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Home , Absolute convergence, Conditional convergence, Improper integral, Lebesgue integration

78 2. THE THEORY

11. Improper Riemann 11.1. Preliminaries. The Riemann is defined for a bounded f and a bounded region Ω. Can the be defined for f that is not bounded in a neighborhood of some point of Ω and (or) not bounded Ω? For example, a f(x)= e−x can be integrated over an unbounded [0, ) using the rule ∞ ∞ b e−x dx = lim e−x dx = lim (1 e−b) = 1 Z0 b→∞ Z0 b→∞ − Note f is integrable on every [0,b] because f is continuous. So, the rule makes sense. Similarly, the function f(x)= x−1/2 is not bounded on [0, 1], but it is continuous on every [a, 1] so it makes sense to define 1 dx 1 dx = lim = lim (2 2√a) = 2 + + Z0 √x a→0 Za √x a→0 − The above Riemann integrals are known as improper Riemann integral. So, a general idea is to reduce the integration domain to a bounded region so that the function in question is integrable on it, and after calculation of the integral over the reduced domain one has to take an appropriate with respect to parameters of the reduced domain in which the later becomes the region in question. The procedure is also often called a regularization of the improper Riemann integral. Does the value of the depend on the regulariza- tion? The answer is not straightforward, especially for higher dimen- sional integrals. Consider the function of two variables: x2 y2 f(x,y)= − (x2 + y2)2 Evidently it is defined everywhere except the origin. One can choose f(0, 0) to be any number. Regardless of the choice, f is not bounded in any neighborhood of the origin. Suppose one wants to integrate this function over Ω= (x,y) x2 + y2 1 , x 0 , y 0 { | ≤ ≥ ≥ } which is the part of the unit disk that lies in the positive quadrant. So, f is not bounded on Ω. In attempt to mimic a one-dimensional improper integral, let us take a subregion Ω Ω a ⊂ Ω = (x,y) a2 x2 + y2 1 , x 0 , y 0 a { | ≤ ≤ ≥ ≥ } 11. IMPROPER RIEMANN INTEGRALS 79

so that Ωa Ω as a 0. Using the polar coordinates it is not difficult to show that→ → f(x,y) dxdy = 0 ZZΩa Alternatively, the result follows from the symmetry argument. The region Ωa is symmetric under the reflection about the line y = x: (x,y) (y,x), whereas the integrand is skew-symmetric, f(x,y) = f(y,x→). Can one conclude that the improper Riemann integral of f over− Ω exists and is equal to zero? It is obvious that Ω can be obtained in many ways as the limit of subregions. For example, consider a collection of expanding subregions Ω Ω which are defined in polar coordinates k ⊂ x = r cos(θ) , y = r sin(θ) by the conditions Ω = (r, θ) a r 1 , β θ π/2 , k = 1, 2,... k { | k ≤ ≤ k ≤ ≤ } where a and β are positive that converge to 0 mono- { k} { k} tonically. These regions coincide with Ωa with a = ak in which a small sector with the angle βk is removed. So, ∞ Ω Ω Ω , Ω = Ω k ⊂ k+1 ⊂ k k[=1

The latter union is a proper mathematical way of saying that Ωk “ap- proaches Ω and coincides with Ω in the limit k ”. Using polar coordinates → ∞ 1 π/2 r2 cos(2θ) 1 f(x,y) dxdy = 4 rdrdθ = sin(2βk)ln(ak) ZZΩk Zak Zβk r 2 The left side is an “0 ”. Its limit may not even exist and, even if it exists, it may have×∞ any value! Indeed, take −c/βk ak = e where c > 0 so that ak 0 monotonically if βk 0 monotonically. Then → → c sin(2β ) lim f(x,y) dxdy = lim k = c k→∞ ZZΩk − k→∞ 2βk − 2 −c/βk If ak = βk, then the limit is 0 and, if ak = e , c> 0, then the limit is . The reader is asked to verify that if the range of the polar angle −∞ in Ωk is restricted to the interval π/2 βk θ π/2, then the limit can be made arbitrary positive number− or≤ + ≤by a suitable choice ∞ of ak. So, the answer really depends on how the improper Riemann integral is regularized! In fact, a similar result can be established for the 80 2. THE LEBESGUE INTEGRATION THEORY integral of f over an unbounded region x2 + y2 1, x 0, y 0 (see Exercises). Naturally, one wants a definitive (or≥ unique)≥ value≥ of the integral, and, for this reason, the definition of the improper Riemann integral requires some amendments in order to be applicable for higher dimensional integrals.

