Math 1B — UCB, Spring 2013 — M. ChristSummary of Lecture 8, Friday 2/8/2013ImproperIntegralsToday’s lecture is based on§7.8 of our text.Please study this section if you have notalready done so. We may return to numerical integration,§7.7, briefly next week, after youhave had a chance to work some problems and to absorb what has already been discussed.•Improper integralsare those which either involve integration over unbounded intervals,or involve integration of unbounded functions.•Examples:R∞1x-1dx,R∞111+x2dx.•Definition.For improper integrals of the formR∞af(x)dx:Z∞af(x)dx=limt→∞Ztaf(x)dx,ifthislimit exists,as a finite number. In this case, the improper integral isconvergent.(Equivalently, we say that the integral converges.) If the limit fails to exist, includingthe case where limt→∞Rtaf(x)dx=±∞, the improper integral isdivergent.There are corresponding definitions for improper integrals of the related formRb-∞f(x)dx;see text.•Example:R∞1x-1dxdiverges.Solution.First consider anyt≥1 and evaluateZt1x-1dx= ln(x)t1= ln(t)-ln(1) = ln(t).Now take the limit ast→ ∞:limt→∞Zt1x-1dx= limt→∞ln(t) =∞.The limit is +∞, so this improper integral diverges. (It is common to writeR∞1x-1dx=+∞.)•Example:R∞011+x2dxconverges.Solution.Fort≥0,Zt011 +x2dx= arctan(x)t0= arctan(t)-arctan(0) = arctan(t).1
A quick review of the graph ofy= arctan(x) reminds us that arctan(t)→π2ast→+∞.Thereforelimt→∞Zt011 +x2dx=π2.Therefore this improper integral converges, andZ∞011 +x2dx=π2.•Example:R∞2511+x2dx: Same analysis as preceding example giveslimt→∞Zt2511 +x2dx=π2-arctan(25).This improper integral converges, andR∞2511+x2dx=π2-arctan(25).•Changing the lower limit of integration from 0 to 25(Affects the numerical value of the convergent integral,But is irrelevant to deciding whether or not the integral converges.