Integration Techniques

Transcription

1 Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u = sin x so tht du = cos x. Then, cos 3 x = cos x cos x = ( sin x) cos x = ( u )du = u 3 u3 + C = sin x 3 sin3 x + C. Exmple.3 Evlute cos 4 x. cos 4 x = (cos x) = +cos x ( ) = 4 ( + cos x + cos x) = 4 ( + cos x + +cos 4x ) = 8 (3 + 4 cos x + cos 4x) = 8 (3x + sin x + 4 sin 4x) + C (x + 8 sin x + sin 4x) + C. = 3

2 Exmple.4 Evlute cos 5 x. Let u = sin x so tht du = cos x. Then, cos 5 x = ( sin x) cos x = ( u ) du = ( u + u 4 )du = u 3 u3 + 5 u5 + C = 5 (5 sin x sin3 x + 3 sin 5 x) + C. Exmple.5 Evlute sin 3 x cos 4 x. Let u = cos x so tht du = sin x. Then, sin 3 x cos 4 x = ( cos x) cos 4 x sin x = ( u )u 4 du = 5 u5 + 7 u7 + C = 5 cos5 x + 7 cos7 x + C. Exmple.6 Evlute sin 3x cos 5x sin 3x cos 5x. = = (sin 3x cos 5x + cos 3x sin 5x + sin 3x cos 5x cos 3x sin 5x) (sin 8x sin x) = ( 8 cos 8x + cos x) + C (4 cos x cos 8x) + C. = 6 Exmple.7 Evlute sin x cos 4 x.

3 sin x cos 4 x = cos x +cos x ( ) = 8 ( cos x)( + cos x) = 8 ( +cos 4x )( + cos x) = 6 ( cos 4x + cos x cos x cos 4x) = 6 ( cos 4x + cos x (cos 6x + cos x)) = 3 ( cos 4x + cos x cos 6x) = 3 (x 4 sin 4x + sin x 6 sin 6x) + C Exmple.8 Evlute sec x. Let u = tn x + sec x so tht du = (sec x + tn x sec x). Then, sec x = sec x(tn x + sec x) tn x + sec x = (sec x + tn x sec x) tn x + sec x = du u = ln u + C = ln(tn x + sec x) + C. Exmple.9 Evlute sec 4 x. Let u = tn x so tht du = sec x. Then, sec 4 x = ( + tn x) sec x = ( + u )du = u + 3 u3 + C = tn x + 3 tn3 x + C.. Integrtion by Prts Theorem. (Integrtion by prt) If u, v re differentible functions, u(x)v (x) = u(x)v(x) u (x)v(x). proof: For every function u, v, (uv) = u v + uv. 3

4 Thus, or u(x)v(x) = u (x)v(x) + u(x)v (x) = u(x)v(x) u(x)v (x) u (x)v(x). Exmple. Evlute xe x. Let u(x) = x, v (x) = e x (so tht we my tke v(x) = e x ). Applying integrtion by prt to the given integrl yields xe x = xe x e x = xe x e x + C. Exmple. Evlute ln x. Let u(x) = ln x, v (x) = (so tht we my tke v(x) = x). Applying integrtion by prt to the given integrl yields ln x Exmple.3 Evlute = x ln x x x = x ln x x + C. π x cos x. Let u(x) = x, v (x) = cos x (so tht we my tke v(x) = sin x). Applying integrtion by prt to the given integrl yields π x cos x = x sin x π π x sin x = π x sin x. To evlute the lst integrl, we pply integrtion by prt gin. This time, we let u(x) = x, v (x) = sin x (so tht we my tke v(x) = cos x). Then, π x cos x = π x sin x = x cos x π π cos x = π sin x π = π. 4

