Does this improper integral converge?

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Discussion Overview

The discussion revolves around the convergence of an improper integral involving a function g(x) that is continuous and non-negative, under the condition that g(x) is less than another function f(x) = 1/x. Participants explore whether the integral of g(x) from a positive constant A to infinity converges, and if it is possible to determine convergence based on the given conditions.

Discussion Character

  • Exploratory, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant poses the question of convergence for the integral of g(x) from A to infinity, given that g(x) is any continuous, non-negative function less than f(x).
  • Another participant suggests using g(x) = 1/x^2 as an example to analyze convergence.
  • A different participant asserts that g(x) does not need to be in the form of 1/(x^p) and acknowledges that functions of the form 1/(x^p) with p > 1 converge, but emphasizes that g(x) could take various forms as long as it meets the criteria.
  • One participant raises the possibility that functions like 1/(x+1) could be considered, indicating that the convergence of g(x) could vary depending on its specific form.

Areas of Agreement / Disagreement

Participants express differing views on the nature of g(x) and its implications for convergence. There is no consensus on whether the integral converges or if it is impossible to determine based on the given conditions.

Contextual Notes

The discussion highlights the dependence on the specific form of g(x) and the conditions under which it is compared to f(x). There are unresolved questions regarding the convergence of non-standard forms of g(x) that meet the criteria.

JoePharos
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Does this improper integral converge?

I have an interesting problem that has been bothering me.

Given:
f(x)=1/x
g(x)= (any continuous, non-negative function)
g(x)<f(x)
A = (a positive constant)

I want to know:
Does the integral of g(x) from A to +infinity converge? Or is it impossible to determine?
 
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Take for example [itex]g(x)=\frac{1}{x^{2}}[/itex] and see what happens...

Daniel.
 
g(x) does NOT need to be in form 1/(x^P)

I already know that for any function of the form 1/(x^p), where p>1, it will converge. Thus, I know 1/(x^2) converges.

HOWEVER, g(x) does NOT need to be in the form of 1/(x^p). g(x) can be ANY non-negative function such that g(x)<f(x). That's a completely different picture.
 
Assumingly, you're asking whether or not a function like 1/(x+1) converges or something? With the criteria you mentioned, the function where f(x) > g(x) and a non-negative function could be either/or convergent/divergent.
 

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