e t dt e t dt = lim e t dt T (1 e T ) = 1


 Ambrose Glenn
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1 Improper Inegrls There re wo ypes of improper inegrls  hose wih infinie limis of inegrion, nd hose wih inegrnds h pproch some poin wihin he limis of inegrion. Firs we will consider inegrls wih infinie limis of inegrion. Infinie Limis of Inegrion Suppose chemicl producion is governed by he differenil equion dp d = e moles per second. If we wn o find ou how much chemicl would be produced were he experimen llowed o run forever, we would like o clcule e d However, since is no number, we cnno jus plug i in s one of he bounds fer evluing he indefinie inegrl. Wh we cn do, is look n indefinie inegrl wih n upper limi T rher hn. This is somehing we cn evlue. Aferwrds, we cn evlue he resul in he limi lim T. Thus, he firs sep in problem of infinie limis of inegrion, is o rewrie he problem in he form of limi. Formlly, we wrie f()d = lim f()d T Exmple Evlue e d Soluion Firs we rewrie he problem e d = lim e d T We evlue he inegrl Evluing he limi we find e d = e T = e T ( e ) = e T e d = lim T ( e T ) = Thus, his re of producion, if producion coninued indefiniely, only mol of chemicl would be produced!
2 Exmple Suppose chemicl producion is governed by moles per second. beginning from =? Soluion Once gin, we wrie dq d = + How much chemicl is genered if producion coninues indefiniely, T d = lim + T + d Using u subsiuion, wih u = +, so du = d, we find T + + d = + du = ln( u ) T = ln(t + ) ln() = ln(t + ) u Thus, d = lim ln(t + ) = + T In his siuion, if producion is llowed o coninue indefiniely, he moun produced grows wihou bound. When he resul of n inegrl is ±, we sy h he inegrl diverges, becuse i does no rech ny rel vlue. When he limi s T is rel vlue, we sy h he inegrl converges. When evluing improper inegrls, i is imporn o se wheher or no here is convergence or divergence, nd if here is convergence, o wh vlue. Wh is i h differs beween he inegrnds of hese wo inegrls h cuses one o converge nd he oher o diverge? In boh of hese cses he inegrnds re lwys posiive over he limis of inegrion. Furhermore, hey boh pproch s T pproches. The difference is h e decys much more quickly hn. Bsed on his observion, we + should be ble o generlize wheher some simple funcions will converge or diverge, bsed on he re which heir inegrnds pproch (for posiive funcions). We should emphsize h if he inregrnd does no decy o zero, hen i is gurneed he inegrl will diverge. For insnce d = lim T T = lim T T = For n rbirry decying exponenil funcion, e α, we hve We find h e α d = e α α T = e αt α + α e α d = lim T α e αt α The conclusion is h ny decying exponenil decys fs enough so h is inegrl o converges. = α
3 We cn lso consider decying funcion of he ype, where p >. We will begin p inegring =, o void division by. If p =, hen we re considering he inegrl d which we lredy observed diverges. For p, we use he power rule so p d = p p T = T p p p When we evlue in he limi h T, we mus consider wheher p > or p <. Firs recll h lim T T p = for p nd For p < lim T p = for p < T ( T p lim T p ) = p becuse p > so h he firs erm grows wihou bound. In he cse h p > ( T p lim T p ) = p p becuse p < so h he firs erm goes o zero. In conclusion { p = p p > p I is noeworhy h for n improper inegrl, moving he lower limi of inegrion finie moun will no ler he inegrl s convergence or divergence, s long s i does no inroduce divison by zero ino he limis of inegrion. This mens h we cn lredy gher lo of informion bou he convergence nd divergence of oher improper inegrls. For exmple, 5 d = 5 d using he summion propery for inegrls. We know h d d diverges, nd h 5 d