5 Exmple.4 Evlute ( + x ) 3/. ( + x ) 3/ = ( + x x ) ( + x ) 3/ = ( + x ) / x x ( + x ) 3/ We will keep the first term unchnged nd del with the second term by integrting by prt. For this purpose, let u(x) = x nd v x (x) = (so tht ( + x ) 3/ v(x) = ). Then, ( + x ) / Exmple.5 Evlute ( + x ) 3/ = ( + x ) + x / ( + x ) / ( + x ) x / = ( + x ) / = /. π e x cos x. Let u(x) = e x, v(x) = sin x. Applying integrtion by prt to the given integrl yields π ex cos x = e x sin x π π ex sin x = π ex sin x. To evlute the lst integrl, we pply integrtion by prt gin. This time, we let u(x) = e x, v(x) = cos x. Then, π ex cos x = π ex sin x = e x cos x π π = e π π ex cos x ex cos x. Rerrnging the eqution yields π e x cos x = (eπ + ). 5

6 Exmple.6 Let n be positive integer nd define Show tht for n. I n = π sin n x. I n = n n I n Let n nd I n = π sin n x = π sin n x sin x. We del with the ltter integrl by integrtion by prt. We let u(x) = sin n x nd v (x) = sin x (so tht we my tke v(x) = cos x). The integrl becomes I n = π sinn x = sin n x cos x π + (n ) π cos x sin n x = (n ) π ( sin x) sin n x (s n ) = (n ) π sinn x (n ) π sinn x = (n )I n (n )I n After rerrngement, it becomes for n. Exmple.7 Evlute ni n = (n )I n π sin 8 x. According to the previous exmple, π sin8 x = I 8 = 7 8 I 6 = I 4 = I = I = π. Exmple.8 Let n be positive integer, evlute π sin n x. 6

7 Note tht I = π If n is positive integer, we hve = π, π sin x =. I n = n n I n = n n 3 n I n 4 = n n 3 n 5 n n n 4 I n 6 ()(3)(5)...(n ) I when n is even = ()(4)(6)...n ()(4)(6)...(n ) I when n is odd (3)(5)(7)...n ()(3)(5)...(n ) π when n is even = ()(4)(6)...n ()(4)(6)...(n ) when n is odd (3)(5)(7)...n Remrk.9 In the previous three exmples, the formul I n = n n I n for n plyed key role. Such kind of formul is clled reduction formul. Definition. Let n be positive integer. Define n! = 3... n. We define! = lso. Remrk. Note tht ()(3)(5)...(n ) ()(4)(6)...n ()()(3)(4)(5)...(n )(n) = () (4) (6)...(n) n! = n () () (3)...(n/) n! = n ( n )! Thus, π sin n x = Exmple. Let n! n ( n when n is n even positive integer )!π n ( n )! when n is n odd positive integer n! I n,m = ( x) n ( + x) m. 7

8 Prove tht for ll positive integers n, m. Let n, m be positive integers. I n,m = I n,m = ( x)n ( + x) m = m + ( x)n ( + x) m+ + = n m + I n,m+ Exmple.3 Evlute for ech positive integer n. n m + I n,m+ n m + ( x)n ( + x) m+ (integrtion by prt) ( x ) n Let n be positive integer. Apply the result in the previous exmple, we hve ( x ) n = ( x)n ( + x) n = I n,n = n n + I n,n+ = n n n + + I n,n+ = n n n n + n + n + 3 I n 3,n+3... (n)(n )(n )...() = (n + )(n + )(n + 3)...(n) I,n n+ = n! (n)! n + = n+ n! (n + )!. Exmple.4 Evlute sec 6 x. Let I n (x) = sec n x which is defined up to ddition of constnt. Then, I n (x) = sec n x sec x = tn x sec n x (n ) sec n x tn x = tn x sec n x (n ) sec n x(sec x ) = tn x sec n x (n )I n (x) + (n )I n (x). 8

