Improper Integrals. Dr. Philippe B. laval Kennesaw State University. September 19, f (x) dx over a finite interval [a, b].

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1 Improper Inegrls Dr. Philippe B. lvl Kennesw Se Universiy Sepember 9, 25 Absrc Noes on improper inegrls. Improper Inegrls. Inroducion In Clculus II, sudens defined he inegrl f (x) over finie inervl [, b]. The funcion f ws ssumed o be coninuous, or les bounded, oherwise he inegrl ws no gurneed o exis. Assuming n niderivive of f could be found, f (x) lwys exised, nd ws number. In his secion, we invesige wh hppens when hese condiions re no me. Definiion (Improper Inegrl) An inegrl is n improper inegrl if eiher he inervl of inegrion is no finie (improper inegrl of ype ) or if he funcion o inegre is no coninuous (no bounded) in he inervl of inegrion (improper inegrl of ype 2). Exmple 2 e x is n improper inegrl of ype since he upper limi of inegrion is infinie. is no coninu- Exmple 3 ous. Exmple 4 x is n improper inegrl of ype 2 becuse x x is n improper inegrl of ypes since he upper limi of inegrion is infinie. I is lso n improper inegrl of ype 2 becuse is no coninuous nd is in he inervl of inegrion. x 2 Exmple 5 2 x 2 is n improper inegrl of ype 2 becuse x 2 coninuous nd. is no

2 Exmple 6 π coninuous π 2. n x is n improper inegrl of ype 2 becuse n x is no We now look how o hndle ech ype of improper inegrl..2 Improper Inegrls of Type These re esy o idenify. Simply look he inervl of inegrion. If eiher he lower limi of inegrion, he upper limi of inegrion or boh re no finie, i will be n improper inegrl of ype. Definiion 7 (improper inegrl of ype ) Improper inegrls of ype re evlued s follows:. If 2. If f (x) exiss for ll, henwedefine f (x) = lim f (x) provided he limi exiss s finie number. In his cse, sid o be convergen (or o converge). Oherwise, o be divergen (or o diverge). f (x) exiss for ll b, henwedefine f (x) = lim f (x) provided he limi exiss s finie number. In his cse, is sid o be convergen (or o converge). Oherwise, sid o be divergen (or o diverge). 3. If boh f (x) nd f (x) = f (x) is f (x) is sid f (x) converge, hen we define f (x) + f (x) The inegrls on he righ re evlued s shown in. nd 2.. f (x) f (x) is 2

3 .3 Improper Inegrls of Type 2 These re more difficul o idenify. One needs o look he inervl of inegrion, nd deermine if he inegrnd is coninuous or no in h inervl. Things o look for re frcions for which he denominor becomes in he inervl of inegrion. Keep in mind h some funcions do no conin frcions explicily like n x, sec x. Definiion 8 (improper inegrl of ype 2) Improper inegrls of ype 2 re evlued s follows:. if f is coninuous on [, b) nd no coninuous b hen we define f (x) = lim f (x) b provided he limi exiss s finie number. In his cse, sid o be convergen (or o converge). Oherwise, o be divergen (or o diverge). f (x) is f (x) is sid 2. if f is coninuous on (, b] nd no coninuous hen we define f (x) = lim + f (x) provided he limi exiss s finie number. In his cse, sid o be convergen (or o converge). Oherwise, o be divergen (or o diverge). 3. If f is no coninuous c where <c<bnd boh c f (x) converge hen we define f (x) is f (x) is sid c f (x) nd f (x) = c f (x) + c f (x) The inegrls on he righ re evlued s shown in. nd 2.. We now look some exmples. 3

4 .4 Exmples Evluing n improper inegrl is relly wo problems. I is n inegrl problem nd limi problem. I is bes o do hem seprely. When breking down n improper inegrl o evlue i, mke sure h ech inegrl is improper only one plce, h plce should be eiher he lower limi of inegrion, or he upper limi of inegrion. Exmple 9 x 2 This is n improper inegrl of ype. We evlue i by finding lim ( Firs, x 2 = ) nd lim ( ) =hence x 2 =. Exmple x This is n improper inegrl of ype. We evlue i by finding lim Firs, x =ln nd lim ln = hence Exmple +x 2 This is n improper inegrl of ype. infinie, we brek i ino wo inegrls. Noe h since he funcion we only need o do +x 2 = +x 2. x diverges. x 2. x Since boh limis of inegrion re +x 2 + +x 2 +x 2 is even, we hve +x 2 = +x 2 ; nd +x 2 = lim +x 2 +x 2 =n x =n n =n 4

