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Thenby definition b ωf (x) dx := limf (x) dx,(6.71)ab→ω aif this limit exists as b → ω, b ∈ [a, ω[.From now on, unless otherwise stated, when studying the improper integral(6.71) we shall assume that the integrand satisfies the hypotheses of Definition 3.Moreover, for the sake of definiteness we shall assume that the singularity (“impropriety”) of the integral arises from only the upper limit of integration. The studyof the case when it arises from the lower limit is carried out word for word in exactlythe same way.From Definition 3, properties of the integral, and properties of the limit, one candraw the following conclusions about properties of an improper integral.3966IntegrationProposition 1 Suppose x → f (x) and x → g(x) are functions defined on an interval [a, ω[ and integrable on every closed interval [a, b] ⊂ [a, ω[.
Suppose theimproper integrals ωf (x) dx,(6.72)aωg(x) dx(6.73)aare defined.Then a) if ω ∈ R and f ∈ R[a, ω], the values of the integral (6.72) are the same,whether it is interpreted as a proper or an improper integral;b) for any λ1 , λ2 ∈ R the function (λ1 f + λ2 g)(x) is integrable in the impropersense on [a, ω[ and the following equality holds: ω ω ω(λ1 f + λ2 g)(x) dx = λ1f (x) dx + λ2g(x) dx;aac) if c ∈ [a, ω[, then ωcf (x) dx =aaωf (x) dx +af (x) dx;cd) if ϕ : [α, γ [ →[a, ω[ is a smooth strictly monotonic mapping with ϕ(α) = aand ϕ(β) → ω as β → γ , β ∈ [α, γ [, then the improper integral of the functiont → (f ◦ ϕ)(t)ϕ (t) over [α, γ [ exists and the following equality holds: γ ωf (x) dx =(f ◦ ϕ)(t)ϕ (t) dt.aαProof Part a) follows from the continuity of the functionbF(b) =f (x) dxaon the closed interval [a, ω] on which f ∈ R[a, ω].Part b) follows from the fact that for b ∈ [a, ω[b(λ1 f + λ2 g)(x) dx = λ1abf (x) dx + λ2abg(x) dx.aPart c) follows from the equalitybf (x) dx =awhich holds for all b, c ∈ [a, ω[.acf (x) dx +bf (x) dx,c6.5 Improper Integrals397Part d) follows from the formula for change of variable in the definite integral: β b=ϕ(β)f (x) dx =(f ◦ ϕ)(t)ϕ (t) dt.a=ϕ(α)αRemark 1 To the properties of the improper integral expressed in Proposition 1 weshould add the very useful rule for integration by parts in an improper integral,which we give in the following formulation:If f, g ∈ C (1) [a, ω[ and the limit limgfx→ω (fx∈[a,ω[· g)(x) exists, then the functionsf · and· g are either both integrable or both nonintegrable in the impropersense on [a, ω[, and when they are integrable the following equality holds: ω ωω f · g (x) dx = (f · g)(x) a −f · g (x) dx,awhereaω(f · g)(x)a = lim (f · g)(x) − (f · g)(a).x→ωx∈[a,ω[Proof This follows from the formula b bb f · g (x) dx = (f · g)α −f · g (x) dxaafor integration by parts in a proper integral.Remark 2 It is clear from part c) of Proposition 1 that the improper integrals ω ωf (x) dx andf (x) dxaceither both converge or both diverge.
Thus, in improper integrals, as in series, convergence is independent of any initial piece of the series or integral.For that reason, when posing the question of convergence of an improper integral,we sometimes omit entirely the limit of integration at which the integral does nothave a singularity.With that convention the results obtained in Examples 1 and 2 can be rewrittenas follows: +∞ dxconverges only for α > 1;the integral dxx αthe integral +0 x α converges only for α < 1.The sign +0 in the last integral shows that the region of integration is containedin x > 0.of variable in this last integral, we immediately find that the integral By a changedxconvergesonly for α < 1.αx0 +0 (x−x0 )3986Integration6.5.2 Convergence of an Improper Integrala.