9 After rerrngement it yields Therefore I n (x) = n (tn x secn x) + n n I n (x) for n >. sec 6 x = I 6 (x) = 5 tn x sec4 x I 4(x) = 5 tn x sec4 x tn x sec x = 5 tn x sec4 x tn x sec x Trigonometric Substitution Exmple.5 Evlute x 4 x. Let x = sin u so tht = cos udu. Then, Exmple.6 Evlute x 4 x = cos udu sin u 4 4 sin u = csc udu = cot u csc u + csc u du cot u + csc u = ln(cot u + csc u) + C = ln + 4 x + C. x x Let x = 3 tn u so tht = 3 sec u. Then, x = 3 sec udu 9 tn u = 9 cos udu = 9 sin u x + C x C. = 9 3 I (x) tn x + C. 3 9

10 Exmple.7 Evlute x 5. x Let x = 5 sec u so tht = 5 tn u sec udu. Then, x 5 x = 5 sec u 5 5 tn u sec udu 5 sec u = 5 tn udu = 5 (sec u )du = 5(tn u u) + C x = 5[ 5 x sec ] + C = x 5 5 cos 5 x + C. Exmple.8 Evlute x + x +. Let x = tn u so tht = sec udu. Then, x + x + = (x + ) + = sec udu tn u + = du = u + C = tn (x + ) + C. Exmple.9 Evlute 9 + 6x 4x.

11 Let x = 5 sin u + so tht = 5 cos udu. Then, 9 + 6x 4x = 5 4(x ) = 5 cos udu 5 5 sin u du = = u + C = sin (x ) 5.4 Digression-Polr Coordintes + C. Definition.3 Let the usul coordinte xes be chosen in the plne. Then the polr coordintes of point P in the plne re (r, θ) if the distnce from P to the origin is r nd the ngle between the ry from the origin to P nd the positive x-xis is θ. Lemm.3 If (x, y) re the usul coordintes of point in the plne nd (r, θ) re the polr coordintes of the sme point. Then, x = r cos θ, nd y = r sin θ. Exmple.3 Let C be the curve in the plne which collects points whose coordintes (r, θ) stisfy r = + cos θ. Describe this curve in the usul rectngulr (Crtesin) coordintes. Now C collects points whose polr coordintes stisfy r = r + r cos θ (Note tht the origin belongs to C.). By the lst lemm, C collects points whose usul rectngulr coordintes stisfy x + y = x + y + x, or x + y = (x + y x). Exmple.33 The curve which collects points whose polr coordintes (r, θ) stisfy r = is the unit circle centered t the origin. Exmple.34 The curve which collects points whose polr coordintes (r, θ) stisfy r = cos θ is the union of two circles touching with ech other t the origin. This curve is indeed one of the p-orbitls in chemistry.

12 Exmple.35 Let C be the curve in the plne which collects points Crtesin coordintes (x, y) stisfy x y = x + y 3. Describe this curve in polr coordintes. By the lst lemm, C collects points whose polr coordintes stisfy or r cos θ r sin θ = r 3 r = 3 cos θ. Note tht this curve is one of the d-orbitls in chemistry. Exmple.36 Let e, l >. The curve which collects points whose polr coordintes (r, θ) stisfy l r = + e cos θ. Suppose tht L is the verticl line which is locted l/e on the right hnd side of the y-xis. Then the curve we re describing is the locus of points P so tht distnce from the origin to P perpendiculr distnce from P to the line L = e. Such curve is clled conic with eccentricity e nd semi-ltus rectum l. Moreover, this conic is clled n ellipse if < e <, prbol if e = nd hyperbol if e >. The origin is one of the foci of this conic. Exmple.37 The curve which collects points whose polr coordintes (r, θ) stisfy r = θ is clled spirl. Theorem.38 Let f be function. C is the curve which collects points whose polr coordintes (r, θ) stisfy r = f(θ). Then the slope of the line tngent to C t the point with polr coordintes (f(), ) is f () sin + f() cos f () cos f() sin. proof: For ech h, the slope of the line joining the points with polr coordintes (f(), ) nd (f( + h), + h) is f( + h) sin( + h) f() sin g( + h) g() = f( + h) cos( + h) f() cos h( + h) h()

