1、Improper Integrals2OverviewBy the definition of definite integral,the definite integrals we have studiedmust satisfy at least the following two conditions:(1)The interval of integration a,b is finite;(2)The integrand f is bounded on a,b.In practice,however,we frequently encounter problems that fail
2、to meet oneor both of these conditions.Hence,it is necessary that we extended the conceptof definite integrals to investigate integrals with infinite interval of integrationor with an unbounded integrand.3Integration on an Infinite Interval(Improper Integrals with Infinite integration Limits 无穷积分无穷积
3、分)Suppose a function f is defined on the interval a,+).If for anyb a,function f is Riemann integrable on the interval a,b,thenlim()bbaf x dx is called the integral of the function f on the infinite interval a,+)or simply the infinite integral of the function f,denoted by()lim().bbaaf x dxf x dx Simi
4、larly,we have()lim(),bbaaf x dxf x dx andlimlim()()()()().abccbcacf x dxf x dxf x dxf x dxf x dx4Integration on an Infinite IntervalGeometric meaning of the infinite integralSuppose that f(x)0,x a,+),since()baf x dx express the area formedby the curve y=f(x),the x-axis and the vertical lines x=a and
5、 x=b,it is easyto see that the area becomes larger when b increases,so that if lim()babf x dxA exists then the area formed by the curve y=f(x),x-axis and x=a exists andis equal to A;if lim()babf x dx is infinity,then the corresponding area isregarded as infinity,in this case,we say that the area doe
6、s not exist.5Integration on an Infinite IntervalEvaluate the area formed by the curve 21,yx the line x=1,and the x-axis.AOxy1bSolution221111limbbdxdxxx 11limbbx1lim 11.bb6Integration on an Infinite IntervalAOxy1bSolution1111limbbdxdxxx lim(lnln1).bbIn this case,the limit is infinite,sothe infinite i
7、ntegral does not exist,and we say that the area does not exist.Evaluate the area formed by the curve 1,yx the line x=1,and the x-axis.7Integration on an Infinite Interval(Convergence of infinite integrals)Consider an infinite integral().af x dx If lim()bbaf x dxA exists then we say that the infinite
8、 integral()af x dx and that A is the value of the infinite integral,denoted by();af x dxA iflim()bbaf x dx does not exist,()af x dx is divergent.then we say thatis convergent8Integration on an Infinite Interval Determining ConvergenceFor what values of p does the integral 11pdxx converge?Whenthe int
9、egral does converge,what is its value?SolutionIf 1,p 1111bpbpdxxxp 11(1)1pbp 1111.1pp b Thus,11limbppbdxdxxx 111lim11pbp b 1,11.,1ppp 9Integration on an Infinite IntervalSolution(continued)because11limpbb 0,1,1.pp Therefore,the integral converges to the value 1(1)p if p 1 and it divergesif p 1.Deter
10、mining ConvergenceFor what values of p does the integral 11pdxx converge?Whenthe integral does converge,what is its value?10Integration on an Infinite IntervalSolution(continued)If p=1,11pdxdxxx 1limbbdxx 1lim(ln)bbx lim(lnln1).bb and the integral diverges.Determining ConvergenceFor what values of p
11、 does the integral 11pdxx converge?Whenthe integral does converge,what is its value?11Integration on an Infinite IntervalFind the integral2.1dxx 02220111dxdxdxxxxSolution0220limlim11baabdxdxxx00lim arctanlim arctanbaabxx().22 Finish.00lim arctanarctandaaxx 00lim arctanarctandbbxx()()().ccf x dxf x d
12、xf x dx12Integrands with Infinite DiscontinuitiesAnother type of improper integral arises when the integrand has a vertical asymptote an infinite discontinuity at a limit of integration or at some point between the limits of integration.Consider the infinite region in the first quadrantthat lies und
