OffSet Logarithmic Integral

**DwProd** · September 29th 2010, 08:48 AM

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 $\forall x\geq 1\; f(x)=\int_2^x \dfrac 1{lnt}\; dt$

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

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

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

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

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

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 $$f(x)-\dfrac x{lnx}$ when x\rightarrow +\infty$
I'm not sure if I can directly say that it is $-\dfrac 2{ln2}$ 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.

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

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 ^^

**zzzoak** · September 29th 2010, 03:22 PM

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

$ \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 $

**DwProd** · September 30th 2010, 03:03 AM

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 $f(x)-\dfrac x{lnx}$ . Does anyone have an idea on how to find it or can help me with the series?

**CaptainBlack** · September 30th 2010, 10:53 PM

You have:

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

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

and:

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

So:

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

if that is any help?

CB

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