges by LCT, becausesin1t1t 1 ast .2. Forp R, 1e ttpdtconverges by LCT becausee ttpt 2 0 asx .So far we considered the convergence of Improper integrals of only non-negative functions.
5 We willnow consider any real valued functions. The following result is :If an Improper integral a|f(t)|dtconverges then af(t)dtconverges ,every absolutely convergent Improper integral is :Suppose a|f(t)|dtconverges and xaf(t)dtexists for allx > a. Since0 f(x)+|f(x)| 2|f(x)|, by comparison test a(f(x)+|f(x)|)dxconverges. This impliesthat a(f(x)+|f(x)| |f(x)|)dxconverges. The converse of the above theorem is not true (see Problem 2).The following result, known asDirichlet test, is very :Letf, g: [a, ) Rbe such that(i)fis decreasing andf(t) 0ast ,(ii)gis continuous and there existsMsuch that xag(t)dt Mfor allx > af(t)g(t) will not present the proof of the above theorem but we will use : integrals sinttdtand costtdtare integrals of the form b f(t)dtare defined similarly.]
6 We say that f(t)dtis convergent if both c f(t)dtand cf(t)dtare convergent for some elementcinRand f(t)dt= c f(t)dt+ cf(t) integral of second kind :Suppose bxf(t)dtexists for allxsuch thata < x b(thefunctionfcould be unbounded on (a, b]). If limx a+ bxf(t)dt=Mfor someM R,then we say thatthe Improper integral (of the second kind) baf(t)dtconverges toMand we write baf(t)dt= :The Improper integral 101tpdtconverges forp <1 and diverges forp test and limit comparison test for Improper integral of the second kind are analogousto those of the first kind.)
7 If an Improper integral is a combination of both first and second kindthen one defines the convergence similar to that of the Improper integral of the kind f(t)dt,Problem 1:Determine the values ofpfor which 0f(x)dxconverges wheref(x) =1 e :LetI1=1 0f(x)dxandI2= 1f(x) have to determine the values ofpfor which theintegralsI1andI2converge. Now one has to see how the functionf(x) behaves in the respectiveintervals and apply the LCT. Since limx 01 e xx= 1, by LCT with1xp 1, we see thatI1is convergentiffp 1<1, ,p <2. Similarly,I2is convergent (by applying LCT with1xp) iffp >1.
8 Therefore 0f(x)dxconverges iff 1< p < 2 :Prove that 1sinxxpdxconverges but not absolutely for 0< p :Let 0< p 1. By Dirichlet s Test, the integral converges. We claim that 1|sinx|xpdxdoes not converge. Since,|sinx| sin2x, we see that|sinxxp| sin2xxp=1 cos Dirichlet s Test, 1cos 2x2xpdxconverges p >0. But 112xpdiverges forp 1. Hence, 1|sinxxp|dxdoes not converge.