11.2. Improper Riemann integrals. Let Ω RN bounded or unbounded. An exhaustion of Ω is a of regions⊂ Ω ∞ such that { k}1 each Ωk is bounded, closed, and contained in Ω; • Ω contains Ω ; • k+1 k the union of all Ωk coincides with Ω except possibly a of • zero. An example of an exhaustion was given in the previous section. Ifa f is Riemann integrable on a closed bounded region Ω, then it can be proved that

lim f(x) dN x = f(x) dN x k→∞ ZΩk ZΩ which is nothing but the continuity property of the Riemann integral. If f is not bounded and/or Ω is not bounded, the improper integral is defined by this property. Definition 11.1. Let Ω ∞ be an exhaustion of Ω. Suppose that { k}1 a function f on Ω is Riemann integrable on each Ωk. Then the function f is said to be Riemann integrable on Ω if the limit

lim f(x) dN x = f(x) dN x k→∞ ZΩk ZΩ exists and is independent of the choice of Ωk. In this case, the value of the limit is called an improper Riemann integral of f over Ω.

11.3. Improper integral of non-negative functions. It is not difficult to find a particular exhaustion and investigate the limit of the sequence of the integrals. Even if the limit is found to exist, one still has to show that it is independent of the choice of the exhaustion in order to prove that the found limit is the value of the improper integral. It is not possible to do so by constructing all possible exhaustions. How then can this be done? Suppose that f(x) is non-negative on Ω. For any exhaustion, the sequence of integrals is monotonically increasing

0 f(x) dN x f(x) dN x ≤ ZΩk ≤ ZΩk+1 11. IMPROPER RIEMANN INTEGRALS 81

by the positivity property of the Riemann integral and that Ωk Ωk+1. The sequence converges only if it is bounded. So, there are on⊂ly two possibilities: either the limit is a number

N lim f(x) d x = If < k→∞ ZΩk ∞ or it is infinite, If = . Suppose that this limit exists, If < . Let 0 ∞ ∞ Ωk be another exhaustion of Ω. Then the sequence of the integrals is{ bounded:} N f(x) d x If 0 ZΩk ≤ 0 because f(x) 0 and Ωk Ω. Since the sequence is also monotonic, it converges ≥ ⊂

N 0 lim f(x) d x = If If k→∞ 0 ZΩk ≤ and its limit cannot exceed If . On the other hand, one can swap the roles of the exhaustions and use the same argument:

N 0 0 f(x) d x If If If ZΩk | | ≤ ⇒ ≤ 0 because Ωk Ω and f(x) 0. Therefore If = If and the value of the limit does not⊂ depend on the≥ choice of the exhaustion. Theorem 11.1. (Improper integral for non-negative functions) Suppose that (i) f(x) 0 , x Ω; ≥ ∀0 ∈ (ii) Ωn and Ωn are exhaustions of Ω; { } { } 0 (iii) f is Riemann integrable on each Ωn and Ωn Then lim f(x) dN x = lim f(x) dN x n→∞ n→∞ 0 ZΩn ZΩn where the limit can also be + . ∞ Suppose that a bounded function f is Riemann integrable on a bounded, closed region Ω, then the non-negative functions 1 f±(x)= f(x) f(x) 0 2| |±  ≥ are also integrable (because the of an integrable func- tion is integrable). The function f+(x) coincides with f(x) whenever f(x) 0 and vanishes otherwise, whereas f (x) coincides with f(x) ≥ − − 82 2. THE LEBESGUE INTEGRATION THEORY whenever f(x) 0 and vanishes otherwise. They can alternatively be written as ≤ f (x) = max f(x), 0 , f (x)= min f(x), 0 + { } − − { } Thus, any Riemann integrable function can be written as the difference of two non-negative integrable functions: f(x) = f (x) f (x) , + − − N N N f(x) d x = f+(x) d x f−(x) d x ZΩ ZΩ − ZΩ and vice versa (integrability of f implies integrability of f and f = ± | | f+ + f− by the linearity of the integral). Using the limit laws, the following theorem can easily be established from the above representation.

Theorem 11.2. Suppose that the improper integrals of f± over Ω exist. Then the improper integral of f over Ω exists and can be com- puted in any exhaustion Ω of Ω. { n} Indeed, by the limit laws and the existence of the improper integral of f±,

N N f(x) d x = lim f+(x) f−(x) d x n→∞ ZΩ ZΩn  −  N N = lim f+(x) d x lim f−(x) d x n→∞ n→∞ ZΩn − ZΩn

N N = f+(x) d x f−(x) d x ZΩ − ZΩ and, by Theorem 11.1 the values of the limits in the right side of the equation do not depend on the choice of the exhaustion (or regulariza- tion) of the integrals.

