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I How does the limit comparison test for integrability go?

  1. Apr 12, 2016 #1
    Hi everybody! I have another question about integrability, especially about the limit comparison test. The script my teacher wrote states:

    (roughly translated from German)
    Limit test: Let -∞ < a < b ≤ ∞ and the functions f: [a,b) → [0,∞) and f: [a,b) → (0,∞) be proper integrable for any c ∈ [a,b). Let lim x↑b f(x)/g(x) =: h exist. Then:
    (i) If h > 0, then is f improper integrable on [a,b) if and only if g is improper integrable on [a,b).
    (ii) If h = 0, then is f improper integrable on [a,b) if g is improper integrable on [a,b).

    So...that statement only makes little sense to me, and I believe there must be at least one mistake there: what's up with that "c"?? I'd like to reformulate that test in two parts (integrability and non integrability) and so in a more logical way for myself. Let me know if that holds (apart from wherever the mistake is with "c"):

    1. Let f: [a,b) → [0,∞) and f: [a,b) → (0,∞) are proper (Riemann?) integrable for any c ∈ [a,b), g is improper integrable on [a,b) and lim x↑b f(x)/g(x) 0, then f is improper integrable on [a,b).

    2. Let f: [a,b) → [0,∞) and f: [a,b) → (0,∞) are proper (Riemann?) integrable for any c ∈ [a,b), g is not improper integrable on [a,b) and lim x↑b f(x)/g(x) > 0, then f is not improper integrable on [a,b).

    This all looks very absurd to me! :/ I also can't seem to find anything about such a test on the internet, that's weird! Does it even exist in the first place?


    Thank you very much in advance for your answers.


    Julien.
     
  2. jcsd
  3. Apr 12, 2016 #2
  4. Apr 12, 2016 #3

    Samy_A

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    Not exactly clear how you want to formulate the adapted version of theorem 17.3.
    If the condition is supposed to become ##\displaystyle \lim_{x \rightarrow +\infty}\frac{|f(x)|}{g(x)}= c \neq 0##, then no. You can easily construct a function f such that the improper integral of ##f## converges, while the improper integral of ##|f|## (and hence of ##g## by theorem 17.3) diverges.
     
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