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 ^^