f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

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1 Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one of these two conditions is not met, we cll the integrl improper. Our usul definition for the vlue for the definite integrl f(x) dx = F (b) F (x) does not hold for such improper integrls; in this section, we will discuss the new definitions needed to evlute them. Two different types of integrls cn qulify s improper. Type I improper integrls hve infinite bounds. In the grph below, we might like to know the re of the region below f(x) from x = to x = : Type II improper integrls include discontinuity on the bounds. For exmple, we might like to know the re of the region below f(x) from x = to x =, but f(x) hs n symptote t x = :

2 Thinking of definite integrls s tools for clculting re, the integrls bove don t seem to mke sense the curves pper to enclose region of infinite re. However, their re nunces to be explored: in the first grph bove, for exmple, it my turn out tht the curve gets 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 We need new definition for integrls like e x dx. If the top bound of integrtion bove were number, we could simply find n ntiderivtive nd evlute it on the upper nd lower bounds of integrtion. However, it doesn t mke sense to evlute t infinity, which is not number to begin with. We solve the problem by using limits: if we wish to evlute we begin by finding e x dx, e x dx, leving b s vrible; then we tke the limit s b :

3 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. f(x) dx =. If f(x) is continuous on (, ), then for ny rel number. 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 similr ide evlute the integrl over region on which the function is continuous, nd then use limits to get to the desired vlue:

4 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. Exmple. Find x dx, if it exists. We begin by noting tht the integrl bove is Type I improper integrl thus we cnnot use 4

5 our old definition for the definite integrl. Insted, we evlute dx dx x x ( x ) ( ) =. Since the limit does not exist, the improper integrl diverges. Exmple. Find e x dx, if it exists. Agin, since the region over which we wish to integrte is infinite, we must use our new definition for Type I improper integrls: e x dx e x dx ( e x ) ( e x ) ( e + e ) = e since lim e = 0 (the denomintor pproches infinity, so the frction itself pproches 0). So in this cse, the function encloses n re of only e over the infinite region from to. Exmple. 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: 5

6 v v dv 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. Since we conclude tht A + B = 0 A = Av + Bv A =, (by equting the v terms) (by equting the v 0 terms). Thus A = nd B =. So the frction decomposes s Now we cn evlute the integrl: v v dv v v = v + (v ). v v dv v + (v ) dv ( ln v + ln v ) ( ln + ln + ln ln ) ( ln + ln ) + ln. 6

7 Since lim ln = nd lim ln =, it is tempting to guess tht the nswer is 0. 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 lim ( ln + ln ) = lim ( ln + ln ) ln c ln d = ln( c d ), we my rewrite the lst expression s Then Thus ln + ln = ln. lim ( ln + ln ) = lim (ln ) = ln lim LR = ln lim = ln = 0. v v dv = lim ( ln + ln ) + ln = 0 + ln = ln. Exmple. Let p be number so tht p > 0. For wht vlues of p does dx converge? xp Let s evlute the integrl using limits; of course, we will need to be creful, since there re two different ntiderivtives for function of the form /x p : ( ) x p dx = + C if p, p + xp+ 7

8 but x Thus we will need two different cses:. Cse : p If p, we hve: dx = ln x + C. b dx xp x p dx ( x p dx p + x p+ ) b ) (x ) ( ( p b p p p ) p b p p ) = ( ) p + lim ( p)b p = ( ) p + lim ( p)b p We need to find ( ) lim ( p)b p = p lim ( b p ). There re two cses, depending upon the vlue for p: () If p >, then b p s b, so nd lim ( ) = 0, ( p)bp x p dx = ( ) p + lim ( p)b p (b) If 0 < p < then p < 0 mens tht b p 0 s b, so p lim ( ) =, bp 8 = p.

9 so tht diverges. x p dx. Cse : p = If p =, the integrl is given by x dx x dx (ln x) b ) (ln b ln ) (ln b) =, so the integrl diverges. x x dx Thus x p dx { converges to p if p > diverges if p. Exmple. Find y dy. While it is tempting to integrte immeditely, we must note tht so the integrl is ctully Type II improper integrl. We cn rewrite it s y is not continuous t y = 0, y dy = 0 y dy + 0 y dy. 9

10 Let s evlute ech integrl seprtely: 0 y dy 0 y y 0 0 (y ) dy 0 ( + ) = dy since lim 0 = 0. Similrly, 0 y dy 0 + y 0 +(y ) dy 0 +( ) =. So our finl nswer is y dy = 6. Exmple. Find 0 x dx. This is Type II improper integrl becuse the function x is not continuous t x = ; in prticulr, the function hs verticl symptote t x =. In order to evlute the integrl, we will need to set up 0 x dx dx 0 x x (sin ) (sin sin 0). 0 0

11 We know tht sin 0 = 0, so we just need to determine lim sin. We cn use the unit circle to see tht in fct, sin = π : So 0 dx sin 0) x (sin = π. 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 n integrl such s 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 develop tool in this section tht my help us to do so.

12 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 [, ).. If 0 f(x) g(x) for ll x, nd. If 0 g(x) f(x) for ll x, nd g(x)dx converges, then g(x)dx diverges, then f(x)dx does s well. f(x)dx does s well.

13 Unfortuntely, the theorem does not help us hndle every cse tht could occur. For instnce, if 0 f(x) g(x) for ll x nd g(x)dx diverges, we gin no knowledge bout the convergence or divergence of f(x)dx: 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 0 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 my be ble to 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 should try to verify your belief by finding smller function g(x) whose integrl diverges. The difficult 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:

14 . Choose comprison function whose convergence or divergence you cn estblish.. Try to incorporte the sine or cosine functions, since 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). Exmple. 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 cn try to use the direct comprison test to determine if the integrl converges. Notice tht the function e x is very similr to the function e x ; it seems resonble tht they should behve similrly, so we will try using e x s our comprison function. Erlier, we sw tht converges. Note tht e x e x when x, so tht Then since e x dx e x = e x e x = e x. e x dx converges nd e x e x, e x dx must converge s well. 4

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