# How does the limit comparison test for integrability go?

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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.

## Answers and Replies

Samy_A
Science Advisor
Homework Helper
I think I found it here, written in a much better way: http://home.iitk.ac.in/~psraj/mth101/lecture_notes/lecture18.pdf (theorem 17.3). May I write the same with |f(x)| instead of f(x) ≥ 0?

Thank you in advance for your help.

Julien.
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.