CREATIVE MATH. 12 (2003), 7 - 10
Improper integrals and Riemann sums
Gabriella Kovacs´
Abstract. The aim of this note is to give simple examples of absolutely convergent improper integrals having integral sums whose limit does not equal the value of the integral.
1. Preliminaries f a, b ⊂ R → R a x(n) Received: 05.04.2003; In revised form: 12.12.2003 Key words and phrases. Riemann integral, improper integral, integral sum, sequence, series. 7 8 Gabriella Kov´acs t on [a, b) of the function f and we denote it by f(x)dx = lim f (x) dx. t→b,tabsolute convergence implies convergence, but not conversely. Nowweprepare some very simple examples of absolutely convergent im- proper integrals with integral sums whose limit differs from the value of the integral. The first example is integral of unbounded function over bounded integration domain. The second one is integral over unbounded domain. 2. The examples 1 1 1 1 Example 1. Denote A = , , , ,... .Considerp>1fixedand 1 2 3 4 0,x∈ (0, 1]\A f :(0, 1] → R,f(x)= 1 . np,x ∈ A = n On every interval [t, 1] ⊂ (0, 1] this function differs from the constant 0 1 function only in a finite number of points, namely in x = with n 1 n ∈ , , ,..., 1 2 3 t where [ ] means integer part. So on every interval 1 1 [t, 1] ⊂ (0, 1] the functionf is Riemann integrable and f (x) dx = 0dx =0. t t 1 It results that f (x) dx = lim f (x) dx = 0, so the improper inte- t→0,t>0 (0,1] t gral f (x) dx is convergent, and since f is nonnegative, it is absolutely (0,1] convergent. n 1 1 f i· n ∈ N ∗ We examine the sequence of integral sums n n , .By i=1 comparing the sum with its first term, since f is nonnegative, we have n 1 1 1 1 n 1 1 f i· ≥ ·f 1· = np−1; p − 1 > 0, so lim f i· = ∞. n→∞ i=1 n n n n i=1 n n Improper integrals and Riemann sums 9 Clearly, this limit of integral sums fails to equal the value of the improper integral. Example 2. Let p>0. Consider a sequence of positive numbers (yn) minorized by p, and the function 0,x∈ [1, ∞) \{1, 2, 3, 4, ...} f :[1, ∞) → R,f(x)= . yn,x= n ∈{1, 2, 3, 4,...} On every interval [1,t] ⊂ [1, ∞) the function f differs from the constant 0 function only in a finite number of points, namely in x ∈{1, 2, 3, ··· , [t]}. So, on every interval [1,t] the function f is Riemann integrable and t t f (x) dx = 0dx =0. 1 1 t It results that f (x) dx = lim f (x) dx =0, so the improper inte- t→∞ [1,∞) 1 gral f (x) dx is convergent, and since f is nonnegative, it is absolutely [1,∞) convergent. ∞ 1 1 Nevertheless, the equality lim f 1+i · = f(x)dx does not n→∞ i n n =1 [1,∞) ∞ 1 1 hold. To showthis, weexamine f 1+i · for n ∈ N∗ fixed: by i=1 n n i ∈ I {jn|j ∈ N∗} summing-up the terms corresponding to n = in this series, ∞ 1 1 1 1 f f i · ≥ f i · since is nonnegative, we obtain n 1+ n n 1+ n = i=1 i∈ In ∞ 1 ∞ 1 ∞ 1 ∞ 1 1 f j y ≥ p ∞ f i · ∞ = n (1 + )= n j+1 n = ,so n 1+ n = . j=1 j=1 j=1 i=1 Choosing the sequence (yn) majorized, we get f bounded. 3. Final remarks The reason why the limit of integral sums does not always equal the value of the convergent improper integral is that in case of improper integrals we perform two limit computing processes – one refines the divisions of an integration domain, the other expands this integration domain – and these processes do not commute unless suitable conditions met. The following result is immediately. Proposition 1. If a, b ∈ R, a < b and the improper integral f (x) dx is (a,b] n b − t b − t convergent, then lim lim f t + i · = f (x) dx holds. t→a,t>a n→∞ n n i=1 (a,b] 10 Gabriella Kov´acs Proof. Apply Definition 1 and then formula (1) for the interval [t, b]bytaking b − t x(n) ξ(n) t i · i ∈{ , ,...,n} i = i = + n , 1 2 . Remark that the process n →∞refines the divisions of the integration b domain of the Riemann integral f (x) dx and t → a, t > a expands the t interval [t, b]to(a, b]. Under the conditions of Proposition 1, to deduce n b − a b − a lim f a + i · = f (x) dx (2) n→∞ n n i=1 (a,b] we have to invert the two limit computing processes contained in Proposition 1 and to commute lim with the computation of the values of f.These, t→a,t>a even if the improper integral is absolutely convergent, may be done only if appropriate conditions are met. Sufficient conditions which ensure the validity of (2) in special cases of convergent improper integrals, and which can be easily checked practically, one can deduce from [1] or [4]. Sufficient conditions for a function f :[a, ∞) → R with convergent im- proper integral on [a, ∞) to verify equality ∞ lim hf (a + i · h)= f (x) dx (3) h→0,h>0 i=1 [a,∞) one can deduce based on [1], [3] or [4]. References [1] Fihtenholt G. M., Differential and Integral Calculus, Editura Tehnic˘a, Bucure¸sti 1964 [2] Kov´acs G., Convergence Criteria for Sequences of Integral Sums in Case of Improper Integrals, First Conference on Applied Mathematics, Univ. Baia Mare, 16-17 Oct. 1998 [3] Panuska R., Two Propositions in Elementary Real Analysis, Matematikai Lapok 32(1981-1984) 1-3, Budapest, 115-125 [4] P´olya Gy., Szeg˝oG.,Problems and Theorems in Analysis,Tank¨onyvkiad´o, Budapest 1980 North University of Baia Mare Department of Mathematics and Computer Science Victoriei 76, 430122 Baia Mare, Romania E-mail address: [email protected]