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Improper integral convergence proof

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [itex] [a, b) [/itex] be an interval in the reals, with [itex] -\infty < a < b \leq \infty [/itex], and let [itex] \alpha: [a,b) \to \mathbb{R} [/itex] be monotone increasing. Suppose that [itex] f: [a,b) \to \mathbb{R} [/itex] is a function such that for each [itex] c \in (a,b) [/itex], [itex] f [/itex] is integrable over [itex] [a,c ] [/itex] with respect to [itex] \alpha [/itex] (in particular f is bounded on every compact subinterval of [itex] [a, b) [/itex], although f may not be bounded on all of [itex] [a, b) [/itex]).

    Now for the actual problem....

    Prove that if [itex] F(x) = \int_a^x f(x) d\alpha[/itex] is a bounded function [itex] [a, b) \to \mathbb{R} [/itex] and [itex] g: [a, b) \to \mathbb{R} [/itex] is non-negative valued, monotone decreasing, integrable with respect to [itex] \alpha [/itex] over each interval [itex] [a, c] [/itex] with a < c < b, and [itex] \lim_{x \to b^{-}} g(x) = 0 [/itex], then

    \int_a^b f(x)g(x)\, d\alpha


    2. Relevant equations

    3. The attempt at a solution
    I've tried putting a uniform bound M on F, and finding [itex] x_0 [/itex] such that [itex] x_0 < x < b [/itex] implies that [itex] 0 \leq g(x) < \frac{\epsilon}{M} [/itex]. However, I can't find a way to bring [itex] g(x) [/itex] "inside" the integral, so to speak. My intuition is that at worst [itex] F [/itex] is oscillating and that multiplying by g will bring it under control to make it convergent, sort of like [itex] \int_1^\infty \frac{\sin x}{x} dx [/itex] but I just can't get any ideas to work, and I'm really stuck.
  2. jcsd
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