Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph below, we might like to know the re below e x from x = to x = : Alterntively, we my need to evlute n integrl over region on which the function hs verticl symptote (in other words, the function is not continuous on the intervl of interest). Below, we might like to know the re below / x from x = to x = ; of course, the function hs verticl symptote t x = : We cll this Type II improper integrl. Thinking of integrls s wy to clculte re, the integrls bove don t seem to mke sense. Don t the curves enclose region of infinite re? Then nswer is, not necessrily. In the first exmple, it my turn out tht the curve gets so close to the x-xis so quickly tht, even though we wish to integrte over n infinite region, the enclosed re is ctully finite. Similr behvior cn occur in the second cse. Type I Improper Integrls How do we evlute n integrl such s e x dx? If the top bound of integrtion were number, we could simply find n ntiderivtive nd plug the bounds in. However, it doesn t mke sense to plug in infinity, which is not number to begin with. We solve the problem by using limits: if we wish to evlute e x dx,
we begin by finding e x dx, leving b s vrible; then we tke the limit s b : If the limit bove is rel (finite) number, then we sy tht the improper integrl converges; this mens tht the function encloses finite re over n infinite region. Otherwise, we sy tht the integrl diverges, tht is the function encloses infinite re over the region. In summry,. If f(x) is continuous on [, ), then if the limit exists. f(x)dx f(x)dx,. If f(x) is continuous on (, ], then if the limit exists.. If f(x) is continuous on (, ), then for ny rel number. f(x)dx = f(x)dx = lim b b f(x)dx + f(x)dx, f(x)dx, Type II Improper Integrls We cn evlute type II improper integrls using limits, s well. If f(x) hs discontinuity on the intervl over which we wish to integrte, then we define the improper integrls s follows:
. If f(x) is continuous on (, b], but discontinuous t x =, then if the limit exists. f(x)dx c + c f(x)dx,. If f(x) is continuous on [, b), but discontinuous t x = b, then if the limit exists. f(x)dx c b c f(x)dx,. If f(x) is continuous on [, b], except t the point x = c with < c < b, then f(x)dx = c f(x)dx + c f(x)dx. Agin, if the limit exist (tht is the limit returns rel number), then we sy tht the improper integrl converges; the improper integrl diverges if the limit does not exist, tht is the function encloses infinite re. Exmples Find dx, if it exists. x By definition, we hve dx x dx x ) ( x ( ) = Since the limit does not exist, the improper integrl diverges. Find e x dx, if it exists.
Agin, using the definition, we hve e x dx e x dx e x dx ( e x ) ( e + e ) = e since lim e = (the denomintor pproches infinity, so the frction itself pproches ). So in this cse, the function encloses n re of only e over the infinite region from to. Find v v dv. Agin, since the intervl over which we wish to integrte is infinite, we will need to evlute the improper integrl using limits: v dv v v v dv. We will need prtil frctions to evlute the integrl: since the denomintor fctors s v v = v(v ), the form of the decomposed frction must be Adding the frctions, we see tht A v + B v. A v + B Av A = v v(v ) + Bv v(v ) Av A + Bv = v(v ) Av A + Bv = = v v v v. 4
Since Av + Bv A =, we conclude tht A + B = A = (by equting the v terms) (by equting the v terms). Thus A = nd B =. So the frction decomposes s Now we cn evlute the integrl: v dv v v v = v + (v ). v v dv v + (v ) dv ( ln v + ln v ) ( ln + ln + ln ln ) ( ln + ln ) + ln. Since lim ln = nd lim ln =, it is tempting to guess tht the nswer is. However, is n indeterminte form, so we ctully do not know the vlue for the limit. Let s fctor the out of the limit: Now since ln c ln d = ln( c d tht So Since the limit lim lim ( ln + ln ) = lim ( ln + ln ) ), let s try rewriting ln + ln s ln. Then lim ( ln + ln ) = lim (ln ) = ln lim. hs the indeterminte form, we cn use L Hopitl s rule to see LR lim =. ln lim = ln =, 5
which mens tht v v dv = lim ( ln + ln ) + ln = + ln = ln. Let p be number so tht p >. For wht vlues of p does dx converge? xp Let s evlute the integrl using limits; we will need two different cses, one when p = nd when when p. Let s begin with the p cse: dx xp ( ( ( ( ( x p dx x p dx b p + x p+ ) ) (x ) p b p p p ) p b p p ) ( p)b p ) p ( p)b p ) + p We need to find lim ( ( p)b p ) = p lim ( b There will be two cses, depending upon the vlue for p:. If p >, then b p s b, so lim ( ) =. ( p)bp p ).. If < p < then p < mens tht b p s b, so p lim ( ) =. bp 6
