Impope Inegls To his poin we hve only consideed inegls f() wih he is of inegion nd b finie nd he inegnd f() bounded (nd in fc coninuous ecep possibly fo finiely mny jump disconinuiies) An inegl hving eihe n infinie i of inegion o n unbounded inegnd is clled impope. Hee e wo emples + 2 The fis hs n infinie domin of inegion nd he inegnd of he second ends o s ppoches he lef end of he domin of inegion. We ll s wih n emple h illuses he ps h you cn fll ino if you e such inegls sloppily. Then we ll see how o e hem cefully. Emple The compuion = 2 = = 2 is wong. In fc, he nswe is idiculous. The inegnd >, so he inegl hs o be 2 posiive. The flw in he gumen is h he fundmenl heoem of clculus, which sys h if F () = f() hen f() = F(b) F(), is pplicble only when F () eiss nd equls f() fo ll b. In his cse F () = does no eis fo =. The given 2 inegl is impope. We ll see le h he coec nswe is +. Emple The ceful wy o e n inegl like h hs bounded inegnd bu hs + 2 domin of inegion h eends o + is o ppoime he inegl by one wih bounded domin of inegion, like R, nd hen ke he i R. Hee is he + 2 y = f() R foml definiion nd some viions. We ll wok hough some emples in deil sholy. c Joel Feldmn. 25. All ighs eseved. Febuy 5, 25
Definiion 2. () If he inegl R f() eiss fo ll R >, hen R f() f() when he i eiss (nd is finie). (b) If he inegl f() eiss fo ll < b, hen when he i eiss (nd is finie). (c) If he inegl R f() f() f() eiss fo ll < R, hen f() c R f() + f() c when boh is eis (nd e finie). Any c cn be used. When he i(s) eis, he inegl is sid o be convegen. Ohewise i is sid o be divegen. The ceful wy o e n inegl like h hs finie domin of inegion bu whose inegnd is unbounded ne one i of inegion is o ppoime he inegl by one wih domin of inegion on which he inegnd is bounded, like, wih >, nd hen ke he i +. Hee is he foml definiion nd some viions. y y = c Joel Feldmn. 25. All ighs eseved. 2 Febuy 5, 25
Definiion 3. () If he inegl f() eiss fo ll < < b, hen when he i eiss (nd is finie). (b) If he inegl T f() f() + f() eiss fo ll < T < b, hen when he i eiss (nd is finie). T f() f() T b (c) Le < c < b. If he inegls T f() nd f() eis fo ll < T < c nd c < < b, hen T f() f() + f() T c c+ when boh i eis (nd e finie). When he i(s) eis, he inegl is sid o be convegen. Ohewise i is sid o be divegen. Ifnineglhsmoehnone souceofimpopiey, foemple ninfiniedominof inegion nd n inegnd wih n unbounded inegnd o muliple infinie disconinuiies, hen you spli i up ino sum of inegls wih single souce of impopiey in ech. Fo he inegl, s whole, o convege evey em in h sum hs o convege. Fo emple, he inegl hs domin of inegion h eends o boh + nd nd, ( 2) 2 in ddiion, hs n inegnd which is singul (i.e. becomes infinie) = 2 nd =. So we would wie ( 2) 2 = ( 2) 2 + ( 2) 2 + ( 2) 2 + 2 ( 2) 2 + 3 2 ( 2) 2 + 3 ( 2) 2 (Hee he bek poin could be eplced by ny numbe sicly less hn ; could be eplced by ny numbe sicly beween nd 2; nd 3 could be eplced by ny numbe sicly bigge hn 2.) On he igh hnd side he fis inegl hs domin of inegion eending o c Joel Feldmn. 25. All ighs eseved. 3 Febuy 5, 25
he second inegl hs n inegnd h becomes unbounded s, he hid inegl hs n inegnd h becomes unbounded s +, he fouh inegl hs n inegnd h becomes unbounded s 2, he fifh inegl hs n inegnd h becomes unbounded s 2+, nd he ls inegl hs domin of inegion eending o +. We e now edy fo some (impon) emples. Emple 4 ( p ) Fi ny p >. The domin of he inegl eends o + nd he inegnd is p p coninuous nd bounded on he whole domin. So he definiion of his inegl is R p p When p, one nideivive of p is p+, nd when p = nd >, one nideivive p+ of p is ln. So p p ] R p if p lnr if p = As R, lnr ends o nd R p ends o when p > (i.e. he eponen is posiive) nd ends o when p < (i.e. he eponen is negive). So { divegen if p = p if p > p Emple 4 Emple 5 ( p ) Agin fi ny p >. The domin of inegion of he inegl is finie, bu he inegnd becomes unbounded s ppoches he lef end,, of he domin of inegion. p p So he definiion of his inegl is p + p When p, one nideivive of p is p+, nd when p = nd >, one nideivive p+ of p is ln. So ] p if p p p + ln if p = c Joel Feldmn. 25. All ighs eseved. 4 Febuy 5, 25
As +, ln ends o nd p ends o when p > (i.e. he eponen is posiive) nd ends o when p < (i.e. he eponen is negive). So { = if p < p p divegen if p Emple 5 Emple 6 ( p ) Ye gin fi p >. This ime he domin of inegion of he inegl eends o p +, nd in ddiion he inegnd becomes unbounded s ppoches he lef end,, p of he domin of inegion. So we hve o spli i up. p = p + We sw, in Emple 5, h he fis inegl diveged wheneve p, nd we lso sw, in Emple 4, h he second inegl diveged wheneve p. So he inegl diveges p fo ll vlues of p. Emple 6 p Emple 7 ( ) This is pey suble emple. I looks fom he skech below h he signed e o he y y = lef of he y is should ecly cncel he e o he igh of he y is mking he vlue of he inegl ecly zeo. Bu boh of he inegls ] ln = + + T T + T ln ] T + ln ln T = T c Joel Feldmn. 25. All ighs eseved. 5 Febuy 5, 25
divege so diveges. Don mke he miske of hinking h =. I is undefined nd fo good eson. Fo emple, we hve jus seen h he e o he igh of he y is is + = + nd h he e o he lef of he y is is (subsiue 7 fo T bove) + 7 = If =, he following i should be. 7 ] + + ln ] + +ln 7 ln ] + +ln(7) ln7 + = ln7 This ppes o give = ln7. Of couse he numbe 7 ws picked ndom. You cn mke be ny numbe ll, by mking suible eplcemen fo 7. Emple 7 Emple 8 (Emple evisied) The ceful compuion of he inegl of Emple is 2 T T T = + + 2 + ] T 2 + ] + Emple 8 Since Emple 9 ( ) + 2 R ] R + cn 2 + 2 cn The inegl conveges nd kes he vlue π. + 2 ] cnr = π 2 cn = π 2 Emple 9 c Joel Feldmn. 25. All ighs eseved. 6 Febuy 5, 25
Emple Fo wh vlues of p does e convege? (ln) p Soluion. Fo e, he denomino (ln) p is neve zeo. So he inegnd is bounded on he enie domin of inegion nd his inegl is impope only becuse he domin of inegion eends o + nd we poceed s usul. We hve e R (ln) p e (ln) p lnr du u p = p (lnr) p wih u = ln, du = ln(lnr) if p = { divegen if p if p > p ] if p jus s in Emple 4, bu wih R eplced by lnr. Emple c Joel Feldmn. 25. All ighs eseved. 7 Febuy 5, 25