Lecture 18 : Improper integrals

We de¯ned

Rb af(t)dtunder the conditions thatfis de¯ned and bounded on the bounded interval [a;b]. In this lecture, we will extend the theory of integration to bounded functions de¯ned on unbounded intervals and also to unbounded functions de¯ned on bounded or unbounded intervals. Improper integral of the ¯rst kind:Supposefis (Riemann) integrable on [a;x] for allx > a, i.e.,Rx af(t)dtexists for allx > a. If limx!1R x af(t)dt=Lfor someL2R;then we say that the improper integral (of the ¯rst kind)R1 af(t)dtconverges toLand we writeR1 af(t)dt=L.

Otherwise, we say that the improper integralR1

af(t)dtdiverges. Observe that the de¯nition of convergence of improper integrals is similar to the one given for series. For example,Rx af(t)dt; x > ais analogous to the partial sum of a series.

Examples :1. The improper integralR1

11 t

2dtconverges, because,Rx

11 t

2dt= 1¡1

x !1 asx! 1:

On the other hand,R1

11 t dtdiverges because limx!1Rx 11 t dt= limx!1logx. In fact, one can show thatR1 11 t pdtconverges to1 p¡1forp >1 and diverges forp·1.

2. Consider

0te¡t2dt:We will use substitution in this example. Note thatRx

0te¡t2dt=1

2 R x2

0e¡sds=1

2 (1¡e¡x2)!1 2 as x! 1:

3. The integral

0sintdtdiverges, because,Rx

0sintdt= 1¡cosx:

We now derive some convergence tests for improper integrals. These tests are similar to those used for series. We do not present the proofs of the following three results as they are similar to the proofs of the corresponding results for series. Theorem 17.1 :Supposefis integrable on[a;x]for allx > awheref(t)¸0for allt > a. If there existsM >0such thatRx af(t)dt·Mfor allx¸athenR1 af(t)dtconverges. This result is similar to the result:Ifan¸0for all n and the partial sumSn·Mfor all n, thenPanconverges.The proofs are also similar. One uses the above theorem to prove the following theorem which is analogous to the comparison test of series. In the following two results we assume thatfandgare integrable on [a;x] for allx > a. Theorem 17.2 : (Comparison test)Suppose0·f(t)·g(t)for allt > a:IfR1 ag(t)dt converges, thenR1 af(t)dtconverges.

Examples :1. The improper integralR1

1cos2t

2dtconverges, because 0·cos2t

2·1

t 2.

2. The improper integral

12+sint

t dtdiverges, because2+sint t ¸1 t >0 for allt >1. Theorem 17.3 : (Limit Comparison Test(LCT))Supposef(t)¸0andg(t)>0for allx > a.

Iflimt!1f(t)

g(t)=cwherec6= 0;then both the integralsR1 af(t)dtandR1 ag(t)dtconverge or both diverge. In casec= 0, then convergence ofR1 ag(t)dtimplies convergence ofR1 af(t)dt.

Examples :1. The integralR1

1sin1 t dtdiverges by LCT, becausesin1 t 1 t !1 ast! 1.

2. Forp2R,R1

1e¡ttpdtconverges by LCT becausee¡ttp

¡2!0 asx! 1.

So far we considered the convergence of improper integrals of only non-negative functions. We will now consider any real valued functions. The following result is anticipated. 2

Theorem 17.4 :If an improper integralR1

ajf(t)jdtconverges thenR1 af(t)dtconverges i.e., every absolutely convergent improper integral is convergent.

Proof :SupposeR1

ajf(t)jdtconverges andRx af(t)dtexists for allx > a. Since

0·f(x)+jf(x)j ·2jf(x)j, by comparison testR1

a(f(x)+jf(x)j)dxconverges. This implies thatR1 The converse of the above theorem is not true (see Problem 2). The following result, known asDirichlet test, is very useful.

Theorem 17.5 :Letf;g: [a;1)!Rbe such that

(i)fis decreasing andf(t)!0ast! 1, (ii)gis continuous and there existsMsuch thatRx ag(t)dt·Mfor allx > a.

ThenR1

af(t)g(t)dtconverges. We will not present the proof of the above theorem but we will use it.

Examples :IntegralsR1

¼sint

t dtandR1

¼cost

t dtare convergent.

Improper integrals of the form

¡1f(t)dtare de¯ned similarly. We say thatR1

¡1f(t)dt

is convergent if bothRc

¡1f(t)dtandR1

cf(t)dtare convergent for some elementcinRandR1

¡1f(t)dt=Rc

¡1f(t)dt+R1

cf(t)dt:

Improper integral of second kind :SupposeRb

xf(t)dtexists for allxsuch thata < x·b(the functionfcould be unbounded on (a;b]). If limx!a+R b xf(t)dt=Mfor someM2R;then we say that the improper integral (of the second kind) Rb af(t)dtconverges toMand we writeRb af(t)dt=M.

Example :The improper integralR1

01 t pdtconverges forp <1 and diverges forp¸1. Comparison test and limit comparison test for improper integral of the second kind are analogous

to those of the ¯rst kind. If an improper integral is a combination of both ¯rst and second kind

then one de¯nes the convergence similar to that of the improper integral of the kindR1

¡1f(t)dt,

Problem 1:Determine the values ofpfor which1R

0f(x)dxconverges wheref(x) =1¡e¡x

x p.

Solution :LetI1=1R

0f(x)dxandI2=1R

1f(x)dx:We have to determine the values ofpfor which the

integralsI1andI2converge. Now one has to see how the functionf(x) behaves in the respective intervals and apply the LCT. Since limx!01¡e¡x x = 1, by LCT with1 x p¡1, we see thatI1is convergent i®p¡1<1,i.e.,p <2. Similarly,I2is convergent (by applying LCT with1 x p) i®p >1. Therefore 1R

0f(x)dxconverges i® 1< p <2.

Problem 2 :Prove thatR1

1sinx x pdxconverges but not absolutely for 0< p·1. Solution :Let 0< p·1. By Dirichlet's Test, the integral converges. We claim that1R

1jsinxj

x pdx does not converge. Since,jsinxj ¸sin2x, we see thatjsinx x pj ¸sin2x x p=1¡cos2x

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