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Unformatted text preview: 7 Improper Integrals, Exp, Log, Arcsin, and the Integral Test for Series We have now attained a good level of understanding of integration of nice functions f over closed intervals [ a,b ]. In practice one often wants to extend the domain of integration and consider unbounded intervals such as [ a, ∞ ) and ( −∞ ,b ]. The simplest nontrivial examples are the infinite trumpets defined by the areas under the graphs of x t for t > 0, i.e., the improper integrals A t = ∞ ∫ 1 1 x t dx. We will see below that A t has a well defined meaning if t > 1, but becomes unbounded for t ≤ 1. One is also interested in integrals of functions f over finite intervals with f being unbounded. The natural examples are given (for t > 0) by B t = 1 ∫ 1 x t dx. Here it turns out that B t is well defined, i.e., has a finite value, if and only if t < 1. In particular, neither A 1 nor B 1 makes sense. Moreover, A t + B t = ∞ ∫ 1 x t dx is unbounded for every t > 0. 7.1 Improper Integrals Let f be a function defined on the interior of a possibly infinite interval J such that either its upper endpoint – call it b , is ∞ or f becomes unbounded as one approaches b . But suppose that the lower endpoint – call it a , is finite and that f ( a ) is defined. In the former case the interval is unbounded, while in the latter case the interval is bounded, but the function is unbounded. We 1 will say that the integral of f over J exists iff the following two conditions hold: (7.1.1) (i) For every u ∈ ( a,b ), f is integrable on [ a,u ]; and (ii) the limit lim u → b,u<b u ∫ a f ( x ) dx exists. When this limit exists, we will call it the integral of f over J and write it symbolically as (7 . 1 . 2) b ∫ a f ( x ) dx. Sometimes we will also say that the integral (7.1.2) converges when it makes sense. Similarly, if a is either −∞ or is a finite point where f becomes un bounded, but with b a finite point where f is defined, one sets (7 . 1 . 3) b ∫ a f ( x ) dx = lim u → a,u>a b ∫ u f ( x ) dx when the limit on the right makes sense. Lemma 7.1 We have ∞ ∫ a f ( x ) dx = c ∫ a f ( x ) dx + ∞ ∫ c f ( x ) dx ∀ c ∈ ( a, ∞ ) and b ∫ −∞ f ( x ) dx = c ∫ −∞ f ( x ) dx + b ∫ c f ( x ) dx ∀ c ∈ ( −∞ ,b ) , whenever the integrals make sense. 2 Proof . We will prove the first additivity formula and leave the other as an exercise for the reader. Pick any c ∈ ( a, ∞ ). or all real numbers u > c , we have by the usual additivity formula, u ∫ a f ( x ) dx = c ∫ a f ( x ) dx + u ∫ c f ( x ) dx. Thus ∞ ∫ a f ( x ) dx = lim u →∞ c ∫ a f ( x ) dx + u ∫ c f ( x ) dx , which equals c ∫ a f ( x ) dx + lim u →∞ u ∫ c f ( x ) dx = c ∫ a f ( x ) dx + ∞ ∫ c f ( x ) dx....
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 Winter '08
 Borodin,A
 Derivative, Improper Integrals, Integrals, Inverse function, Logarithm

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