An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples.

Transcription

1 Improper Inegrls To his poin we hve only considered inegrls f(x) wih he is of inegrion nd b finie nd he inegrnd f(x) bounded (nd in fc coninuous excep possibly for finiely mny jump disconinuiies) An inegrl hving eiher n infinie i of inegrion or n unbounded inegrnd is clled improper. Here re wo exmples +x 2 The firs hs n infinie domin of inegrion nd he inegrnd of he second ends o s x pproches he lef end of he domin of inegrion. We ll sr wih n exmple h illusres he rps h you cn fll ino if you re such inegrls sloppily. Then we ll see how o re hem crefully. Exmple The compuion x = x2 x = = 2 is wrong. In fc, he nswer is ridiculous. The inegrnd >, so he inegrl hs o be x 2 posiive. The flw in he rgumen is h he fundmenl heorem of clculus, which sys h if F (x) = f(x) hen f(x) = F(b) F(), is pplicble only when F (x) exiss nd equls f(x) for ll x b. In his cse F (x) = does no exis for x =. The given x 2 inegrl is improper. We ll see ler h he correc nswer is +. Exmple The creful wy o re n inegrl like h hs bounded inegrnd bu hs +x 2 domin of inegrion h exends o + is o pproxime he inegrl by one wih bounded domin of inegrion, like, nd hen ke he i R. Here is he +x 2 y = f(x) R x forml definiion nd some vriions. We ll work hrough some exmples in deil shorly. c Joel Feldmn. 25. All righs reserved. Mrch 6, 25

2 Definiion 2. () If he inegrl f(x) exiss for ll R >, hen f(x) f(x) when he i exiss (nd is finie). (b) If he inegrl r f(x) exiss for ll r < b, hen when he i exiss (nd is finie). (c) If he inegrl r f(x) f(x) r r f(x) exiss for ll r < R, hen f(x) r c r f(x) + f(x) c when boh is exis (nd re finie). Any c cn be used. When he i(s) exis, he inegrl is sid o be convergen. Oherwise i is sid o be divergen. The creful wy o re n inegrl like h hs finie domin of inegrion bu x whose inegrnd is unbounded ner one i of inegrion is o pproxime he inegrl by one wih domin of inegrion on which he inegrnd is bounded, like, wih >, x nd hen ke he i +. Here is he forml definiion nd some vriions. y y = x x c Joel Feldmn. 25. All righs reserved. 2 Mrch 6, 25

3 Definiion 3. () If he inegrl f(x) exiss for ll < < b, hen when he i exiss (nd is finie). (b) If he inegrl T f(x) f(x) + f(x) exiss for ll < T < b, hen when he i exiss (nd is finie). T f(x) f(x) T b (c) Le < c < b. If he inegrls T f(x) nd f(x) exis for ll < T < c nd c < < b, hen T f(x) f(x) + f(x) T c c+ when boh i exis (nd re finie). When he i(s) exis, he inegrl is sid o be convergen. Oherwise i is sid o be divergen. Ifninegrlhsmorehnone sourceofimpropriey, forexmple ninfiniedominof inegrion nd n inegrnd wih n unbounded inegrnd or muliple infinie disconinuiies, hen you spli i up ino sum of inegrls wih single source of impropriey in ech. For he inegrl, s whole, o converge every erm in h sum hs o converge. For exmple, he inegrl hs domin of inegrion h exends o boh + nd nd, (x 2)x 2 in ddiion, hs n inegrnd which is singulr (i.e. becomes infinie) x = 2 nd x =. So we would wrie (x 2)x 2 = (x 2)x 2 + (x 2)x 2 + (x 2)x (x 2)x (x 2)x (x 2)x 2 (Here he brek poin could be replced by ny number sricly less hn ; could be replced by ny number sricly beween nd 2; nd 3 could be replced by ny number sricly bigger hn 2.) On he righ hnd side he firs inegrl hs domin of inegrion exending o c Joel Feldmn. 25. All righs reserved. 3 Mrch 6, 25

