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Improper Integrals

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Improper Integrals Improper Integrals. The concept of Riemann integrals as developed in previous chapter requires that the range of integration is finite and the integrand remains bounded on that domain. if either (or both) of these assumptions is not satisfied it is necessary to attach a new interpretation to the integral Definition 3.1. In case the integrand f becomes infinite in the interval a ≤ 푥 ≤ 푏, That is f has points of infinite discontinuity (singular points) in [a, b] or the limits of integration a or b (or both becomes infinite, the 푏 symbol ( ) ∫푎 푓푑푥 is called an improper or infinite or generalised integral. Thus, ∞ 푑푥 ∞ 푑푥 1 푑푥 ∞ 푑푥 ∫ , ∫ . ∫ , ∫ 1 푥2 −∞ 1+푥2 0 푥(1−푥) −1 푥(1−푥) are examples of improper integrals. The integrals which are not improper are called proper integral , thus 1 푠푖푛푥 ∫ 푑푥 is a proper integral. 0 푥 Integration of Unbounded Function with finite limits of integration. Definition 3.2. Let a function f be defined in a interval [a, b] everywhere except possible at finite number of points. (i) Convergence at left –end. Let a be the only points of infinite discontinuity of f so that according to assumption made in the last section, the integral 푏 exists ∫푎+휆 푓푑푥 ∀ 휆, 0 < 휆 < 푏 − 푎. 2 푏 is defined as the 푇ℎ푒 푖푚푝푟표푝푒푟 푖푛푡푒푔푟푎푙 ∫푎 푓푑푥 푏 푙푖푚 ∫ 푓푑푥 so that, 휆→0+ 푎+휆 푏 푏 ∫ 푓푑푥 = 푙푖푚 ∫ 푓푑푥 . 푎 휆→0 푎+휆 푏 Limit exists and is finite, the improper integral is said If this ∫푎 푓푑푥 to converge at (a) if otherwise, it is called divergent. 퐍퐨퐭퐞. For any c , 푎 < 푐 < 푏 푏 푐 푏 + . ∫푎 푓푑푥 = ∫푎 푓푑푥 ∫푐 푓푑푥 Then, 푏 푐 ∫푎 푓푑푥 푎푛푑 ∫푎 푓푑푥 푏 converges and diverges together and is proper. ∫푐 푓푑푥 (ii) Convergence at right-end. Let b be the only point of infinite discontinuity the improper integral is then defined by the relation 푏 푏−휇 ∫ 푓푑푥 = lim ∫ 푓푑푥 0< 휇 < 푏 − 푎 . 푎 휇→0+ 푎 If the limit exists, the improper integral is said to be convergent at 푏. Otherwise is called divergent. 퐍퐨퐭퐞 ∶ For the same reason as above, 푏 푏 and converges and diverges together . ∫푐 푓푑푥 ∫푎 푓푑푥 ∀ 푐, 푎 < 푐 < 푏 (iii) Convergence at both the end points. If the end points a and b are the only points of infinite discontinuity of f , then for any point c, a < c < b, 푏 푐 푏 ∫푎 푓푑푥 = ∫푎 푓푑푥 + ∫푐 푓푑푥 If both the integrals are convergent as by case (i) and (ii),then 3 푏 is convergent , otherwise it is divergent. The improper integral ∫푎 푓푑푥 is also defined as: 푏 푏−휇 ∫ 푓푑푥 = 푙푖푚 ∫ 푓푑푥. 