Improper Integrals R b In this section, we will extend the concept of the definite integral a f (x)dx to functions with an infinite discontinuity and to infinite intervals. I That is integrals of the type Z 1 1 Z 1 1 Z 1 1 A) 3 dx B) 3 dx C) 2 1 x 0 x −∞ 4 + x 1 I Note that the function f (x) = x3 has a discontinuity at x = 0 and the F.T.C. does not apply to B. I Note that the limits of integration for integrals A and C describe intervals that are infinite in length and the F.T.C. does not apply. Annette Pilkington Improper Integrals Infinite Intervals An Improper Integral of Type 1 R t (a) If a f (x)dx exists for every number t ≥ a, then Z 1 Z t f (x)dx = lim f (x)dx a t!1 a provided that limit exists and is finite. R b (c) If t f (x)dx exists for every number t ≤ b, then Z b Z b f (x)dx = lim f (x)dx −∞ t→−∞ t provided that limit exists and is finite. R 1 R b The improper integrals a f (x)dx and −∞ f (x)dx are called Convergent if the corresponding limit exists and is finite and divergent if the limit does not exists. R 1 R a (c) If (for any value of a) both a f (x)dx and −∞ f (x)dx are convergent, then we define Z 1 Z a Z 1 f (x)dx = f (x)dx + f (x)dx −∞ −∞ a If f (x) ≥ 0, we can give theAnnette definite Pilkington integralImproper above Integrals an area interpretation. Infinite Intervals Z 1 Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t!1 a −∞ t→−∞ t If f (x) ≥ 0, we can give the definite integral above an area interpretation; namely that if the improper integral converges, the area under the curve on the infinite interval is finite. Example Determine whether the following integrals converge or diverge: Z 1 1 Z 1 1 dx; 3 dx; 1 x 1 x R 1 1 R t I By definition 1 x dx = limt!1 1 1=x dx ˛t I = limt!1 ln x˛ = limt!1(ln t − ln 1) ˛1 I = limt!1 ln t = 1 R 1 1 I The integral 1 x dx diverges. Annette Pilkington Improper Integrals Infinite Intervals Z 1 Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t!1 a −∞ t→−∞ t Example Determine whether the following integrals converge or diverge: Z 1 1 Z 1 1 dx; 3 dx; 1 x 1 x R 1 1 R t 1 I By definition 1 x3 dx = limt!1 1 x3 dx −2 ˛t x ˛ I = limt!1 −2 ˛1 “ −1 1 ” I = limt!1 2t2 + 2 1 1 I = 0 + 2 = 2 . R 1 1 1 I The integral 1 x3 dx converges to 2 . Annette Pilkington Improper Integrals Area Interpretation 1.0 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 5 10 15 20 25 30 5 10 15 20 25 30 R 1 1 Since 1 x dx diverges, the area under the curve y = 1=x on the interval [1; 1) (shown on the left above) is not finite. R 1 1 3 Since 1 x3 dx converges, the area under the curve y = 1=x on the interval [1; 1) (shown on the right above) is finite. Annette Pilkington Improper Integrals Infinite Intervals Z 1 Z t Z b Z b f (x)dx = lim f (x)dx f (x)dx = lim f (x)dx a t!1 a −∞ t→−∞ t Example Determine whether the following integral converges or diverges: R 0 x −∞ e dx R 0 x R 0 x I By definition −∞ e dx = limt→−∞ t e dx 0 x ˛ I = limt→−∞ e ˛ ˛t 0 t I = limt→−∞(e − e ) I = 1 − 0 = 1. R 0 x I The integral −∞ e dx converges to 1. Annette Pilkington Improper Integrals Infinite Intervals R 1 R a If (for any value of a) both a f (x)dx and −∞ f (x)dx are convergent, then we define Z 1 Z a Z 1 f (x)dx = f (x)dx + f (x)dx −∞ −∞ a Example Determine whether the following integral converges or diverges: Z 1 1 dx −∞ 4 + x2 R 1 1 I Since this is a continuous function, we can calculate 0 4+x2 dx and R 0 1 −∞ 4+x2 dx ˛t R 1 1 R t 1 1 −1 x ˛ I 0 4+x2 dx = limt!1 0 4+x2 dx = limt!1 2 tan 2 ˛0 1 −1 