The function $\mathrm{Li}_2(x)=\int_2^x\frac{dt}{\log^2t}$, its inverse and summation I am reading the more understandable mathematics in the section Preliminary Results of a paper in which the authors give a explanation of facts for the logarithmic integral and its inverse. In this post I ask about similar equations but now for a different logarithmic integral function, now I consider $\mathrm{Li}_2(x)=\int_2^x\frac{dt}{\log^2t}$, and its inverse $\mathrm{Li}_2^{-1}(x)$. My goal is understand more these facts (the genuine computations from the authors), and refresh basics about analysis. I know the relation between a function that has inverse, its graph, the function $x$ and the inverse function (I write this because could be useful to find the asyptotic that I ask in my question, I don't know how deduce it now).
I can write and know that
$$\mathrm{Li}_2(x)\sim\frac{(1+o(1))x}{\log^2x}\quad\text{  as }x\to\infty,$$
but, 

Question 1. Can you give the correspoding and similar asymptotic for $\mathrm{Li}_2^{-1}(x)$, as $x\to\infty$? Here, as I've said $\mathrm{Li}_2^{-1}(x)$ is the inverse function of the logarithmic integral $\int_2^x\frac{dt}{\log^2t}$. 

I know that by differentiation of a inverse function and by the Fundamental Theorem of Calculus I can write my explanation for a similar equation for $\mathrm{Li}_2(x)$ (following the equation that the authors write for the logarithmic integral)
$$\left(\mathrm{Li}_2^{-1}\right)'\left(\mathrm{Li}_2(x)\right)=\frac{1}{\frac{d}{dx}\left(\int_2^x\frac{dt}{\log^2t}\right)}=\log^2x,$$
and combining this last with the composition with $\mathrm{Li}^{-1}_2(x)$ one has the coresponding equation and compute $\left(\mathrm{Li}_2^{-1}\right)'(x)$ as 
$$\left(\mathrm{Li}^{-1}_2\right)'\left(\mathrm{Li}_2\left(\mathrm{Li}_2^{-1}(x)\right)\right)=\log^2\left(\mathrm{Li}_2^{-1}(x)\right).$$

Question 2. Can you use the asymptotic that you've computed as answer of Question 1, to get an asymptotic as $x\to\infty$ for $\log^2\left(\mathrm{Li}_2^{-1}(x)\right)$?

I believe, following similar computations that authors give for their logarithmic integral function, that now I with $\mathrm{Li}_2(x)$ can write 
$$\sum_{1<m<n}\mathrm{Li}^{-1}_2(x)=\int_2^{\mathrm{Li}_2^{-1}(n)}\frac{tdt}{\log^2t}+O\left(\sum_{1\leq m\leq n}\log^2 m\right).$$

Question 3 (Now this question is optional, thus 
  it is not necessary to answer). I say that I've asked to me if previous identity is right when I write the similar statement for which I believe that holds. Can you refute previous statement? Can you prove it using previous answers to my questions (I believe that with these the authors can prove their identity), or at least give a reasonable approximation for other methods (I say perhaps partial summation)?

 A: $$\mathrm{Li}_2(x)\sim\frac{(1+o(1))x}{\log^2x}\quad\text{  as }x\to\infty,$$
Means that 
$$\log(\mathrm{Li}_2(x))\sim \log\left(\frac{(1+o(1))x}{\log^2x}\right)\quad\text{  as }x\to\infty,$$
therefore we also have 
$$\log(\mathrm{Li}_2(x))\sim \log((1+o(1))x)-\log(\log^2x)\quad\text{  as }x\to\infty,$$
$$\log(\mathrm{Li}_2(x))\sim \log(x)-2\log(\log x)+\log(1+o(1)) \quad\text{  as }x\to\infty,$$
$$\log(\mathrm{Li}_2(x))\sim \log(x)-2\log(\log x)+o(1) \sim (1+o(1))\log(x) \quad\text{  as }x\to\infty,$$
$$\log^2(\mathrm{Li}_2(x)) \sim (1+o(1))\log^2(x) \quad\text{  as }x\to\infty,$$
This means that 
$$\mathrm{Li}_2(x)\sim\frac{(1+o(1))x}{\log^2x} \sim \frac{(1+o(1))x}{\log^2(\mathrm{Li}_2(x))}\quad\text{  as }x\to\infty,$$
And hence $$(1+o(1))x \sim \mathrm{Li}_2(x) \log^2(\mathrm{Li}_2(x))$$
Rewriting gives $x \sim (1+o(1))\mathrm{Li}_2(x) \log^2(\mathrm{Li}_2(x))$.
Substitute $x=\mathrm{Li}_2^{-1}(y)$ in to get 
$$\mathrm{Li}_2^{-1}(y) \sim (1+o(1))y \log^2(y)$$

This also means that:
$$\log(\mathrm{Li}_2^{-1}(y)) \sim \log((1+o(1))y \log^2(y)) \sim (1+o(1))\log(y)$$
$$\log^2(\mathrm{Li}_2^{-1}(y)) \sim \log((1+o(1))y \log^2(y)) \sim (1+o(1))\log^2(y)$$
