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Homework Help: If integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) converge

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data
    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


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


    3. The attempt at a solution
    Well I think that there is ..
    But i have no clue on how to solve it....
    Thank you.
     
  2. jcsd
  3. Feb 14, 2010 #2

    Dick

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    Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

    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.
     
  4. Feb 15, 2010 #3
    Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

    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?
     
  5. Feb 15, 2010 #4
    Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

    notice that you function is undefined at x=0 and the inequality you propose is incorrect for 0<x<1
     
  6. Feb 15, 2010 #5
    Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

    sorry it is completely incorrect
     
  7. Feb 15, 2010 #6
    Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

    i have a feeling it is true since f(x)^p depends continously on p. are you sure this is all the info?
     
  8. Feb 15, 2010 #7
    Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

    yes but I dont 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]
     
  9. Feb 15, 2010 #8
    Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

    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
     
  10. Feb 15, 2010 #9

    Dick

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    Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

    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: Feb 15, 2010
  11. Feb 16, 2010 #10
    Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

    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.
     
  12. Feb 16, 2010 #11

    Dick

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    Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

    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.
     
  13. Feb 16, 2010 #12
    Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

    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.
     
  14. Feb 16, 2010 #13

    Dick

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    Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

    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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