13、er the curve from x=0to x=1.1/yx First,we find the area of the portion from a to 1,112aadxxx 22 aThen we find the limit of this area as0.a 100limlim(22)2.aaadxax where0.a 13Integrands with Infinite Discontinuities(Improper Integrals with Infinite Discontinuities无界函数的积分无界函数的积分)Integrals of functions
14、that become infinite at a point whit in the interval of integration are improper integrals.1.If f(x)is continuous on(a,b,then()lim().bbaccaf x dxf x dx 2.If f(x)is continuous on a,b),then()lim().bcaacbf x dxf x dx 3.If f(x)is continuous on ,then,)(,a cc b()()().bcbaacf x dxf x dxf x dxIn part 1 and
15、2,if the limit is the value of the improper converges and the limit is the value of the improper integral.If the limit fails to exist,the improper integraldiverges.In part 3,the integral on the left-hand side of the equation converges if both integrals on the right-hand side have values;otherwise it
16、 diverges.14Integrands with Infinite Discontinuities A Divergent Improper Integral Investigate the convergence of 101.1dxx SolutionThe integrand f(x)=1/(1 x)is continuous on 0,1)but becomesinfinite as1.x We evaluate the integral as00111limlimln 11bbbbdxxx 1limln(1)0.bb The limit is infinite,so the i
17、ntegral diverges.15Integrands with Infinite Discontinuities Infinite Discontinuity at an Interior PointEvaluate32/30.(1)dxx Solution The integrand has a infinite discontinuous point at x=1,thus16Integrands with Infinite DiscontinuitiesSolution(continued)Next,we evaluate each improper integral on the
18、 right-hand side of this equation.12/30(1)dxx 2/301lim(1)bbdxx 1/301lim3(1)bbx 1/31lim 3(1)33.bb 3132/32/32/3001(1)(1)(1)dxdxdxxxx Infinite Discontinuity at an Interior PointEvaluate32/30.(1)dxx 17Integrands with Infinite DiscontinuitiesSolution(continued)We conclude that332/3033 2.(1)dxx and32/31(1
19、)dxx 32/31lim(1)ccdxx 31/31lim3(1)ccx 1/31/331lim 3(31)3(1)3 2.cc Finish.Infinite Discontinuity at an Interior PointEvaluate32/30.(1)dxx 18Integrands with Infinite Discontinuities Find the integral220(0).adxaax Solution Since a is a singular point of the integrand,the integral has infinitediscontinu
20、ous point.2222000limaadxdxaxax 00lim arcsinaxa .2 By the definition,we have19Integrands with Infinite Discontinuities Discuss the convergence of the integral(,0).()bpadxab pxa Solution Since a is a singular point of the integrand,it is an improper integrands with infinite discontinuities.ln()ln()ln.
21、bbaadxxabaxa Because does not exist,is divergent if p=1.0lim ln ()bpadxxa When ,1p 100()limlim()1bpbpaadxxaxap When p=1,for any 0 we have20Integrands with Infinite DiscontinuitiesSolution(continued)100()limlim()1bpbpaadxxaxap 110()lim11ppbapp 1(),1,1,1.pbappp Hence,converges if p 1;diverges if p 1.(
22、)bpadxxa Finish.Discuss the convergence of the integral(,0).()bpadxab pxa 21Improper Integrals Find the integral1.1dxx x Solution This integral is an infinite integral.the integrand,it is also an improper integrands with infinite discontinuities.2112111dxdxdxxxxxxxWe can rewrite it in the form 22211
23、100limlim2 arctan(1)11dxdxxxxxx Since 0lim 2 arctan12 arctan 2.421222 arctan(1),(1)1txdxtdtxCt txx Since x=1 is a singular point of22Improper Integrals22lim11bbdxdxxxxx Solution(continued)lim2arctan12bb.2 Then 11dxxx .Finish.Find the integral1.1dxx x 23Improper Integrals The improper integral10,(0,)
24、.xxedx Note:The improper integral 10,(0,)xxedx is convergent.The value of the improper integral is a function of the parameter,denoted by10(),(0,).xxedx Gamma function0(1)xx edx 000 xxxxedxx exedx 10().xxedx (1)()The function satisfies the following recurrence relation:24Improper Integrals(1)()Taking ,nN we have(1)()!(1).nnnnL LSince 00(1)1,xxedxe we obtain that(1)!,nnor0!.nxnx edx
