# If 1 we write the power series for 1 1 1 1 1 2 3 1 2

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If 𝛼 = −1 , we write the power series for (1 + ?) −1 : (1 + ?) −1 = 1 − ? + ? 2 − ? 3 … = 1 + ? + ? 2 + ? 3 Here we get our usual geometric progression where ? = −? .
Page 62 of 74 Chapter 6: Improper Integrals The Riemann integral is defined only when the interval is bounded and closed and the function is bounded. In this chapter we will deal with the situation where at least one of these fail. We denote by ? any interval, bounded or unbounded, open or closed at either end. We will often write ? = 〈?, ?〉 . Recall that these angular brackets mean that the end points could either be included or excluded. Recall that there are 9 different types of intervals. The function ? will always be assumed to be ??? (?) . Our aim is to define, if possible: ?(?) ?? ? ? Definition 6.1: Let ? ∈ ℛ ??? (?) , ? = 〈?, ?〉 . i. We say that ? is integrable at ? in the improper sense if ∃𝑝 ∈ ? s.t. lim ?→? + ?(?) ?? ? ? exists. We will denote this limit by ?(?) ?? ? ? . ii. We say that ? is integrable at ? in the improper sense if ∃𝑝 ∈ ? s.t. lim ?→? ?(?) ?? ? ? exists. We will denote this limit by ?(?) ?? ? ? . iii. We say that ? is integrable on ? in the improper sense if both (i) and (ii) hold for the same 𝑝 . We will then write: ?(?) ?? ? ? = ∫ ?(?) ?? ? ? + ∫ ?(?) ?? ? ? If, for example, ? = −∞, then “? → ? + ” means that “? → −∞”. Likewise, if ? = +∞ , then “? → ? ” means that “? → +∞” . If limits (i) and (ii) exist for one 𝑝 , then they exist for any other 𝑝 . Moreover, the resulting ?(?) ?? ? ? does not depend on the choice of 𝑝 . Choose a 𝑝 that is convenient.
Page 63 of 74 Linear properties, the order property and the domain splitting property carry over to improper integrals without problems. With the two remaining properties, triangle inequality and Cauchy-Schwarz inequality, the situation is trickier as it is not clear whether the integrals in question exist in the improper sense. The notion of an Improper integral is an extension of the notion of the Riemann Integral: If ? is closed and bounded, ? = [?, ?] , then the Riemann Integral coincides with the Improper Integral. Also, in this case, ??? (?) = ℛ(?) . Example 6.1: Examine the existence and evaluate, if possible, ∫ ? 𝛼 ?? 1 0 Here, 𝛼 ∈ ℝ . Here we are integrating the function ?: (0,1] → ℝ , ?(?) = ? 𝛼 over the interval (0,1] . Of course, here ? ∈ ℛ ??? (?) because ? ∈ 𝒞(?) . Case 1: 𝛼 ≥ 0 In this case our function has a continuous extension to ? = 0 . For 𝛼 > 0 we set ?(0) = 0 . For 𝛼 = 0 we set ?(0) = 1 . This gives a function on [0,1] and this extended function happens to be continuous on [0,1] . for 𝛼 = 0 for 𝛼 > 0 ? = ? 𝛼 ? = ? 𝛼 1 1 1 1 0 0 ? ? ? ? We should understand how Definition 6.1 works for particular intervals of ? : Say, let ? = [1, +∞) . Then the existence of the improper integral ?(?) ?? +∞ 1 means that lim ?→+∞ ?(?) ?? ? 1 exists. If ? = (0,1] , then the existence of the improper integral ?(?) ?? 1 0 means that lim ?→0 + ?(?) ?? 1 ? exists.
Page 64 of 74 Hence, the integral ∫ ? 𝛼 ?? 1 0 exists as a Riemann Integral and: ∫ ? 𝛼 ?? 1 0 = [ ? 𝛼+1 𝛼 + 1 ] 0 1 = 1 𝛼 + 1 Case 2: 𝛼 < 0 We cannot define ?(0) by continuity. In order to establish the existence of ∫ ? 𝛼 ?? 1 0 as an improper integral we start by choosing a 𝑝 ∈ (0,1] . It is convenient to choose 𝑝 = 1 . We need to look at: lim ?→0 + ∫ ? 𝛼 ??