4 he second inegrl hs n inegrnd h becomes unbounded s x, he hird inegrl hs n inegrnd h becomes unbounded s x +, he fourh inegrl hs n inegrnd h becomes unbounded s x 2, he fifh inegrl hs n inegrnd h becomes unbounded s x 2+, nd he ls inegrl hs domin of inegrion exending o +. We re now redy for some (imporn) exmples. Exmple 4 ( x p ) Fix ny p >. The domin of he inegrl exends o + nd he inegrnd is x p x p coninuous nd bounded on he whole domin. So he definiion of his inegrl is R x p x p When p, one niderivive of x p is x p+, nd when p = nd x >, one niderivive p+ of x p is lnx. So x p p [ ] R p if p lnr if p = As R, lnr ends o nd R p ends o when p > (i.e. he exponen is posiive) nd ends o when p < (i.e. he exponen is negive). So { divergen if p x = p if p > p Exmple 4 Exmple 5 ( x p ) Agin fix ny p >. The domin of inegrion of he inegrl is finie, bu he inegrnd becomes unbounded s x pproches he lef end,, of he domin of inegrion. x p x p So he definiion of his inegrl is x p + x p When p, one niderivive of x p is x p+, nd when p = nd x >, one niderivive p+ of x p is lnx. So ] [ p if p x p p + ln if p = c Joel Feldmn. 25. All righs reserved. 4 Mrch 6, 25

5 As +, ln ends o nd p ends o when p > (i.e. he exponen is posiive) nd ends o when p < (i.e. he exponen is negive). So { x = if p < p p divergen if p Exmple 5 Exmple 6 ( x p ) Ye gin fix p >. This ime he domin of inegrion of he inegrl exends o x p +, nd in ddiion he inegrnd becomes unbounded s x pproches he lef end,, x p of he domin of inegrion. So we hve o spli i up. x p = x p + We sw, in Exmple 5, h he firs inegrl diverged whenever p, nd we lso sw, in Exmple 4, h he second inegrl diverged whenever p. So he inegrl diverges x p for ll vlues of p. Exmple 6 x p Exmple 7 ( ) x This is prey suble exmple. I looks from he skech below h he signed re o he y y = x x lef of he y xis should excly cncel he re o he righ of he y xis mking he vlue of he inegrl excly zero. Bu boh of he inegrls x [ ] lnx = + x + x T T x + x T [ ln x ] T + ln ln T = T c Joel Feldmn. 25. All righs reserved. 5 Mrch 6, 25

6 diverge so diverges. x Don mke he miske of hinking h =. I is undefined nd for good reson. For exmple, we hve jus seen h he re o he righ of he y xis is + x = + nd h he re o he lef of he y xis is (subsiue 7 for T bove) + 7 x = If =, he following i should be. [ 7 ] + x + [ ln ] [ x + +ln 7 ln ] + +ln(7) ln7 + = ln7 This ppers o give = ln7. Of course he number 7 ws picked rndom. You cn mke be ny number ll, by mking suible replcemen for 7. Exmple 7 Exmple 8 (Exmple revisied) The creful compuion of he inegrl of Exmple is x2 T [ T T = + x + x2 + ] T x 2 + [ ] + x Exmple 8 Since Exmple 9 ( ) +x 2 r r [ ] R +x rcn x 2 +x 2 r [ rcn x The inegrl converges nd kes he vlue π. +x 2 ] r rcnr = π 2 r rcnr = π 2 Exmple 9 c Joel Feldmn. 25. All righs reserved. 6 Mrch 6, 25

7 Exmple For wh vlues of p does e converge? x(lnx) p Soluion. For x e, he denominor x(lnx) p is never zero. So he inegrnd is bounded on he enire domin of inegrion nd his inegrl is improper only becuse he domin of inegrion exends o + nd we proceed s usul. We hve e x(lnx) p e x(lnx) p lnr du wih u = lnx, du = u p x [ ] (lnr) p if p p ln(lnr) if p = = { divergen if p if p > p jus s in Exmple 4, bu wih R replced by lnr. Exmple Exmple (he gmm funcion) The gmm funcion Γ(x) is defined by he improper inegrl Γ() = x e x We shll now compue Γ(n) for ll nurl numbers n. To ge sred, we ll compue [ ] R Γ() = e x e x e x = nd Γ(2) = xe x [ xe x R + xe x ] e x (by inegrion by prs wih u = x, dv = e x, v = e x, du = ) [ ] R xe x e x = c Joel Feldmn. 25. All righs reserved. 7 Mrch 6, 25