Now if p =, then b dx x x dx (ln x) b ) (ln b ln ) (ln b) =, so the integrl diverges. Thus x x dx x p dx { converges to p if p > diverges if p. Find y dy. While it is tempting to integrte immeditely, we must note tht so the integrl is ctully improper. We cn rewrite it s y is not continuous t y =, y dy = y dy + y dy. Let s evlute ech integrl seprtely: since lim =. y dy y (y dy ) ( + ) = 7
Similrly, y dy + y +(y ) dy +( ) =. So the finl nswer is Find x dx. y dy = 6. This is n improper integrl becuse the function x is not continuous t x = ; in prticulr, there is verticl symptote t x =. In order to evlute the integrl, we will need to set up dx dx x x x (sin ) (sin sin ). We know tht sin =, so we just need to determine lim sin. We cn use the unit circle to see tht in fct, sin = π : So dx sin ) x (sin = π. 8
Testing for Convergence or Divergence We hve noted erlier in the course tht certin integrls cnnot be evluted using elementry methods; for exmple, we cn not hope evlute e x dx becuse we do not know how to find n ntiderivtive of e x. However, we my still be ble to determine whether or not the integrl converges; we will lern two tools in this section tht my help us to do so. Direct Comprison Test If the grph of f(x) lies underneth the grph of g(x), then the re below f(x) must be less thn the re below g(x): In prticulr, if g(x) converges (encloses finite re) then so does f(x). Alterntely, if the grph of f(x) lies bove the grph of g(x), then the re below f(x) must be more thn the re below g(x): So if g(x) diverges (encloses n infinite re), f(x) must s well. These observtions led to the direct comprison test: Theorem... Let f(x) nd g(x) be continuous on [, ). 9
. If f(x) g(x) for ll x, nd g(x)dx converges, then f(x)dx does s well.. If g(x) f(x) for ll x, nd g(x)dx diverges, then f(x)dx does s well. Unfortuntely, the theorem does not help us hndle every cse tht could occur. For instnce, if f(x) g(x) for ll x nd or divergence of f(x)dx: g(x)dx diverges, we gin no knowledge bout the convergence Since f(x) g(x), the fct tht g(x) bounds region whose re is infinite does not inform us s to the size of the re under f(x). On the other hnd, if g(x) f(x) for ll x, nd we lern nothing bout the convergence or divergence of g(x)dx converges, then once gin f(x)dx: Since g(x) f(x), the fct tht g(x) bounds region of finite re does not help us determine the size of the re bounded by f(x). To sum up the bove informtion, if you suspect tht n improper integrl of f(x) converges, you verify your clim by finding lrger function g(x) whose integrl converges. Alterntely, if
you believe tht n improper integrl of f(x) diverges, you need to find smller function g(x) whose integrl diverges. The hrd prt in using the bove theorems is tht you will need to find functions to use in the comprison; this is not lwys esy or obvious. The following tips my help you to find comprison functions:. Choose comprison function whose convergence or divergence you cn estblish. sin x nd cos x for ny choice of x.. Incresing the size of frction s denomintor while holding the numertor constnt will decrese the size of the frction 4. Decresing the size of frction s denomintor while holding the numertor constnt will increse the size of the frction 5. It is often helpful to simplify some portion of function if tht portion is known to be bounded by constnt (see bove) Exmples Does e x dx converge or diverge? Notice tht we do not know how to find n ntiderivtive of e x, so even if we knew tht the integrl converges, we would not be ble to find its ctul vlue; the best we cn hope for is to be ble to determine convergence or divergence of the improper integrl. We will need to use the direct comprison test to determine convergence, so we will need to be ble to compre e x = with nother function whose convergence or divergence we cn e estblish. It is best to choose x function similr to the originl one for esy comprison. Erlier, we found tht e x dx converges; it is similr to the current function of interest nd so might be helpful. Note tht since e x e x when x, then e x = e x = e x. e x Then since e x dx converges nd e x e x, e x dx must converge s well.