13 where g(t) = f(t) sin t nd h(t) = f(t) cos t for ll t. By the Cuchy men vlue theorem, there exists c between nd + h such tht g( + h) g() h( + h) h() = g (c) h (c) which tends to g ()/h () s h tends to zero. Therefore the slope of the line tngent to C t the point with polr coordintes (f(), ) is f( + h) sin( + h) f() sin lim h f( + h) cos( + h) f() cos = g () h () = f () sin + f() cos f () cos f() sin. Exmple.39 Let e, l >, nd C is the conic collecting points whose polr coordintes (r, θ) stisfy l r = + e cos θ. Find the slope of the line tngent to C t point lying on the positive y-xis. Let so tht f(θ) = f (θ) = l + e cos θ for θ π le sin θ ( + e cos θ) for θ π. Hence f(π/) = l nd f (π/) = le. By the previous theorem, the slope of the line tngent to C t point lying on the positive y-xis is f (π/) sin π/ + f(π/) cos π/ f (π/) cos π/ f(π/) sin π/ = e. Remrk.4 Let f be function so tht f(t) for ll t b. Then, the re of the region which collects points whose polr coordintes (r, θ) stisfy is r f(θ) nd θ b b (f(θ)) dθ. Exmple.4 Evlute the re bounded by the prbol with its focus t the origin nd semi-ltus rectum l; nd the line through the origin with slope tn α. The given prbol bounds the points whose polr coordintes (r, θ) stisfy r l + cos θ nd θ π. 3

14 Hence the re of the given region is α α π ( l + cos θ ) dθ. We let u = tn θ, so tht du = sec θ. Then α α π α = l = l 8 = l 4 ( l + cos θ ) dθ ( cos θ dθ ) α π α α π tn α/ cot α/ sec 4 θ dθ ( + u )du = l 4 [tn α + 3 tn3 α + cot α + 3 cot3 α ] = l 3 sin 3 α Exmple.4 Compute the re of the region which collects points so tht their polr coordintes (r, θ) stisfy The re of the given region is π/3 π/3 = π/3 = π/3 π/3 = π/3 π/3 = r mx{, + cos θ}. ( + cos θ) dθ π/3 ( + 4 cos θ + 4 cos θ)dθ ( + 4 cos θ + ( + cos θ))dθ (3 + 4 cos θ + cos θ)dθ (3θ + 4 sin θ + sin θ) π/3 π/3 = π Lemm.43 The re of the region collecting points whose polr coordintes (r, θ) stisfy g(θ) r f(θ) nd θ b is b (f(θ) g(θ) )dθ. Exmple.44 Compute the re of the region collecting points whose polr coordintes (r, θ) stisfy + 3 r + cos θ. 4

15 The re of the given region is π/6 = π/6 [( + cos θ) ( + 3) ]dθ π/6 π/6 ( cos θ + cos θ)dθ = 6 ( + 3)π Prtil Frctions Exmple.45 Evlute x. Since x =, we will try finding two numbers A nd B such (x )(x + ) tht (x )(x + ) = A x + B x + if possible. If such numbers A nd B re there, they stisfy (A + B)x + (A B) = (x )(x + ) (x )(x + ) for ll x. Tht is, we need A + B = nd A B =. The choice A = / nd B = / does the job. Thus, Finlly, (x )(x + ) = ( x x + ). x = ( x x + ) = (ln(x ) ln(x + )) + C = ln x x + + C. Remrk.46 In the previous exmple, we split frction into sum of severl simpler frctions. Such technique is clled the technique of prtil frctions. Exmple.47 Evlute sec x. 5