푎 휆→0+ 푎+휆 휇→0+ The improper integral exists if the limit exists. (iv) Convergence at Interior points. If an interior point c, a < c < b, is the only point of infinite discontinuity of f, we get 푏 푐 푏 (1) ∫푎 푓푑푥 = ∫푎 푓푑푥 + ∫푐 푓푑푥 푏 exists of the both integral on R.H.S of the improper integral ∫푎 푓푑푥 (1) are exists. Example 3.1. Examine the convergence of: 1 푑푥 1 푑푥 2 푑푥 (푖) ∫ (ii) ∫ (iii) ∫ . 0 푥2 0 √푖−푥 6 2푥−푥2 (i) 0 is the point of infinite discontinuity of integrand [0, 1]. Thus, 1 푑푥 1 푑푥 ∫ = 푙푖푚 ∫ , o < 휆 < 1 0 푥2 휆→0+ 휆 푥2 1 = 푙푖푚 ( − 1) = ∞ 휆→0+ 휆 Thus the proper integral is divergent. (ii) Home Assignment. Home Assignment 4 풃 Comparison Tests for Convergence At ‘a’ of . ∫풂 풇풅풙 Theorem 3.1. A necessary and sufficient condition for the 푏 convergence of the improper integral at ‘a’ where f is positive in ∫푎 푓푑푥 [a, b]. This is , ∃ a psoitive number 푀, independent of 휆 , such that 푏 < M, 0< . ∫푎+휆 푓푑푥 휆 < 푏 − 푎 푏 Proof. We know that the improper integral ‘a’ if for 0 ∫푎 푓푑푥 converges at 푏 + 0 < tends to finite limit as 휆 < 푏 − 푎, ∫푎+휆 푓푑푥 휆 → 0 . 푏 Since f is positive in [ is 푎 + 휆, 푏], the positive function of 휆 , ∫푎+휆 푓푑푥 monotonic incereasing as λ , decreases and will therefore tend to a finite limit iff it is bounded above , This is, ∃ a positive number 푀 independent of 휆, such that 푏 < M, 0 . ∫푎+휆 푓푑푥 < 휆 < 푏 − 푎 Hence the theorem is proved. Note. If no such number M exists, the monotonic increasing function 푏 + is not bounded above and therefore tend to + and ∫푎+휆 푓푑푥 ∞ as λ → 0 , 푏 hence the improper integral diverges to + ∫푎 푓푑푥 ∞. Comparison Test. Theorem 3.2: If f and g are two positive functions and ‘a’ is only singular point of f and g on [a. b], such that f(x) ≤ 푔(푥), 푓표푟 푎푙푙 푥 ∈ [푎, 푏] 푏 푏 ( ) converges . i ∫푎 푓푑푥 converges, if ∫푎 푔푑푥 5 푏 푏 diverges, if converges . (ii) ∫푎 푔푑푥 ∫푎 푓푑푥 Proof. Since f and g are two positive functions on [a, b] and ‘a’ is only singular point of f and g. Therefore 푓 and 푔 are bound in [푎 + 휆, 푏], for all 0 < 휆 < 푏 − 푎. Also Since, f(x) ≤ 푔(푥), 푓표푟 푎푙푙 푥 ∈ [푎, 푏], implies 푏 푏 (i) ∫푎+휆 푓푑푥 ≤ ∫푎+휆 푔푑푥 푏 (1) Suppose be convergent, so that for all ∫푎 푔푑푥 ∃ 푚 > 0 , 푠푢푐ℎ 푡ℎ푎푡 휆, 0 < 휆 < 푏 − 푎, 푏 < m . ∫푎+휆 푓푑푥 From (i) we have 푏 < m , for all ∫푎+휆 푓푑푥 휆 , 0 < 휆 < 푏 − 푎 . 푏 Hence is convergent. ∫푎 푓푑푥 푏 (2) Now suppose is divergent then the positive function ∫푎 푓푑푥 푏 is not bounded above. ∫푎+휆 푓푑푥 푏 Therefore from (i) it follows that the positive function is not ∫푎+휆 푔푑푥 bounded above. 푏 Hence is divergent. This completes the Theorem. ∫푎+휆 푔푑푥 Comparison Test (limit form). Theorem 3.3. If f and g are two positive functions [a, b] and ‘a’ is the only singular point of f and g in [a, b], such that 푓(푥) 푙푖푚 = l where ‘l’ is a non – zero finite number. 