5 Thus +x 2 = lim ( n ) = π 2 I follows h +x 2 = π 2 + π 2 = π Exmple 2 π 2 sec x This is n improper inegrl of ype 2 becuse sec x is no coninuous π 2.We evlue i by finding Firs, lim π 2 sec x. sec x =ln sec x +nx =ln sec +n π 2 nd lim (ln sec +n ) = hence sec x diverges. π 2 Exmple 3 π sec2 x This is n improper inegrl of ype 2, sec 2 x is no coninuous π 2.Thus, π sec 2 x = Firs, we evlue π 2 sec2 x. π 2 π 2 sec 2 x = sec 2 x + lim π 2 π π 2 sec 2 x sec 2 x Therefore, sec 2 x =n n =n π 2 sec 2 x = lim (n ) π 2 = I follows h π 2 sec2 x diverges, herefore, π sec2 x lso diverges. 5

6 Remrk 4 If we hd filed o see h he bove inegrl is improper, nd hd evlued i using he Fundmenl Theorem of Clculus, we would hve obined compleely differen (nd wrong) nswer. π sec 2 x =nπ n =(his is no correc) Exmple 5 x 2 This inegrl is improper for severl resons. Firs, he inervl of inegrion is no finie. The inegrnd is lso no coninuous. To evlue i, we brek i so h ech inegrl is improper only one plce, h plce being one of he limis of inegrion, s follows: x 2 = x 2 + x 2 + x 2 +. We hen evlue ech improper inegrl. The reder will verify h x2 i diverges. Severl imporn resuls bou improper inegrls re worh remembering, hey will be used wih infinie series. The proof of hese resuls is lef s n exercise. Theorem 6 x p converges if p>, i diverges oherwise. Proof. This is n improper inegrl of ype since he upper limi of inegrion is infinie. Thus, we need o evlue lim x p. When p =, wehve lredy seen h he inegrl diverges. Le us ssume h p. Firs, we evlue he inegrl. x p = = x p x p p = p p p The sign of p is imporn. ( When p > h is when p<, p is in he p numeror. Therefore, lim p ) = hus he inegrl diverges. p When( p< h is when p>, p is relly in he denominor so h p lim p ) = nd herefore p p x p converges. In conclusion, we hve looked he following cses: 6

7 Cse : p =. In his cse, he inegrl diverges. Cse 2: p<. In his cse, he inegrl diverges. Cse 3: p>. In his cse, he inegrl converges. Theorem 7 x p converges if p<, i diverges oherwise. Proof. See problems..5 Comprison Theorem for Improper Inegrls Someimes n improper inegrl is oo difficul o evlue. One echnique is o compre i wih known inegrl. The heorem below shows us how o do his. Theorem 8 Suppose h f nd g re wo coninuous funcions for x such h g (x) f (x). Then, he following is rue:. If 2. If f (x) converges hen g (x) lso converges. g (x) diverges, hen f (x) lso diverges. The heorem is no oo difficul o undersnd if we hink bou he inegrl in erms of res. Since boh funcions re posiive, he inegrls simply represen he re of he region below heir grph. Le A f be he re of he region below he grph of f. Use similr noion for A g.if g (x) f (x), hen A g A f. Pr of he heorem is simply sying h if A f is finie, so is A g ; his should be obvious from he inequliy. Pr 2 sys h if A g is infinie, so is A f. Exmple 9 Sudy he convergence of e x2 We cnno evlue he inegrl direcly, e x2 does no hve n niderivive. We noe h x x 2 x x 2 x e x2 e x 7

8 Now, e x = lim e x ( = lim e e ) = e nd herefore converges. I follows h heorem. e x2 converges by he comprison Remrk 2 When using he comprison heorem, he following inequliies re useful: x 2 x x nd ln x x e x.6 Things o know Bebleoellifninegrlisimproperornondwhypeiis. Be ble o ell if n improper inegrl converges or diverges. If i converges, be ble o find wh i converges o. Be ble o wrie n improper inegrl s limi of definie inegrl(s). Reled problems ssigned: Pge 436: #, 3, 5, 7,, 3, 5, 7, 9, 2, 23, 25, 29, 39, 4, 57. Discuss he convergence of e x Prove heorem 7. 8