4 is jus some finie moun. If we subrc some finie moun from diverging inegrl, he resul will sill be somehing h diverges. Thus, wihou ny compuion we cn deduce h d 5 diverges. In sense, his mens h he funcion s behvior for smll inpu vlues hs no influence over he convergence of such n inegrl  convergence is reled solely o he re which he funcion decys o zero s inpus grow lrger. Now suppose h we re fced wih more compliced inegrl, somehing like ( + e ) d We cn wrie ( ) + e d = d + e d Since e d is jus posiive number, we cn deduce h ( + e ) d > d = Thus, we cn conclude h our inegrl diverges, since i is lrger hn n inegrl diverging o. A very similr ide o his one leds us o he comprison es. The Comprison Tes Suppose f(x) g(x) for ll x... f(x)dx converges if g(x)dx diverges if g(x)dx converges. f(x)dx diverges. In summry, if some posiive funcion f(x) is lwys less hn or equl o noher posiive funcion g(x), hen is inegrl will be less, so if g(x) converges, hen f(x) mus converge s well. Similrly, if f(x) diverges, nd g(x) is greer hn or equl o i, hen i mus lso diverge, s he inegrl will be greer. Exmple 3 Deermine wheher or no d converges or diverges. + e Soluion Firs we cn noe h becuse he inegrnd is lwys posiive, he inegrl mus be greer hn. Also, for ll > we hve + e < e
5 so by he comprison es, we hve h < + e < e = Alhough we don know o wh exc vlue, we cn conclude h his inegrl converges. Exmple 4 Deermine wheher or no x + dx converges or diverges. If i converges, find o wh vlue. Soluion This funcion looks like, which is divergen, so we suspec h his inegrl x should lso diverge. However, i is no cler how o use he comprison es in his cse, so le us rewrie T dx = lim x + T x + dx Now, using subsiuion u = x +, so du = dx, nd x + dx = + u du Now h his inegrl is wrien in more fmilir form, we cn see h i diverges, using he fc h u du = Exmple 5 Deermine wheher or no dx (x ) / converges or diverges. If i converges, find o wh vlue. Soluion Using he comprison es, (x ) > / x / where he ler of hese erms is he inegrnd for divergen inegrl under hese limis. Thus, he inegrl in quesion diverges. Exmple 6 Deermine wheher or no (3x 5) dx converges or diverges. If i converges, find o wh vlue.
6 Soluion This inegrnd hs he form of, so we suspec h i should converge. Using x subsiuion u = 3x 5 so du = dx. Evluing he improper inegrl 3 (3x 5) dx = u du = 3 3 u + c = 3 3x 5 + c Now we see h (3x 5) dx = 3 Thus, we finlly conclude ( dx = lim (3x 5) T Infinie Inegrnds 3x 5 T = 3 3 3T T 5 + ) = 3 3 We cn lso consider funcions he pproch infiniy some poin, nd look heir definie inegrl on n inervl conining h poin. For now we will resric ourselves o funcions h pproch s x, bu here is no reson we cnno generlize nd consider ny poin where he funcions pproch. Once gin we ll rewrie he inegrl using limi, csing i in form we cn evlue. We will look he limi s he lower limi of inegrion pproches zero from he righ, so h we cn evlue n inegrl where he inegrnd is lwys finie, bu pproches he originl inegrl. Wrie Exmple Evlue Soluion We find h Exmple Evlue Soluion We find h x / dx x / dx = lim + d = lim + x dx f(x)dx = lim + f(x)dx x / dx = lim + x = lim +( ) = d = lim ln( ) + = lim +(ln() ln() = Jus s wih infinie limis of inegrion, we cn generlize he resul for p d We ve seen h if p = hen his inegrl diverges. If p hen d = lim p + p ( p d = lim p + p = lim + p p p
7 If p <, he inegrl converges, nd if p >, he inegrl diverges. We cn summrize hese resuls s { p = p p p < p Jus s before, we cn use he comprison es o evlue he convergence or divergence of inegrls wih inegrnds h pproch. Consider he following exmple Exmple 3 Deermine wheher or no d converges or diverges + 3 Soluion We will use he comprison es. For we hve 3 > so h + 3 < + = Thus + 3 d < d = =
4.8 Improper Integrals
4.8 Improper Inegrls Well you ve mde i hrough ll he inegrion echniques. Congrs! Unforunely for us, we sill need o cover one more inegrl. They re clled Improper Inegrls. A his poin, we ve only del wih inegrls
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