11. Absolute and conditional convergence of improper integrals Cauchy criterion for the convergence of an improper integral 3.9A/00:00 (10:08) Theorem (Cauchy criterion for the convergence of an im- proper integral). Let the function f be defined on the interval [a; b) and there exists the R c R b integral a f(x) dx for any point c 2 (a; b). The improper integral a f(x) dx converges if and only if the following condition is satisfied: 8 " > 0 9 B 2 (a; b) 8 c0; c00; B < c0 < c00 < b; Z c00 f(x) dx < ". (1) c0 Proof. R c Let us introduce the auxiliary function Φ(c) = a f(x) dx. According to the definition of an improper integral, the convergence of the integral R b a f(x) dx is equivalent to the existence of the limit of the function Φ(c) as c ! b − 0. By virtue of the Cauchy criterion for the existence of a function limit, the limit of Φ(c) as c ! b − 0 exists if and only if the following condition is satisfied: 8 " > 0 9 B 2 (a; b) 8 c0; c00; B < c0 < c00 < b; jΦ(c00) − Φ(c0)j < ". (2) Let us transform the difference Φ(c00) − Φ(c0): Z c00 Z c0 Φ(c00) − Φ(c0) = f(x) dx − f(x) dx = a a Z c0 Z c00 Z c0 Z c00 = f(x) dx + f(x) dx − f(x) dx = f(x) dx. a c0 a c0 After substituting the found expression for the difference Φ(c00) − Φ(c0) into condition (2), we obtain condition (1). 2 Thus, condition (1) is equivalent to the existence of the limit limc!b−0 Φ(c) and the existence of this limit is equivalent to the convergence of the inte- R b gral a f(x) dx, therefore the condition (1) is necessary and sufficient for the convergence of this integral. Absolute convergence of improper integrals 3.9A/10:08 (06:46) Definition. Let the function f be defined on the interval [a; b). The improper integral R b R b a f(x) dx is called absolutely convergent if the integral a jf(x)j dx converges. Theorem (on the convergence of an absolutely convergent integral). R b If the improper integral a f(x) dx absolutely converges, then it converges. Remark. The converse is not true: we will show later that a convergent improper in- tegral is not necessarily absolutely convergent. Thus, the property of absolute convergence is stronger than the property of usual convergence. Proof. R b We are given that the integral a jf(x)j dx converges, and we need to prove R b that the integral a f(x) dx converges. R b Since the integral a jf(x)j dx converges, by the necessary condition of the Cauchy criterion for improper integrals, we get 8 " > 0 9 B 2 (a; b) 8 c0; c00; B < c0 < c00 < b; Z c00 jf(x)j dx < ". (3) c0 Since c0 < c00, the last inequality in condition (3) can be rewritten without specifying the external absolute value sign on the left-hand side: Z c00 jf(x)j dx < ". (4) c0 Recall the property of the integral of the absolute value of a function: Z c00 Z c00 f(x) dx ≤ jf(x)j dx. (5) c0 c0 Estimates (4) and (5) imply the following estimate: Z c00 f(x) dx < ". (6) c0 11. Absolute and conditional convergence of improper integrals 3 Thus, we can use (6) as the last inequality in condition (3): Z c00 0 00 0 00 8 " > 0 9 B 2 (a; b) 8 c ; c ; B < c < c < b; f(x) dx < ". c0 This means, due to the sufficient condition of the Cauchy criterion for R b improper integrals, that the integral a f(x) dx converges. Properties of improper integrals of non-negative functions Criterion for the convergence of improper integrals of non-negative functions 3.9A/16:54 (10:00) In this section, we consider improper integrals of non-negative functions. Since the absolute value of the function is non-negative, all the results ob- tained in this section can also be used to study the absolute convergence of improper integrals of functions taking both negative and positive values. Theorem (criterion for the convergence of improper inte- grals of non-negative functions). Let a function f be defined on [a; b) and f(x) ≥ 0 for any value x 2 [a; b). R c Suppose that, for any c 2 [a; b), there exists an