The Cauchy CriterionBy Definition 3, the convergence of the improper integral (6.71) is equivalent to theexistence of a limit for the function bF(b) =f (x) dx(6.74)aas b → ω, b ∈ [a, ω[.This relation is the reason why the following proposition holds.Proposition 2 (Cauchy criterion for convergence of an improper integral) If thefunction x → f (x) is defined on the interval ω[a, ω[ and integrable on every closedinterval [a, b] ⊂ [a, ω[, then the integral a f (x) dx converges if and only if forevery ε > 0 there exists B ∈ [a, ω[ such that the relation b 2<εf(x)dxb1holds for any b1 , b2 ∈ [a, ω[ satisfying B < b1 and B < b2 .Proof As a matter of fact, we haveb2b1b2f (x) dx =af (x) dx −b1f (x) dx = F(b2 ) − F(b1 ),aand therefore the condition is simply the Cauchy criterion for the existence of a limitfor the function F(b) as b → ω, b ∈ [a, ω[.b.
Absolute Convergence of an Improper IntegralωDefinition4 The improper integral a f (x) dx converges absolutely if the integralωa |f |(x) dx converges.Because of the inequality b b 2 2≤f(x)dx|f|(x)dx b1b1and Proposition 2, we can conclude that if an integral converges absolutely, then itconverges.The study of absolute convergence reduces to the study of convergence of integrals of nonnegative functions. But in this case we have the following proposition.6.5 Improper Integrals399Proposition 3 If a function f satisfies the hypotheses of Definition 3 and f (x) ≥ 0on [a, ω[, then the improper integral (6.71) exists if and only if the function (6.74)is bounded on [a, ω[.Proof Indeed, if f (x) ≥ 0 on [a, ω[, then the function (6.74) is nondecreasing on[a, ω[, and therefore it has a limit as b → ω, b ∈ [a, ω[, if and only if it is bounded.
As an example of the use of this proposition, we consider the following corollaryof it.Corollary 1 (Integral test for convergence of a series) If the function x → f (x)is defined on the interval [1, +∞[, nonnegative, nonincreasing, and integrable oneach closed interval [1, b] ⊂ [1, +∞[, then the series∞f (n) = f (1) + f (2) + · · ·n=1and the integral+∞f (x) dx1either both converge or both diverge.Proof It follows from the hypotheses that the inequalities n+1f (n + 1) ≤f (x) dx ≤ f (n)nhold for any n ∈ N. After summing these inequalities, we obtainkn=1k+1f (n + 1) ≤f (x) dx ≤1kf (n)n=1orsk+1 − f (1) ≤ F(k + 1) ≤ sk ,bwhere sk = n=1 f (n) and F(b) = 1 f (x) dx.
Since sk and F(b) are nondecreasing functions of their arguments, these inequalities prove the proposition.kIn particular, one can say that the result of Example 1 is equivalent to the assertionthat the series∞1n=1nαconverges only for α > 1.The most frequently used corollary of Proposition 3 is the following theorem.4006IntegrationTheorem 1 (Comparison theorem) Suppose the functions x → f (x) and x → g(x)are defined on the interval [a, ω[ and integrable on any closed interval [a, b] ⊂[a, ω[.If0 ≤ f (x) ≤ g(x)on [a, ω[, then convergence of the integral (6.73) implies convergence of (6.72) andthe inequality ω ωf (x) dx ≤g(x) dxaaholds.
Divergence of the integral (6.72) implies divergence of (6.73).Proof From the hypotheses of the theorem and the inequalities for proper Riemannintegrals we haveF(b) =bf (x) dx ≤abg(x) dx = G(b)afor any b ∈ [a, ω[. Since both functions F and G are nondecreasing on [a, ω[, thetheorem follows from this inequality and Proposition 3.Remark 3 If instead of satisfying the inequalities 0 ≤ f (x) ≤ g(x) the functions fand g in the theorem are known to be nonnegative and of the same order as x → ω,x ∈ [a, ω[, that is, there are positive constants c1 and c2 such thatc1 f (x) ≤ g(x) ≤ c2 f (x),then by the linearity of the improper integral and theorem just proved, in this casewe can conclude that the integrals (6.72) and (6.73) either both converge or bothdiverge.Example 4 The integralconverges, since+∞√x dx√1 + x4√x1∼ 3/2√1 + x4 xas x → +∞.Example 5 The integral1+∞cos xdxx26.5 Improper Integrals401converges absolutely, since cos x 1 x2 ≤ x2for x ≥ 1.