16 Let u = sin x so tht du = cos x. Then, sec x = cos x sin x = du u = ( u + + u )du = ( ln u + ln + u ) + C = + sin x ln sin x + C Exmple.48 Evlute x 3 x Note tht we hve Hence, x 3 = x(x ) + x, x 3 x = x(x ) x + x x = x + x + x + (x )(x + ) = x + ( x + x + ). x 3 x = [x + ( x + x + )] = (x + ln x + ln x + ) + C = (x + ln x ) + C. Exmple.49 Evlute x + x(x ) 3 We will try finding three numbers A nd B such tht x + x(x ) 3 = A x + B (x ) 3 6

17 if possible. If such numbers A nd B re there, they stisfy x + x(x ) 3 = A(x )3 + Bx x(x ) 3 for ll x. Tht is, we need A(x ) 3 + Bx = x + for ll x. Putting x = yields B = nd putting x = yields A =. But it is obvious tht (x ) 3 + x x +!! We conclude tht such numbers A nd B re not there. To rectify the sitution, we will see if there re numbers A, B, C, D such tht x + x(x ) 3 = A x + B (x ) 3 + C (x ) + D x. In other words, or x + x(x ) 3 = A(x )3 + Bx + Cx(x ) + Dx(x ) x(x ) 3 x + = A(x ) 3 + Bx + Cx(x ) + Dx(x ). We find tht the choices A =, B =, C =, D = do the job. Hence, Finlly, x + x(x ) 3 = x + (x ) 3 (x ) + x. x + x(x ) 3 = ( x + (x ) 3 (x ) + = ln x = ln Exmple.5 Evlute x ) (x ) + + ln(x ) + C x x x + x (x ) + C. x (x + )(x + 4). We will find numbers A, B, C, D such tht These numbers stisfy x (x + )(x + 4) = Ax + B x + + Cx + D x + 4. x (x + )(x + 4) = (Ax + B)(x + 4) + (Cx + D)(x + ) (x + )(x + 4) 7

18 or x = (A + C)x 3 + (B + D)x + (4A + C)x + 4B + D. The choice A = /3, B =, C = /3, D = does the job. Hence,.6 Numericl Integrtion x (x + )(x + 4) = x 3 ( x + x x + 4 ) = 6 ln + x x C. Theorem.5 (Mid-point Rule) Let [, b] be prtitioned by i = + i n (b ) for i running from to n. Then, b f(x) b n n i= f( i + i ). proof: This is just Riemnn sum in which we choose to evlute f t the midpoint of ech of the subintervls. Lemm.5 b f(x) f() + f(b) (b ). proof: If f(x) for x b, we pproximte the region bounded by the grph of f, the x-xis nd the lines y =, y = b by the trpezium whose vertices re (, f()), (, ), (b, ), (b, f(b)). The re of this trpezium is (b )(f() + f(b))/. In generl let K be number so tht f(x)+k for x b. According to the lst prgrph, b K)/. Therefore b (f(x) + K) is pproximted by (b )(f() + f(b) + f(x) is pproximted by (b )(f() + f(b))/. Theorem.53 (Trpezoidl Rule) Let [, b] be prtitioned by i = + i n (b ) for i running from to n. Then, proof: b f(x) b n [f( ) + f( ) f( n ) + f( n )]. 8

19 b f(x) = n i i= i f(x) n i= ( i i )(f( i ) + f( i )) by the lst lemm = n b i= n (f( i ) + f( i )) = b n [f() + n i= f( i) + f(b)]. Theorem.54 If f hs continuous second derivtive, there exists c in [, b] such tht b f(x) b n [f( (b )3 ) + f( ) f( n ) + f( n )] = n f (c). proof: We will work out the proof in cse [, b] is subdivided into one subintervl. b f(x) = (x + b )f(x) b b (x + b )f (x) = b (f() + f(b)) b (x + b )f (x) = b (f() + f(b)) [ (x + b (b ) ) ]f (x) b + b 8 [ (x + b ) = b (f() + f(b)) + b [ (x + b (b ) ) ]f (x) 8 Now, observe tht if M is the mximum of f over [, b], then so tht (b ) 3 b [ (x + b (b ) ) ]f (x) 8 M b [ (x + b (b ) ) ] 8 = M[ 6 (x + b (b )3 ) x] b 8 = M(b ) 3 /, b [ (x + b ) Similrly if m is the minimum of f over [, b], then m (b ) 3 b [ (x + b ) (b ) ]f (x) M. 8 (b ) ]f (x). 8 Apply the intermedite vlue theorem to the continuous function f, we see tht there is number c in [, b] such tht (b ) 3 b [ (x + b ) 9 (b ) ]f (x) = f (c). 8 (b ) ]f (x) 8

