Proving Convergence of f(x) Integrals in [a,∞)

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Homework Help Overview

The discussion revolves around the convergence of integrals of a positive and continuous function f(x) defined on the interval [a, ∞). The original poster seeks to prove or disprove the statement that if the integral of f(x) converges, then there exists a constant c such that the integral of f(x)^p converges for all p in the range [c, 1].

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the validity of the original statement by considering specific functions and their behavior under different powers. Some suggest looking for counterexamples, while others question the assumptions made regarding the function's properties and the implications of the convergence of the integrals.

Discussion Status

The discussion is active, with participants presenting examples and counterexamples to challenge the original claim. There is a mix of agreement and disagreement on the interpretations of the function's behavior, particularly concerning the limits and inequalities involved.

Contextual Notes

Participants note that the function f(x) is positive and continuous, but there are concerns about its behavior near certain points, such as x=0. Additionally, the implications of the convergence criteria for different values of p are under scrutiny.

no_alone
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Homework Statement


f(x) positive and continuous in [a,[tex]\infty[/tex])
prove of disprove:
if [tex]\int_a^\infty f(x)dx[/tex] converge there is a 0<c<1 so that
[tex]\int_a^\infty f(x)^p dx[/tex] converge for every c[tex]\leq[/tex]p[tex]\leq[/tex]1


Homework Equations


every thing in calculus 1+2 no two vars ..


The Attempt at a Solution


Well I think that there is ..
But i have no clue on how to solve it...
Thank you.
 
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I don't think it's true. Try find an example of an f(x) that just barely converges and f(x)^p diverges for all 0<p<1.
 


Ok
what about
[tex]\frac{1}{xln^2(x)}[/tex]
for every p<1
[tex]\frac{1}{x^pln^{2/p}(x)} > \frac{1}{x}[/tex]

But i always have problem proving that for every p an z lim(x^p) > lim(ln^z(x))
Can you help me with this?
 


notice that you function is undefined at x=0 and the inequality you propose is incorrect for 0<x<1
 


sorry it is completely incorrect
 


i have a feeling it is true since f(x)^p depends continously on p. are you sure this is all the info?
 


rsa58 said:
notice that you function is undefined at x=0 and the inequality you propose is incorrect for 0<x<1
yes but I don't care about when 0<x<1 because its in a half closed section when f(X) is defined and continuous ,
I only look at infinity, and in infinity ,

for every p<1


[tex] \frac{1}{x^pln^{2/p}(x)} > \frac{1}{x}[/tex]
 


incorrect try x=3. the inequality is actually backwards. plus your function cannot include p. the result must hold for ALL p. since f is positive you can use the ration
 


no_alone said:
Ok
what about
[tex]\frac{1}{xln^2(x)}[/tex]
for every p<1
[tex]\frac{1}{x^pln^{2/p}(x)} > \frac{1}{x}[/tex]

But i always have problem proving that for every p an z lim(x^p) > lim(ln^z(x))
Can you help me with this?

That's a great choice! It would be enough to show that in the limit ln(x)<x^p for all p>0, wouldn't it? Just use l'Hopital on the limit of x^p/ln(x).
 
Last edited:
  • #10


Thank you very much dick..
It's not really homework question,
Its a example test question .
This question really on my mind because it was really against my intuition ,
And my usually my intuition is true ,(my problem is proving it)
Can you supply another f(X) that is good?,
Thank you.
 
  • #11


You're welcome. 'Intuition' can go a little wrong when you are dealing with an 'infinite' part in the problem. Guess I don't have a ready stock of problems. You might scan through other forum posts looking for interesting problems. There are some in there. And feel free to help the poster if you've got a good idea on how to solve it.
 
  • #12


Dick said:
You're welcome. 'Intuition' can go a little wrong when you are dealing with an 'infinite' part in the problem. Guess I don't have a ready stock of problems. You might scan through other forum posts looking for interesting problems. There are some in there. And feel free to help the poster if you've got a good idea on how to solve it.
Sorry dick but I think you misunderstood me,
I asked if you can give me another counter example beside this one I supplied?
Thank you.
 
  • #13


no_alone said:
Sorry dick but I think you misunderstood me,
I asked if you can give me another counter example beside this one I supplied?
Thank you.

Nothing pops into my head. There's 1/(log(x)^n*x) for n>1. But that's just a variation on the same theme.
 

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