integral a f(x) dx. Then the R b improper integral a f(x) dx converges if and only if the set of values of all R c integrals a f(x) dx is bounded from above: Z c 9 M > 0 8 c 2 [a; b) f(x) dx ≤ M. (7) a Proof. R c We introduce an auxiliary function Φ(c) = a f(x) dx. The inequality f(x) ≥ 0, which holds, by condition, for all x 2 [a; b), 00 R c 0 00 0 00 implies the inequality c0 f(x) dx ≥ 0 for any c ; c 2 [a; b) such that c < c . Therefore, for c0 < c00, we have Z c00 Z c0 Z c00 Φ(c00) = f(x) dx = f(x) dx + f(x) dx = a a c0 Z c00 = Φ(c0) + f(x) dx ≥ Φ(c0). c0 We obtain that, for all c0 < c00, the estimate Φ(c0) ≤ Φ(c00) is true. This means that the function Φ(c) is non-decreasing on the interval [a; b). 4 1. Sufficiency. Given: condition (7) is satisfied. Prove: the integral R b a f(x) dx converges. Condition (7) means that the function Φ(c) is bounded from above on the interval [a; b). Thus, the function Φ(c) is non-decreasing and bounded from above on the interval [a; b), therefore, by virtue of the theorem on the limit of a monotonous and upper-bounded function, there exists a limit of the function Φ(c) as c ! b − 0 (equal to supc2[a;b) Φ(c)). It remains to note that the existence of this limit is equivalent to the convergence of the improper R b integral a f(x) dx. The sufficiency is proven. R b 2. Necessity. Given: the integral a f(x) dx converges. Prove: condi- tion (7) is satisfied. R b As noted above, if the integral a f(x) dx converges, then there exists a limit limc!b−0 Φ(c) = M. Since the function Φ(c) is non-decreasing, we obtain that, for all c0; c00 2 [a; b) such that c0 < c00, the inequality holds: Φ(c0) ≤ Φ(c00). In this inequality, we pass to the limit as c00 ! b − 0. By virtue of the theorem on passing to the limit in the inequalities, the following inequality holds for any c0 2 [a; b): Φ(c0) ≤ M. Substituting the definition of the function Φ into this inequality, we obtain the inequality from condition (7). The comparison test 3.9A/26:54 (10:35) Theorem (the comparison test for improper integrals of non-negative functions). Let the functions f and g be defined on the interval [a; b) and the double inequality 0 ≤ f(x) ≤ g(x) holds for any x 2 [a; b). Suppose that, for any R c R c c 2 [a; b), there exist integrals a f(x) dx and a g(x) dx. Then the following two statements are true. R b 1. If the improper integral a g(x) dx converges, then the integral R b a f(x) dx also converges. R b R b 2. If the improper integral a f(x) dx diverges, then the integral a g(x) dx also diverges. Proof. R b 1. Let the integral a g(x) dx converge. Then, by the necessary condition of the previous criterion, we obtain 11. Absolute and conditional convergence of improper integrals 5 Z c 9 M > 0 8 c 2 [a; b) g(x) dx ≤ M. (8) a Since f(x) ≤ g(x) for all x 2 [a; b), a similar inequality holds for the proper integrals: Z c Z c f(x) dx ≤ g(x) dx. (9) a a Combining estimates (8) and (9), we obtain Z c Z c f(x) dx ≤ g(x) dx ≤ M. a a R c Thus, condition (8) is also satisfied for the integral a f(x) dx. Therefore, R b by virtue of a sufficient part of the previous criterion, the integral a f(x) dx converges. R b 2. Let the integral a f(x) dx diverge. R b If we assume that the integral a g(x) dx converges, then by the already R b proved statement 1, the integral a f(x) dx should also converge. But this contradicts the condition. Therefore, the assumption made is false and the R b integral a g(x) dx diverges. Remark. Obviously, the theorem remains valid if the functions f and g satisfy the double inequality 0 ≤ f(x) ≤ Cg(x) with some constant C > 0 on the interval [a; b). Corollary of the comparison test 3.9A/37:29 (04:25) Corollary. Let the functions f and g be defined on the interval [a; b) and be non- negative on this interval. Let f(x) ∼ g(x) as x ! b − 0. Then the integrals R b R b a f(x) dx and a g(x) dx either both converge or both diverge.