Corollary 11.1. Let Ωn be an exhaustion of Ω. Suppose that f and its absolute f are integrable{ } on each Ω and | | n lim f(x) dN x = f(x) dN x< n→∞ ZΩn | | ZΩ | | ∞ Then the improper integral of f over Ω exists and

f(x) dN x = lim f(x) dN x n→∞ ZΩ ZΩn In other words, if the improper integral of the absolute value of f converges in any particular regularization, then the improper integral 11. IMPROPER RIEMANN INTEGRALS 83 of f exists and can be computed in any suitable regularization. Indeed, since 0 f (x) f(x) ≤ ± ≤| | It is concluded that monotonic sequences of integrals of f± over Ωn are bounded:

N N N 0 f±(x) d x f(x) d x f(x) d x< ≤ ZΩn ≤ ZΩn | | ≤ ZΩ | | ∞ and, hence, converge. By Theorem 11.1 the limits are independent of the choice of Ωn. By Theorem 11.2, the improper integral of f over Ω exists (it is independent of regularization).

11.4. Absolutely and conditionally convergent integrals. If the limit

lim f(x) dN x n→∞ ZΩn exists for some exhaustion (regularization) Ωn , it is called convergent in this exhaustion (regularization). If the integral{ } of the absolute value also converges in this exhaustion

N lim f(x) d x = If < n→∞ ZΩn | | ∞ then the integral of f over Ω is said to be absolutely convergent, and if If = , then the integral of f is said to be conditionally convergent in the exhaustion∞ Ω . In the previous section, it was shown that { n} the of the Riemann integral implies the • existence of the improper Riemann integral (the value of the integral does not depend on the regularization). This terminology is due to the analogy with absolutely and condi- tionally convergent . The region Ω can be viewed as the union of shells obtained by carving out Ωk from Ωk+1: ∞ Ω= S , S = Ω Ω k k k \ k−1 k[=1 where Ω0 is an empty set by definition. Note that Ωk contains its boundary so that Ω Ω does not contain its inner boundary. The k \ k−1 shell Sk is closed and (the missing boundary is added by taking the closure of Ω Ω ). Suppose that f is integrable on each k \ k−1 Sk. Then ∞ f(x) dN x = f(x) dN x Z Z Ω Xk=1 Sk 84 2. THE LEBESGUE INTEGRATION THEORY

If the series of integrals of the absolute value converges ∞ N N f(x) d x = f(x) d x = If < Z | | Z | | ∞ Ω Xk=1 Sk then the series for the integral of f converges absolutely and, hence, converges because

f(x) dN x f(x) dN x Z ≤ Z | | Sk Sk

Recall that every absolutely converges.

The comparison test for the absolute convergence. It has been shown that the absolute convergence of a Riemann integral in a particular regularization is sufficient to establish the existence of the improper Riemann integral, just like the absolute convergence of a series implies the convergence of the series. It is therefore important to establish some tests for absolute convergence of Riemann integrals. Let f and g be Riemann integrable on Ω, and f(x) g(x) , x Ω | |≤ ∀ ∈ Then f(x) dN x g(x) dN x ZΩ | | ≤ ZΩ If now f and g are not integrable in the proper sense, then for any exhaustion Ω { n} f(x) dN x g(x) dN x ZΩn | | ≤ ZΩn If the improper integral of g converges, then the integral f converges absolutely because

g(x) dN x g(x) dN x< ZΩn ≤ ZΩ ∞ lim f(x) dN x g(x) dN x n→∞ ⇒ ZΩn | | ≤ ZΩ By Theorem 11.2 the limit does not depend on the choice of the ex- haustion and the improper integral of f exists. Theorem 11.3. (Comparison test for absolute convergence) If the absolute value of a function f is bounded on Ω by a function whose improper Riemann integral over Ω exists, the improper integral of f over Ω also exists (it converges absolutely). 11. IMPROPER RIEMANN INTEGRALS 85