8 For he ls equliy, we used h x xe x =. Now we move on o generl n, using he sme ype of compuion s we jus used o evlue Γ(2). For ny nurl number n, Γ(n+) = x n e x [ x n e x R + x n e x ] nx n e x (by inegrion by prs wih u = x n, dv = e x, v = e x, du = nx n ) n x n e x = nγ(n) To ge o he hird row, we used h x x n e x =. Now h we know Γ(2) = nd Γ(n+) = nγ(n), for ll n N, we cn compue ll of he Γ(n) s. Γ(2) = Γ(3) = Γ(2+)= 2Γ(2) = 2 Γ(4) = Γ(3+)= 3Γ(3) = 3 2 Γ(5) = Γ(4+)= 4Γ(4) = Γ(n) = (n) (n 2) = (n)! Exmple Convergence Tess for Improper Inegrls I is very common o encouner inegrls h re oo compliced o evlue explicily. Numericl pproximion schemes, evlued by compuer, re ofen used insed. You wn o be sure h les he inegrl converges before feeding i ino compuer. Forunely i is usully possible o deermine wheher or no n improper inegrl converges even when you cnno evlue i explicily. For pedgogicl purposes, we re going o concenre on he problem of deermining wheher or no n inegrl f(x) converges, when f(x) hs no singulriies for x. Recll h he firs sep in nlyzing ny improper inegrl is o wrie i s sum of inegrls ech of hs only single source of impropriey eiher domin of inegrion h exends o +, or domin of inegrion h exends o, or n inegrnd which is singulr one end of he domin of inegrion. So we re now going o consider only he firs of hese hree possibiliies. Bu he echniques h we re bou o see hve obvious nlogues for he oher wo possibiliies. c Joel Feldmn. 25. All righs reserved. 8 Mrch 6, 25

9 Now le s sr. Imgine h we hve n improper inegrl f(x), h f(x) hs no singulriies for x nd h f(x) is compliced enough h we cnno evlue he inegrl explicily. The ide is find noher improper inegrl g(x) wih g(x) simple enough h we cn evlue he inegrl g(x) explicily, or les deermine esily wheher or no g(x) converges, nd wih g(x) behving enough like f(x) for lrge x h he inegrl if nd only if g(x) converges. f(x) converges So fr, his is prey vgue sregy. Here is heorem which srs o mke i more precise. Theorem 2 (Comprison). Le be rel number. Le f nd g be funcions h re defined nd coninuous for ll x nd ssume h g(x) for ll x. () If f(x) g(x) for ll x nd if g(x) converges hen f(x) converges. (b) If f(x) g(x) for ll x nd if g(x) diverges hen f(x) diverges. We will no prove his heorem, bu, hopefully, he following supporing rgumens should les pper resonble o you. If g(x) converges, hen he re of { (x,y) } x, y g(x) is finie. When f(x) g(x), he region { (x,y) } x, y f(x) is conined inside { } (x,y) x, y g(x) nd so lso hs finie re. Consequenly he res of boh he regions { (x,y) } { } x, y f(x) nd (x,y) x, f(x) y re finie oo. (Look grph of f(x) nd grph of f(x).) The inegrl f(x) represens he signed re of he union of { (x,y) } x, y f(x) nd { } (x,y) x, f(x) y. If g(x) diverges, hen he re of { (x,y) x, y g(x) } is infinie. When f(x) g(x), he region { (x,y) x, y f(x) } conins he region { (x,y) x, y g(x) } nd so lso hs infinie re. Exmple 3 ( e x2 ) We cnno evlue he inegrl e x2 explicily. Bu we would sill like o ell if i is finie or no. So we wn o find noher inegrl h we cn compue nd h we cn compre o e x2. To do so we pick n inegrnd h looks like e x2, bu whose indefinie inegrl we know like e x. When x, we hve x 2 x nd hence e x2 e x. The inegrl [ ] R e x e x e x ] [e e R = e c Joel Feldmn. 25. All righs reserved. 9 Mrch 6, 25

10 converges. So, by Theorem 2, wih =, f(x) = e x2 nd g(x) = e x, he inegrl e x2 converges oo. Exmple 3 Exmple 4 ( /2 e x2 ) Of course he inegrl is quie similr o he inegrl e /2 e x2 x2 of Exmple 3. Bu we cnno jus repe he rgumen of Exmple 3 becuse i is no rue h e x2 e x when < x <. In fc, for < x <, x 2 < x so h e x2 > e x. However he difference beween nd e /2 e x2 x2 is excly which is clerly well defined /2 e x2 finie number. So we would expec h should be he sum of he proper inegrl /2 e x2 inegrl nd he convergen inegrl /2 e x2 e x2 nd so should be convergen inegrl. This is indeed he cse. The Theorem below provides he jusificion. Exmple 4 Theorem 5. Le < c nd le he funcion f(x) be coninuous for ll x. Then he improper inegrl f(x) converges if nd only if he improper inegrl f(x) converges. c Proof. By definiion he improper inegrl f(x) converges if nd only if he i [ c f(x) = c f(x) + f(x) + c c ] f(x) f(x) exiss nd is finie. (Remember h, in compuing he i, c f(x) is finie consn independen of R nd so cn be pulled ou of he i.) Bu h is he cse if nd only if he i f(x) exiss nd is finie, which in urn is he cse if nd only if he c inegrl f(x) converges. c Exmple 6 Does he inegrl x x 2 +x converge or diverge? Soluion. Our firs sk is o idenify he poenil sources of impropriey for his inegrl. Cerinly he domin of inegrion exends o +. Bu we mus lso check o see if he inegrnd conins ny singulriies. On he domin of inegrion x so he denominor is never zero nd he inegrnd is coninuous. So he only problem is +. c Joel Feldmn. 25. All righs reserved. Mrch 6, 25

