Chap 11 Improper Integrals

# Chap 11 Improper Integrals - FIRST YEAR CALCULUS W W L CHEN...

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Unformatted text preview: FIRST YEAR CALCULUS W W L CHEN c W W L Chen, 1994, 2005. This chapter is available free to all individuals, on the understanding that it is not to be used for ﬁnancial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Chapter 11 IMPROPER INTEGRALS 11.1. Introduction To deﬁne the Riemann integral B f (x) dx, A it is crucial that [A, B ] is a ﬁnite interval, and that the function f (x) is bounded in [A, B ]. On the other hand, these two conditions are not suﬃcient to guarantee that the Riemann integral exists. There are also many integrals that we have performed and which fail one or both of these requirements. Example 11.1.1. When we write 1 0 1 2 1/2 dx = x 3 3x1/2 1 = 0 2 , 3 we may not even realize that the integrand is not deﬁned at x = 0. Example 11.1.2. When we write ∞ 1 1 1 dx = − x2 x ∞ = 1, 1 we may even wave our hands and say, “Well, when x = ∞, we have 1/x = 1/∞ = 0.” Chapter 11 : Improper Integrals page 1 of 4 First Year Calculus c W W L Chen, 1994, 2005 11.2. Unbounded Integrands Example 11.2.1. Let us return to Example 11.1.1, and try to attach some meaning to the integral 1 0 1 dx. 3x1/2 For every positive δ ∈ R satisfying δ ≤ 1, consider the integral 1 F (δ ) = δ 1 dx. 3x1/2 Then F (δ ) = 2 1 /2 x 3 1 = δ 2 2 1/2 −δ . 33 Clearly F (δ ) → 2/3 as δ → 0+. The real meaning of the integral in question is therefore given by 1 0 1 dx = lim δ →0+ 3x1/2 1 δ 1 dx. 3x1/2 The example suggests the following strategy. We wish to study the “integral” B f (x) dx. A Definition. Suppose that the function f (x) is not deﬁned at the point x = A. Suppose further that for every positive δ ∈ R satisfying δ ≤ B − A, the Riemann integral B F (δ ) = A+δ f (x) dx exists, and that F (δ ) converges to a limit L as δ → 0+. Then we write B L= A f (x) dx and call this an improper integral. Definition. Suppose that the function f (x) is not deﬁned at the point x = B . Suppose further that for every positive δ ∈ R satisfying δ ≤ B − A, the Riemann integral B −δ F (δ ) = A f (x) dx exists, and that F (δ ) converges to a limit L as δ → 0+. Then we write B L= A f (x) dx and call this an improper integral. Chapter 11 : Improper Integrals page 2 of 4 First Year Calculus c W W L Chen, 1994, 2005 Definition. Suppose that the function f (x) is not deﬁned at the point x = C ∈ (A, B ). Suppose further that for every positive δ ∈ R satisfying δ ≤ min{C − A, B − C }, the Riemann integrals C −δ B f (x) dx A and C +δ f (x) dx exist, and that their sum F (δ ) converges to a limit L as δ → 0+. Then we write B L= A f (x) dx and call this an improper integral. Example 11.2.2. One of the most famous improper integrals is the function li(X ), deﬁned for every real number X > 1 by X li(X ) = 0 1 dx. log x Note that one needs to study the behaviour of the integrand at x = 0 and x = 1 very carefully. This function arises from the study of the distribution of prime numbers, and is a good approximation to the function π (X ) which is the number of prime numbers p satisfying 2 ≤ p ≤ X . In fact, the famous Prime number theorem states that li(X ) →1 π (X ) as X → ∞. 11.3. Unbounded Intervals Example 11.3.1. Let us return to Example 11.1.2, and try to attach some meaning to the integral ∞ 1 1 dx. x2 For every Y ∈ R satisfying Y ≥ 1, consider the integral Y F (Y ) = 1 1 dx. x2 Then Y F (Y ) = 1 1 1 dx = − 2 x x Y =1− 1 1 . Y Clearly F (Y ) → 1 as Y → +∞. The real meaning of the integral in question is therefore given by ∞ 1 1 dx = lim Y →+∞ x2 Y 1 1 dx. x2 The example suggests the following strategy. We wish to study the “integral” ∞ f (x) dx. A Chapter 11 : Improper Integrals page 3 of 4 First Year Calculus c W W L Chen, 1994, 2005 Definition. Suppose that for every Y ∈ R satisfying Y ≥ A, the Riemann integral Y F (Y ) = A f (x) dx exists, and that F (Y ) converges to a limit L as Y → +∞. Then we write ∞ L= A f (x) dx and call this an improper integral. The reader is left to formulate suitable deﬁnitions for the improper integrals B ∞ f (x) dx −∞ and −∞ f (x) dx. Problems for Chapter 11 1. For each of the following, determine whether the improper integral exists, and if so, ﬁnd its value: 5 4 ∞ 1 x x cos x − sin x √ a) dx b) dx dx c) 2 x2 x−3 0x 3 0 1 2. Try to give an interpretation for the improper integral −1 1 dx = 0. x Chapter 11 : Improper Integrals page 4 of 4 ...
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