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F(x) continuous, lim integral x to 2x of f(t) = 0 does integral of f(x)dx converge

  1. Dec 15, 2009 #1
    1. The problem statement, all variables and given/known data
    True Or False
    if f(x) continuous in [tex][a,\infty][/tex] and [tex]\lim_{x\to\infty}\int_x^{2x}f(t)dt = 0 [/tex]
    Then [tex]\int_a^\infty f(x)dx [/tex] converge


    2. Relevant equations
    Anything from calc 1 and 2


    3. The attempt at a solution
    Actually I'm really stuck..
    My main motive is to try and show that if [tex]\lim_{x\to\infty}\int_x^{2x}f(t)dt = 0 [/tex] there is no way that Cauchy test can fail,
    I assume Cauchy test fail and that [tex]\lim_{x\to\infty}\int_x^{2x}f(t)dt = 0 [/tex] and try to show that there is no option for this to happen.
    First I tried on f(x) that can be positive and negative .. didn't made it, Then I tried on f(x) that is only positive also didn't made it..
    Maybe its false, But I didn't found any function that does not converge and also [tex]\lim_{x\to\infty}\int_x^{2x}f(t)dt = 0 [/tex]
    Thank you.
     
  2. jcsd
  3. Dec 15, 2009 #2

    Dick

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    Re: f(x) continuous, lim integral x to 2x of f(t) = 0 does integral of f(x)dx converg

    Well, let's try and make such a function. Suppose we make the integral over [a,2a]~1/2, the integral over [2a,4a]~1/3, the integral over [4a,8a]~1/4. Then integral over all t should diverge like a harmonic series, right? Now, what must the values of such a function look like? Over [a,2a] it should be about 1/(2a). Over [2a,4a] it should be about 1/(3*(2a)). Over [4a,8a] like about 1/(4*(4a)). Etc, etc. That sort of looks like 1/(log(t)*t) to me.
     
    Last edited: Dec 15, 2009
  4. Dec 16, 2009 #3
    Re: f(x) continuous, lim integral x to 2x of f(t) = 0 does integral of f(x)dx converg

    Thank you Dick
    I didn't thought about series all I had in mind were function..
    I tried and tried , even when I tried to build a function like you did by [a,2a],[4a,8a], I did it by building it by other function, I never imagine building a function with a series
    Thank you.
     
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