11 Our second sk is o develop some inuiion. As he only problem is h he domin of inegrion exends o infiniy, wheher or no he inegrl converges will be deermined by he behvior of he inegrnd for very lrge x. When x is very lrge x x 2 so h he denominor x 2 +x x 2 nd he inegrnd x x =. By Exmple 4, wih p = x 2 +x x 2 x 3 /2, 3/2 x x 2 +x he inegrl converges. So we would expec h converges oo. x 3/2 Our finl sk is o verify h our inuiion is correc. To do so, we wn o pply pr () of Theorem 2 wih f(x) = x nd g(x) being, or possibly some consn imes x 2 +x x 3/2. We will be ble o pply Theorem 2. if we cn show h x is smller hn some x 3/2 x 2 +x consn imes on he domin of inegrion, x. Here goes. x 3/2 x = x 2 +x > x 2 = x 2 +x < x x x = 2 x 2 +x < x = 2 x 3/2 So Theorem 2. nd Exmple 4, wih p = 3 /2 do indeed show h he inegrl converges. Exmple 6 x x 2 +x There is vrin of he comprison Theorem 2 h is ofen esier o pply hn Theorem 2 iself nd h lso fis well wih he sor of inuiion h we developed in Exmple 6. Le s briefly review wh hppened In Exmple 6. We firs idenified he problem x s s he lrge x s. (The inegrl ws improper only becuse he domin of inegrion exended o +.) Our inegrnd, f(x) = x, ws relively compliced bu x 2 +x we found h, for lrge x, f(x) behved like he simple funcion g(x) =. We knew x 3/2 h he inegrl of g(x), over he domin of inegrion of ineres, converged. We hen used Theorem 2 o show h he inegrl of f(x) converged oo. A good wy o formlize f(x) behves like g(x) for lrge x is o require h he i f(x) x g(x) exiss nd is nonzero number. Suppose h his is he cse nd cll he i L. Then mus pproch L s x ends o + nd, in priculr, here mus be number c such f(x) g(x) h f(x) lies beween L nd 2L, h is f(x) lies beween L g(x) nd 2Lg(x), for ll x c. g(x) 2 2 Consequenly, he inegrl of f(x) converges if nd only if he inegrl of g(x) converges, by Theorems 2 nd 5. These considerions led o he following vrin of Theorem 2. c Joel Feldmn. 25. All righs reserved. Mrch 6, 25

12 Theorem 7 ( Limiing comprison). Le < <. Le f nd g be funcions h re defined nd coninuous for ll x nd ssume h g(x) for ll x. () If g(x) converges nd he i f(x) x g(x) exiss (in his cse i is llowed o be ny rel number, including ), hen f(x) converges. (b) If g(x) diverges nd he i exiss nd is nonzero, hen f(x) x g(x) f(x) diverges. There re obvious nlogs of his heorem for he oher ypes of improper inegrls oo. Here is n exmple of how Theorem 7 is used. Exmple 8 Does he inegrl x+sin x converge or diverge? e x +x 2 Soluion. Our firs sk is o idenify he poenil sources of impropriey for his inegrl. The domin of inegrion exends o +. On he domin of inegrion he denominor is never zero so he inegrnd is coninuous. Thus he only problem is +. Our second sk is o develop some inuiion bou he behvior of he inegrnd for very lrge x. When x is very lrge e x x 2, so h he denominor e x +x 2 x 2 nd sinx x, so h he numeror x+sinx x nd he inegrnd x+sinx e x +x 2 x x 2 = x. Since diverges, we would expec x+sin x o diverge oo. x e x +x 2 Our finl sk is o verify h our inuiion is correc. To do so, we se f(x) = x+sinx e x +x 2 g(x) = x c Joel Feldmn. 25. All righs reserved. 2 Mrch 6, 25

13 nd compue f(x) x g(x) x+sinx x e x +x 2 x x (+sinx/x)x (e x /x 2 +)x 2 x x +sinx/x e x /x 2 + = Since g(x) = diverges, by Exmple 4 wih p =, Theorem 7.b now ells us x h f(x) = x+sin x diverges oo. e x +x 2 Exmple 8 c Joel Feldmn. 25. All righs reserved. 3 Mrch 6, 25