20 Consequently b f(x) b )3 (f() + f(b)) = (b f (c). In generl, we just prtition [, b] into subintervls nd pply the previous nlysis to the subintervls individully. Corollry.55 If f is concve upwrd, trpezoidl rule overestimtes integrls of f. If f is concve downwrd, trpezoidl rule underestimtes integrls of f. Exmple.56 Estimte the number of subintervls in which [, ] is subdivided, so tht the error of pproximting e x by trpezoidl rule is less thn.5. If f(x) = e x for ll x. Then f (x) = ( + x )e x for ll x. Thus, f (x) 6 for ll x. We mke N to be n integer so tht ( ) 3 N 6 <.5 or N > in order tht the error of pproximting e x by trpezoidl rule is less thn.5. Lemm.57 b proof: Observe tht the qudrtic function f(x) b + b [f() + 4f( 6 ) + f(b)]. f() f( + b [ ) + f(b) (b ) ](x + b f(b) f() ) + [ ](x + b b ) + f( + b ) + b + b evlutes t,, b to be f(), f( ) nd f(b) respectively. It is n pproximtion of f over the intervl [, b]. Thus b f(x) is pproximted by the integrl of this qudrtic function over [, b], which turns out to be b + b [f() + 4f( 6 ) + f(b)]. Theorem.58 (Simpson Rule) Let [, b] be prtitioned by i = + i n (b ) for i running from to n. Then b f(x) b 3 n [f()+4( )+f( )+4f( 3 )+f( 4 )+...4f( n )+f( n )].

21 proof: Apply the previous lemm, we obtin b f(x) = f(x) + 4 f(x) n n f(x) b 3 n [f( ) + 4f( ) + f( ) + f( ) + 4f( 3 ) + f( 4 ) f( n ) + 4f( n ) + f( n )] = b 3 n [f() + 4( ) + f( ) + 4f( 3 ) + f( 4 ) +...4f( n ) + f( n )]. Remrk.59 [, b] must be subdivided into n even number of subintervls in order to pply Simpson rule to b f(x). Theorem.6 Let [, b] be prtitioned by i = + i n (b ) for i running from to n. f is function which hs continuous forth derivtive so tht f (4) (x) M for ll x b. Then b f(x) b 6n [f() + 4( ) + f( ) + 4f( 3 ) + f( 4 ) +...4f( n ) + f( n ) M(b ) ( b 8 n )4. proof: omitted Exmple.6 Estimte the number of subintervls in which [, ] is subdivided, so tht the error of pproximting by Simpson rule is less thn e x.5. If f(x) = e x for ll x. Then f (4) (x) = (6x 4 48x + )e x so tht f (4) (x) 76 for ll x. We mke N to be n integer so tht ( ) 5 76 <.5 8N 4 or N in order tht the error of pproximting e x by Simpson rule is less thn.5..7 Improper Integrl Definition.6 Let f be continuous function defined on [, + ). Define the improper integrl to be + f(x) = M lim f(x). M + We lso sy tht the improper integrl + f(x) converges if the limit bove exists. Otherwise we sy tht this improper integrl diverges.