t −1 1 π π I = limt!1 2 (tan 2 − tan 0) = 2 ( 2 − 0) = 4 : ˛0 R 0 1 R 0 1 1 −1 x ˛ I −∞ 4+x2 dx = limt→−∞ t 4+x2 dx = limt→−∞ 2 (tan 2 ) ˛t 1 −1 −1 t 1 (−π) π I = limt→−∞ 2 (tan 0 − tan 2 ) = 2 (0 − 2 ) = 4 : R 1 1 I The integral −∞ 4+x2 dx converges and is equal to Z 0 1 Z 1 1 π π π dx + dx = + = : −∞ 4 + x2 0 4 + x2 4 4 2 Annette Pilkington Improper Integrals Infinite Intervals Theorem Z 1 1 p dx is convergent if p > 1 and divergent if p ≤ 1 1 x Proof I We've verified this for p = 1 above. If p 6= 1 1−p ˛t “ 1−p ” R 1 1 R t 1 x ˛ t 1 I 1 xp dx = limt!1 1 xp dx = limt!1 1−p = limt!1 1−p − 1−p ˛1 “ t1−p 1 ” 1 I If p > 1, limt!1 1−p − 1−p = − 1−p and the integral converges. “ t1−p 1 ” t1−p I If p < 1, limt!1 1−p − 1−p does not exist since 1−p ! 1 as t ! 1 and the integral diverges. Annette Pilkington Improper Integrals Functions with infinite discontinuities Improper integrals of Type 2 (a) If f is continuous on [a; b) and is discontinuous at b, then Z b Z t f (x)dx = lim f (x)dx a t!b− a if that limit exists and is finite. (b) If f is continuous on (a; b] and is discontinuous at a, then Z b Z b f (x)dx = lim f (x)dx a t!a+ t if that limit exists and is finite. R b The improper integral a f (x)dx is called convergent if the corresponding limit exists and Divergent if the limit does not exist. R c (c) If f has a discontinuity at c, where a < c < b, and both a f (x)dx and R b c f (x)dx are convergent, then we define Z b Z c Z b f (x)dx = f (x)dx + f (x)dx a a c Annette Pilkington Improper Integrals Functions with infinite discontinuities If f is continuous on [a; b) and is discontinuous at b, then Z b Z t f (x)dx = lim f (x)dx a t!b− a if that limit exists and is finite. Example Determine whether the following integral converges or diverges Z 2 1 dx 0 x − 2 1 I The function f (x) = x−2 is continuous on [0; 2) and is discontinuous at 2. Therefore, we can calculate the integral. ˛t R 2 1 R t 1 ˛ I 0 x−2 dx = limt!2− 0 x−2 dx = limt!2− ln jx − 2j ˛0 I = limt!2− (ln jt − 2j − ln j − 2j) which does not exist since ln jt − 2j ! −∞ as t ! 2−. R 2 1 I Therefore the improper integral 0 x−2 dx diverges. Annette Pilkington Improper Integrals Functions with infinite discontinuities If f is continuous on (a; b] and is discontinuous at a, then Z b Z b f (x)dx = lim f (x)dx a t!a+ t if that limit exists and is finite. Example Determine whether the following integral converges or diverges Z 1 1 dx 0 x2 1 I The function f (x) = x2 is continuous on (0; 1] and is discontinuous at 0. Therefore, we can calculate the integral. 1 R 1 1 R 1 1 −1 ˛ + + ˛ I 0 x2 dx = limt!0 t x2 dx = limt!0 x ˛t (−1) (1) I = limt!0+ (−1 − t ) == limt!0+ ( t − 1) which does not exist since 1 + t ! 1 as t ! 0 . R 1 1 I Therefore the improper integral 0 x2 dx diverges. Annette Pilkington Improper Integrals Functions with infinite discontinuities Theorem It is not difficult to show that Z 1 1 p dx is divergent if p ≥ 1 and convergent if p < 1 0 x Annette Pilkington Improper Integrals Functions with infinite discontinuities R c If f has a discontinuity at c, where a < c < b, and both a f (x)dx and R b c f (x)dx are convergent, then we define Z b Z c Z b f (x)dx = f (x)dx + f (x)dx a a c Example determine if the following integral converges or diverges and if it converges find its value. Z 4 1 dx 0 (x − 2)2 1 I The function (x−2)2 has a discontinuity at x = 2. Therefore we must R 2 1 R 4 1 check if both improper integrals 0 (x−2)2 dx and 2 (x−2)2 dx converge or diverge.