푥→푎+ 푔(푥) 6 푏 푏 Then, the two integrals converges and diverges ∫푎 푓푑푥 and ∫푎 푔푑푥 together at ‘a’. Proof. Evidently, 1 > 0. Let 휀 be positive number such that 1 − 휀 > 0 . 푓(푥) Since, 푙푖푚 = 1. 푥→푎+ 푔(푥) Therefore there exists a nbd of ] a, c [ , a < c < b , such that for all x ∈ ]푎, 푐[ 푓(푥) | − 푙| < 휀 푔(푥) or (푙 − 휀)푔(푥) < 푓(푥) < (푙 + 휀)푔(푥) . This implies that (푙 − 휀)푔(푥) < 푓(푥) (2) and 푓(푥) < (푙 + 휀)푔(푥) (3) ∀ 푥 ∈]푎, 푐[ 푏 If converges, then from (i) ∫푎 푓푑푥 푏 also converges at . ∫푎 푔(푥)푑푥 푎 푏 If t (ii) ∫푎 푓푑푥 diverges , hen from 푏 . ∫푎 푓푑푥 diverges at 푎 푓(푥) 푏 If in the above Theorem, 퐿푖푚 → 0 and ∫ 푔푑푥 푐표푛푣푒푟푔푒푠, then 푥→푎+ 푔(푥) 푎 7 푏 converges and if ∫푎 푓푑푥 푓(푥) 푏 푏 퐿푖푚 → ∞ and ∫ 푔푑푥 diverges, then, ∫ 푓푑푥 also diverges. 푥→푎+ 푔(푥) 푎 푎 Useful Comparison Integral. 푏 푑푥 Theorem 3.4. The improper integral ∫ Converges if and only 푎 (푥−푎)푛 if n < 1. Proof. It is proper integral if n ≤ 0 and improper for all other values of n, ‘푎’ being only singular point of the integrand. Now for n ≠ 1 푏 푑푥 푏 푑푥 ∫ = 푙푖푚 ∫ , 푎 (푥−푎)푛 휆→0+ 푎+휆 (푥−푎)푛 1 −푛+1 −푛+1 = 푙푖푚 [(푏 − 푎) -휆 ] 휆→0+ −푛+1 1 [(푏 − 푎)−푛+1, 푖푓 푛 < 1 −푛+1 = { ∞ 푖푓 푛 > 1 . Also for n = 1 푏 푑푥 퐿푖푚 ∫ = 푙푖푚 [푙표푔(푏 − 푎) − 푙표푔휆] = ∞. 휆→0+ 푎+휆 푥−푎 휆→0+ 푏 푑푥 Thus, ∫ converges for n < 1 . 푎 (푥−푎)푛 푏 푑푥 Note. A similar result holds for convergence of ∫ 푎푡 푏. 푎 (푏−푥) Example 3.2. Test the convergence of 1 푑푥 휋/2 푠푖푛푥 (i) ∫ (ii) ∫ dx 0 √1−푥3 0 푥푝 Solution. Let f(x) = 1 √1−푥3 8 1 = √(1−푥)(1+푥+푥2) 1 1 = 1 . 1 (1+푥+푥2)2 (1−푥)2 1 Clearly, 1 is a bounded function. (1+푥+푥2)2 Let M be its upper bound, then , 1 1 푀 1 . 1 ≤ 1 , x ∈ [1, 0]. (1+푥+푥2)2 (1−푥)2 (1−푥)2 1 푚푑푥 1 Also since ∫ is convergent as n = < 1. 0 1 2 (1−푥)2 1 Therefore, is convergent. √1−푥3 (i) For 푝 ≤ 1, it is a proper integral for 푝 > 1 , it is an improper integral 0 being the point of infinite discontinuity 푠푖푛푥 1 푠푖푛푥 Now = ( ) 푥푝 푥푝−1 푥 푠푖푛푥 푠푖푛푥 The function is bounded and ≤ 1. 푥 푥 푠푖푛푥 1 Therefore, ≤ 푥푝 푥푝−1 휋/2 푑푥 Also ∫ converges only if 푝 − 1 < 1 or 푝 < 2 . 0 푥푝−1 휋 푠푖푛푥 Therefore by comparison test ∫2 dx converges for p < 2 and 0 푥푝 diverges for p ≥ 2. Note. If 푙푖푚 [(푥 − 푎)푛푓 (푥)] exists and is non- zero finite, then, the 푥→0+ 푏 integral ∫푎 푓푑푥 convergs iff 푛 < 1. Example 3.3. Find the values of 푚 and 푛 for which the following integrals converges. 9 1 −푚푥 푛 (i) dx . ∫0 푒 푥 1 1 (ii) ∫ (푙표푔 )푚dx . 0 푥 Solution (i) Let 푘 be positive number greater than 1, Then, 푒−푚푥푥푛 ≤ k푥푛, ∀ x ∈ [0 ,1] and m ; 1 1 푑푥 Also ∫ 푥푛 = ∫ converges for – 푛 < 1 , 0 0 푥−푛 that is, n > -1 only .
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