22 Remrk.63 Similrly, we define f(x) = + f( x) nd + f(x) = f(x) + + f(x) (here the choice of the intermedite point cn be modified without ffecting the sum of those two integrls on the right). Exmple.64 Evlute if it converges. + e x For ech positive number M, M e x = e x M = e M. So, M lim e x = lim M + M + e M =. We conclude tht the improper integrl converges nd its vlue is. + e x Exmple.65 Let p be number. Evlute if it converges. + x p For ech positive number M, M = { x p p x p M = p ( M p ) p ln x M = ln M p =

23 Hence, M lim M + x p = Consequently, the improper integrl + + p p p > x p converges only when p >. In this cse, its vlue is /(p ). Theorem.66 (Comprison Test) Let f nd g be functions such tht f(x) g(x) for x. If. If proof: omitted. + + f(x) converges, then g(x) diverges, then + + g(x) converges lso. f(x) diverges lso. Remrk.67 Comprison test remins vlid for the other two types of improper integrls. Exmple.68 Determine if converges. + e x Since the exponentil function is incresing, e x e x for x. We knew tht + e x converges. Therefore, + converges by comprison test. Finlly, e x converges lso. + e x = e x + + e x 3

24 Theorem.69 If f is function nd converges. So is + + f(x) f(x). proof: Let f + = mx{f, } nd f = mx{ f, }. Since f + f nd + f(x) converges by hypothesis, + converges by comprison test. Similrly, + f + (x) f (x) converges. Now, f = f + f nd both the integrls of f + nd f from to + converge. We conclude tht converges. + f(x) Exmple.7 Determine if converges. Since sin z for ll z, + sin x x sin x x x. We sw tht + converges. Hence, + x sin x x converges by comprison test. Consequently, + sin x x 4

25 Exmple.7 Let n be positive integer. Evlute if it converges. + x n e x For ech M >, M x n e x = x n e x M + n M x n e x (integrtion by prt) = M n e M + n M x n e x. Note tht lim M n e M = (why?). We tke limit s M tends to + on M + both sides of the eqution, we see tht for ech n, if + x n e x converges, so is + x n e x. Now, + e x converges by previous exmple. It implies tht + xe x converges. Use the result we hve just derived gin, + x e x converges. Repet this rgument for n times, we conclude tht converges. Hence, + + x n e x = n x n e x + So if I n = + x n e x, I n = ni n. Finlly, I n = ni n = n(n )I n... = n(n )(n )...()I = n!. x n e x. Definition.7 Suppose tht f is continuous function on (, b]. The improper integrl is defined to be b lim y + f(x) b y f(x). We sy tht such n improper integrl converges if the limit bove exists. Otherwise, we sy tht the improper integrl diverges. 5

26 Exmple.73 Evlute if it converges. x p For ech y >, y x p = p ( y p ) p ln y p = Its limit s y + exists only when p <. We see tht the improper integrl converges if p < (in this cse, the integrl is p ) nd it diverges if p. Definition.74 Suppose tht f is function continuous everywhere except t c. If < c < b, define the improper integrl of f from to b to be the sum of the improper integrls b f(x) = c x p f(x) + b c f(x). We sy tht such n improper integrl converges if both terms on the right hnd side converge. Exmple.75 Evlute if it converges. x Since x is discontinuous t. is defined s x x = x + x. Now, the integrl on the right hnd side diverges. We conclude tht the originl integrl diverges. Remrk.76 The comprison test remins vlid for this new kind of improper integrls. x 6

27 Exmple.77 Let n be positive integer. Evlute if it converges. x n For every x <, x n = x + x x + x x x n x n = ( x) + x( x) + x ( x) x n ( x) = x + x + x x n > x = n x, nd n x = n diverges. Therefore, diverges by the comprison test. x n du u Exmple.78 Evlute if it is convergent. π/ (csc x) 3/ Since sin x < x for ll < x < π/, (csc x) 3/ > > for ll < x < π/. x3/ But π/ diverges, x3/ so tht π/ (csc x) 3/ diverges by comprison. 7