It should be pointed out that in the Riemann theory the integrabil- ity of f does not imply integrability of f (e.g., f(x)=1if x is rational, and f|(x|)= 1 otherwise so that f(x) = 1). In the above theorem, it is assumed that− f is integrable on| any| region of an exhaustion so that its improper integral can at least be defined. Suppose Ω = RN and f is a continuous function. Clearly it is integrable on any ball Ωn = Bn (that is, x n, n = 1, 2,...). So the existence of the improper integral would depend| |≤ on how fast f fall off as x . | |→∞ Proposition 11.1. Let f be a continuous function such that M f(x) , x R | |≤ x p | |≥ | | for some constants M and R, and p > N, then the improper integral of f over the whole exists and

f(x) dN x = lim f(x) dN x< . ZRN n→∞ Z|x|≤n ∞ Consider the case N = 2. The integral over the whole plane is split into the integral over the disk B and the rest of the plane R2 B . R \ R Since the integral over BR is a regular integral, one has to investigate the convergence of the improper integral over the rest of the plane. Since f 0, if it converges in a particular regularization, then it con- ≥ verges in any other regularization to the same value. Let Ωn be an annulus R x n. Then ≤| |≤ 2π n 2 M 2 M f(x) d x p d x = p rdrdθ Z n | | ≤ Z n x Z Z r Ω Ω | | 0 R 2πM 1 1 = p 2 Rp−2 − np−2  − The right side converges if p > N = 2 when n . Therefore the integral of f converges absolutely and, hence, the→ improper ∞ integral of f exists by the comparison test (it can be computed in any suitable regularization). Similarly, suppose f is not bounded in any neighborhood of a par- ticular point, and it is continuous otherwise. Without loss of generality, the singular point can be chosen to be the origin x = 0 (values of f(x) becomes infinitely large as x approaches 0). Then it is integrable on Ω = Ω B where B is a ball of radius a, x

Proposition 11.2. Suppose that f is not bounded in any ball Ba and continuous on Ω Ba where Ω is a bounded region with a piecewise smooth boundary. If \ M f(x) , x R | |≤ x p | |≤ | | for some constants M and R, and p < N, then the improper integral of f exists and

f(x) dN x = lim f(x) dN x. + ZΩ a→0 ZΩ\Ba A proof of this assertion can also be done by using spherical coor- dinates in RN . For example, if N = 2, the using the polar coordinates 2π R 2 M 2 M f(x) d x p d x = p rdrdθ Z | | ≤ Z x Z0 Za r a≤|x|≤R a≤|x|≤R | | 2πM = R2−p a2−p 2 p − −   The right side converges in the limit a 0+ if p < 2 = N and so does the left side. Therefore the integral of→f converges absolutely and, hence, the improper integral of f exists by the comparison test.

Remark. In the Lebesgue integration theory, all integrals (proper and improper) are treated in the same footing, and the integrability of the absolute value of a function implies integrability of the function. In this sense, the Lebesgue integrals are often called absolutely convergent integrals.

11.5. Conditionally convergent integrals. Let the integral of f over Ω be conditionally convergent, that is, it converges in some regularization, but the integral of the absolute value diverges. In this case, the integrals of f± must diverge, and the value of a conditionally convergent integral is an indeterminate form “ ” which happens to be a number in a particular regularization:∞−∞

N N N lim f(x) d x = lim f+(x) d x f−(x) d x n→∞ n→∞ ZΩn ZΩn − ZΩn  Note that the of the integral of f = f + f implies that | | + − either the integral of f+, or f−, or both diverge because f± 0. The of the integral of f (the existence of≥ the limit in the left side) is only possible when integrals of f± diverge. 11. IMPROPER RIEMANN INTEGRALS 87

The integrals of f± resemble the (divergent) series of positive and negative terms of a conditionally convergent series. The sum of such a series depends on the arrangement of terms (the order in which the terms are added). In the case of conditionally convergent integrals, the value depends on the choice of the exhaustion (or regularization). In other words, by choosing a suitable exhaustion one can always make the difference of the integrals of f+ and f− over Ωn to be convergent to any desired number even though both the sequences diverges to + . This is illustrated with the following example. ∞ Consider the improper integral ∞ sin(x) bn sin(x) dx = lim dx Z0 x n→∞ Z0 x where bn is positive, monotonically increasing, unbounded sequence. Here the{ } integrand extended to x = 0 by continuity (the integrand approaches 1 as x 0+). In particular, let us take → bn = πn, n = 1, 2,... This regularization corresponds to the exhaustion:

Ωn = [0, πn] The corresponding shells are intervals S =[π(n 1), πn] , Ω = Ω S n − n n−1 ∪ n If the limit exists, then is equal to the sum of the series ∞ sin(x) ∞ πn sin(x) dx = dx Z x Z x 0 Xn=1 π(n−1)

