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Thread: OffSet Logarithmic Integral

  1. #1
    Sep 2010

    OffSet Logarithmic Integral

    Hello, I've posted a thread several threads ago about some "intro" questions to a problem and I'm now trying to deal with the last questions of the problem itself.
    We are studying $\displaystyle $\forall x\geq 1\; f(x)=\int_2^x \dfrac 1{lnt}\; dt$$

    What I've managed (some of it with your help) to prove :

    $\displaystyle $\int_2^x \dfrac 1{({lnt})^2}\; dt=o(f(x))$$

    $\displaystyle $f(x)=\dfrac x{lnx}-\dfrac 2{ln2}+\displaystyle\int_{2}^{x} \dfrac 1{({lnt})^2} \, \mathrm dt$$

    $\displaystyle $f(x)\sim_{x\to+\infty}\dfrac x{lnx}$$

    $\displaystyle $f(x)\sim_{x\to 1} ln(x-1)$$

    I've also proven that f is a bijection of ]1,+∞[ on ]-∞,+∞[

    Here are the questions that are a problem to me : find an equivalent of $\displaystyle $f(x)-\dfrac x{lnx}$ when x\rightarrow +\infty$
    I'm not sure if I can directly say that it is $\displaystyle $-\dfrac 2{ln2}$$ directly because of the second relation I've written.

    Then I've pretty much have some questions about the bijection which are to study the convergence of those series with different alphas.

    $\displaystyle $\Displaystyle \sum_{n\geq 3} \dfrac 1{({f(n)})^\alpha} $$
    Which is, I believe a series of Bertrand (which I've already studied so it's all good) and
    $\displaystyle $\Displaystyle \sum_{n\geq 0} \dfrac 1{({{f}^{-1}(n)})^\alpha} $$

    If you've got any sort of help with the equivalent and the series of the bijection (which I am totally unable to find because I believe that we cannot express the bijection of f with "classic" functions.

    Thank you very much and sorry for my bad Latex, I'm a beginner ^^
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  2. #2
    Senior Member
    Mar 2010
    $f(x)=\dfrac x{lnx}-\dfrac 2{ln2}+\displaystyle\int_{2}^{x} \dfrac t{({lnt})^2} \, \mathrm dt$

    \displaystyle\int_{2}^{x} \dfrac t{({lnt})^2}\mathrm dt=\frac {1}{2}\frac {t^2}{ln^2t} |_2^x+\displaystyle\int_{2}^{x} \dfrac {t^2}{({lnt})^3}\mathrm dt

    This is not useful because free term goes to infinity.
    May be this will better
    \displaystyle\int_{2}^{x} \frac {\mathrm dt}{lnt}=t \; ln(lnt)|_2^x-\displaystyle\int_{2}^{x} ln(lnt)\mathrm dt
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  3. #3
    Sep 2010
    Hello and thanks for your answer. I don't know how this can help me though since I'm trying to find the equivalent of $\displaystyle $f(x)-\dfrac x{lnx}$$. Does anyone have an idea on how to find it or can help me with the series?
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  4. #4
    Grand Panjandrum
    Nov 2005
    You have:

    $\displaystyle \displaystyle f(x)=\int_2^x \dfrac{1}{\ln(t)}dt$

    $\displaystyle \displaystyle f(x)=\dfrac{x}{\ln(x)}-\dfrac{2}{\ln(2)}+\int_2^x \dfrac{1}{(\ln(t))^2}dt$


    $\displaystyle \displaystyle \dfrac{1}{(\ln(t))^2}dt=o(f(x))$


    $\displaystyle \displaystyle f(x)-\dfrac{x}{\ln(x)}=-\dfrac{2}{\ln(2)}+\int_2^x \dfrac{1}{(\ln(t))^2}dt=-\dfrac{2}{\ln(2)}+o(f(x))$

    if that is any help?

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