Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.


 Drusilla Lambert
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1 Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied: The intervl [, ] my e infinite with either = or = + or oth, eg, d The function my hve one or more infinite discontinuities verticl symptotes somewhere on [, ], eg, in the integrl d the integrnd hs infinite n discontinuity t Such integrls re clled improper integrls ecuse they do not stisfy the definition of the definite integrl Nonetheless, we will see tht under certin conditions we cn mke sense numericlly of such improper integrls We will discuss ech type of violtion seprtely Type : Improper Integrls on Infinite Intervls We strt with definition tht descries how we cn mke sense of definite integrls on infinite intervls The definition is divided into three prts depending on the type of intervl DEFINITION 7 Improper Integrls on Infinite Intervls There re three types of infinite intervls to consider Intervls of the form [, : If f is continuous on [,, then f f d provided the limit eists If limit eists, we sy the improper integrl converges Otherwise it diverges Intervls of the form, ]: If If f is continuous on, ], then provided the limit eists f d = lim f d
2 mth 3 improper integrls: type c Intervls of the form, : If f is continuous on,, then we define f d = c lim f d + lim f d, provided oth limits eist Here c cn e ny convenient rel numer often Two Simple Emples EXAMPLE 7 Let s strt with n esy emple Is d convergent? SOLUTION Using Definition 7, we evlute the pproprite limit: + = The improper integrl converges to Geometriclly, this mens tht the re under the curve f = on the infinitely long intervl [, is squre unit see Figure 7 EXAMPLE 7 Here s similr emple Is d convergent? SOLUTION The grph of this function see Figure 7 is remrkly similr to the previous emple Using Definition 7, we evlute the pproprite limit: ln [ln ] = + The nturl log function increses without ound s increses so the limit does not eist We sy tht the improper integrl diverges or does not eist Geometriclly, this mens tht the re under the curve f = on the infinitely long intervl [, is unounded s increses EXAMPLE 73 Here s more interesting emple Is d convergent? + SOLUTION First notice tht this function is continuous on the intervl [, since the denomintor is only t = nd Use prtil frctions for the integrtion Compring the numertors + = A + B + A + A + B = + f = Figure 7: The re under the curve f = on the infinitely long intervl [, is squre unit f = Figure 7: The re under the curve f = on the infinitely long intervl [, is unounded or infinite, despite the fct tht the grph seems similr to tht of s: = A + B constnts: = A B = Using Definition 7, we evlute the pproprite limit: + + ln + ln + + = ln lim + ln + ln + = ln lim ln l Ho ln ln = ln So the improper integrl converges Notice how we simplified the nturl log epression first using log rules to chnge difference of logs into log of quotient
3 3 EXAMPLE 74 Does d eist converge? 4 SOLUTION This time we pply Definition 7 prt nd use minisustitution to do the integrl Use u = 4 nd du = d 4 / 4 The integrl diverges 4 / [ [ 4 / ] ] = + Doule Troule Some improper integrls hve infinite upper nd lower limits According to Definition 7 we must rek the integrl into two pieces nd evlute ech seprtely Only if oth integrls converge does the entire integrl converge EXAMPLE 75 Does d converge? + 4 SOLUTION First we split the integrl into two pieces, sy t = Then using Definition 7 we hve + 4 d = + 4 d d Net we must evlute ech improper integrl You should recognize the integrnd s leding to n inverse tngent function rctn = [ = π 4 rctn rctn π ] Hint: With more complicted improper integrls, it often mkes sense to determine the ntiderivtive first Then do the evlution with the pproprite limits In this cse recll tht + u du = rctn u + c In this emple, = nd u =, so du = d Similrly, the other piece of the integrl is rctn rctn rctn = [ π ] Putting this ll together, + 4 d = The integrl converges = π d d = π 4 + π 4 = π Alredy these emples illustrte the importnce of eing le to evlute limits t infinity with confidence EXAMPLE 76 Here s similr prolem Does d converge? +
4 mth 3 improper integrls: type 4 SOLUTION We split the integrl into two pieces t = + d = + d + + d Net we must evlute ech improper integrl You should recognize this s sustitution integrl ln + [ln ln + ] = Since this piece of the totl integrl diverges, y Definition 7 the entire improper integrl diverges There is no need to try to evlute the second piece EXAMPLE 77 Does e d converge? SOLUTION This is n integrtion y prts prolem Let u = so du = d nd then dv = e d so v = e Then e d = e e d = e e = e 3 u = + nd du = d So + d = u du = ln u + c = ln + + c Rememer, lim ln = + Returning to the improper integrl e e d = lim e = lim [ e ] Notice tht lim e hs the form so we put it into form nd use l Hôpitl s rule lim e = l Ho lim e e = Putting ll our work together, e [ e ] = = The integrl converges YOU TRY IT 7 Determine which of these integrls converge If the integrl converges, determine its vlue e ln 5 d 4 d c 4 d d + 3 d Answers to you try it 7 NOT in 3 order: 6, diverges, ln 3, nd 4 YOU TRY IT 7 Prove tht if n >, then d converges to n n YOU TRY IT 73 Prove tht if n <, then d diverges Note: When n is negtive, n rememer tht n d = n d where n is now positive numer Putting the lst two eercises together with Emple 7 you hve proven the following result: THEOREM 7 The ppower Test For ny rel numer p p d = { p, if p > diverges, if p YOU TRY IT 74 Evlute ech of these integrls or determine tht it diverges y using Theorem 7 You should not need to do ny ntidifferentition d d /3 c 7 d d d 5 YOU TRY IT 75 Griel s Horn The infinitelylong solid formed y revolving the curve f = out the is over the intervl [, is clled Griel s Horn Do you know why? Show tht the volume is finite See Figure 73 f = Figure 73: The re under the curve f = is rotted round the is
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