This is an because the integrand is positive on S2k−1 and negative on S2k. It follows from the inequality sin(x) sin(x) sin(x) | | | | | | , n> 1 πn ≤ x ≤ π(n 1) − that 2 2 πn sin(x) an , an = | | dx > 0 πn ≤ ≤ π(n 1) Z x − π(n−1) and ∞ ∞ sin(x) n+1 dx = ( 1) an Z x − 0 Xn=1 The sequence a is positive and converges to 0 monotonically because { n} 2 a a n+1 ≤ πn ≤ n 88 2. THE LEBESGUE INTEGRATION THEORY

By the , the series converges. However, by the comparison test: 2 n 1 πn sin(x) n ∞ sin(x) | | = ak | | dx = π k ≤ Z x ⇒ Z x ∞ Xk=1 0 Xk=1 0 1 the series and the integral do not converge absolutely because n = . So, the integral is only conditionally convergent and does notP ex- ist∞ in the sense of Definition 11.1. Its value depends on the choice of regularization. The latter follows from the Riemann theorem about alternating series. In particular, the sum can be made equal to any desired number by a suitable rearrangement. In this example, a re- arrangement correspond to a different exhaustion made of the same 0 set of intervals Sn. Consider a rearrangement Sn of the sequence of intervals S and put { } { n} Ω0 = S0 , Ω0 = Ω0 S0 1 1 n+1 n ∪ n+1 0 0 So, Ωn is a collection of any n intervals from Sn and Ωn+1 is obtained by adding any of remaining intervals in the{ collection} S . In other { n} words, the order in which the intervals from Sn are added to obtain an exhaustion is changed, but { } ∞ ∞ Ω = Ω0 = [0, ) . n n ∞ n[=1 n[=1 0 The function is still integrable on any finite collection of intervals Ωn. Therefore in this exhaustion (regularization) ∞ sin(x) ∞ sin(x) dx = dx Z x Z 0 x 0 Xn=1 Sn The series in the left side is a rearrangement of the alternating series ( 1)n+1a . Its sum can be any number, or , or the series may n − n ±∞ notP even converge. The above example admits a generalization. Let f(x) be continuous on Ω. Then let Ω+ be a where f(x) 0 and Ω− be a subset + − ±≥ ± where f(x) 0. Then Ω = Ω Ω . Let Ωn be exhaustions of Ω , respectively.≤ Consider the following∪ exhaustion{ } of Ω Ω = Ω+, Ω−, Ω+, Ω−, { n} { 1 1 2 2 ···} Then the limit of integrals of f over Ωn is the sum of an alternating series ∞ lim f(x) dN x = f(x) dN x + f(x) dN x + − n→∞ Z n Z Z  Ω Xn=1 Sn Sn 11. IMPROPER RIEMANN INTEGRALS 89

± ± ± ± ± where S1 = Ω1 and Ωn+1 = Ωn Sn . If the integral does not converge absolutely, then one can rearrange∪ the series so that it can converge to any desired number. This rearrangement corresponds to a new exhaus- ± tion (a different order of selecting shells Sn in the new exhaustion).

11.6. Exercises.

1. Consider the improper integral

x2 y2 2 − 2 2 dxdy ZZΩ (x + y ) where Ω= (x,y) x2 + y2 1 , x 0 , y 0 { | ≥ ≥ ≥ }

Take an exhaustion Ωn which is a rectangle in the polar coordinates (r, θ) [1,a ] [α ,{ π/2} β ] where α and β are positive and tend ∈ n × n − n n n to 0 monotonically, while 1

2. For the integrand in Problem 1, find f±(x,y) and show that for any exhaustion the improper integrals of f± diverge

f±(x,y) dxdy = ZZΩ ∞

3. Prove Proposition 11.1 for N = 3 using spherical coordinates.

4. Prove Proposition 11.2 for N = 3 using spherical coordinates.

5. Let f(t) be a continuous function and t 0. Then for any ball B RN ≥ a ⊂ a N/2 N N−1 2π f( x ) d x = σN f(r) r dr, σN = dS = , ZBa | | Z0 Z|x|=1 Γ(N/2) where σN is the surface of the unit sphere x = 1. The assertion is proved by converting the integral to spherical| coordinates| in RN . Use this result to prove Propositions 11.1 and 11.2. 90 2. THE LEBESGUE INTEGRATION THEORY

6. Let p and q be positive . Do the following improper in- tegrals exist in the sense of Definition 11.1? ∞ sin2(x) (i) p dx Z0 x ∞ cosq(x) (ii) p dx , Z1 x 1 1 (iii) sinp dx Z0 x