Consequently,+∞1 +∞ +∞ cos x cos x 1 dx ≤≤dxdx = 1.22xxx211Example 6 The integral+∞e−x dx21−x 2converges, since e< e−x for x > 1 and+∞e−x dx <21+∞11e−x dx = .eExample 7 The integral+∞dxln xdiverges, since11>ln x xfor sufficiently large values of x.Example 8 The Euler integralπ/2ln sin x dx0converges, since1| ln sin x| ∼ | ln x| < √xas x → +0.Example 9 The elliptic integral01#dx(1 − x 2 )(1 − k 2 x 2 )4026Integrationconverges for 0 ≤ k 2 < 1, since 1 − x 2 1 − k 2 x 2 ∼ 2 1 − k 2 (1 − x)1/2as x → 1 − 0.Example 10 The integralϕ0dθ√cos θ − cos ϕconverges, since#&cos θ − cos ϕ =2 sinϕ−θ #ϕ+θsin∼ sin ϕ(ϕ − θ )1/222as θ → ϕ − 0.Example 11 The integral'LT =2gϕ00dψsin2 ϕ20sin2ϕ2converges for 0 < ϕ0 < π since as ψ → ϕ0 − 0 we have&ϕ0ψ #− sin2 ∼ sin ϕ0 (ϕ0 − ψ)1/2 .sin222(6.75)(6.76)Relation (6.75) expresses the dependence of the period of oscillations of a pendulum on its length L and its initial angle of displacement, measured from the radius itoccupies at the lowest point of its trajectory.
Formula (6.75) is an elementary versionof formula (6.65) of the preceding section.A pendulum can be thought of, for example, as consisting of a weightless rod,one end of which is attached by a hinge while the other end, to which a point massis attached, is free.In that case one can speak of arbitrary initial angles ϕ0 ∈ [0, π]. For ϕ0 = 0 andϕ0 = π , the pendulum will not oscillate at all, being in a state of stable equilibriumin the first case and unstable equilibrium in the second.It is interesting to note that (6.75) and (6.76) easily imply that T → ∞ as ϕ0 →π − 0, that is, the period of oscillation of a pendulum increases without bound as itsinitial position approaches the upper (unstable) equilibrium position.c. Conditional Convergence of an Improper IntegralDefinition 5 If an improper integral converges but not absolutely, we say that itconverges conditionally.6.5 Improper Integrals403Example 12 Using Remark 1, by the formula for integration by parts in an improperintegral, we find that+∞π/2 +∞ +∞sin xcos xcos xcos x +∞−dx=−dx,dx = −2xx π/2xx2π/2π/2provided the last integral converges.
But, as we saw in Example 5, this integralconverges, and hence the integral+∞π/2sin xdxx(6.77)also converges.At the same time, the integral (6.77) is not absolutely convergent. Indeed, forb ∈ [π/2, +∞[ we have b sin x sin2 x1 b dx 1 b cos 2x dx ≥dx=−dx. x x2 π/2 x2 π/2 xπ/2π/2bThe integral+∞π/2(6.78)cos 2xdx,xas can be verified through integration by parts, is convergent, so that as b → +∞,the difference on the right-hand side of relation (6.78) tends to +∞. Thus, by estimate (6.78), the integral (6.77) is not absolutely convergent.We now give a special convergence test for improper integrals based on the second mean-value theorem and hence essentially on the same formula for integrationby parts.Proposition 4 (Abel–Dirichlet test for convergence of an integral) Let x → f (x)and x → g(x) be functions defined on an interval [a, ω[ and integrable on everyclosed interval [a, b] ⊂ [a, ω[.
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