# Can you give an understandable strategy to compute the asymptotic behaviour of $\sum_{2\leq n\leq x}\frac{\Lambda(n)\operatorname{Li}(n)}{n}$?

First I am looking a discussion about if the statements that I've deduced will find a good asymptotic formula, what of them you can discard as non useful,or one better than other, and ... second my main goal is obtain the asymptotic behaviour as $$x\to\infty$$ for $$\sum_{2\leq n\leq x}\frac{\Lambda(n)\operatorname{Li}(n)}{n},$$ where $$\Lambda(n)$$ is von Mangoldt function and $$\operatorname{Li}(x)=\int_2^x\frac{dt}{\log t}$$ is the logarithmic integral, I don't know if this exercise is in the literature but I want to solve reasoning what strategies (basically analysis, by summation, inversion formulas...these kind of tools, when I am looking to compute the asymptotic behaviour) and unconditionally, this is without assumption of conjectures and without special (high) methods in analytic number theory and too I want to know what is the meaning of research such solution, and it return us at the... First.

When I've used Abel's summation formula I can deduce two statements $$\sum_{2\leq n\leq x}\frac{\Lambda(n)}{n}\cdot \operatorname{Li}(n)=(\log x)\cdot \operatorname{Li}(x)-(x-2)+O\left(\operatorname{Li}(x)\right),$$ here I've used a consequence of Shapiro's Tauberian theorem $$\sum_{n\leq x}\frac{\Lambda(n)}{n}=\log x+O(1)$$, and it is know that $$O\left(\operatorname{Li}(x)\right)=O\left(\frac{x}{\log x}\right)$$. On the other hand $$\sum_{2\leq n\leq x}\Lambda(n)\cdot\frac{\operatorname{Li}(n)}{n}=\psi(x)\cdot\frac{\operatorname{Li}(x)}{x}-\int_2^x\frac{\psi(t)}{t\log t}dt+\int_2^x\frac{\psi(t)\operatorname{Li}(t)}{t^2}dt,$$ here $$\psi(x)=\sum_{n\leq x}\Lambda(n)$$.

Question 1. Now I am in the situation that I've used an asymptotic for $$\operatorname{Li}(t)$$ in the second integral of last identity, or perhaps I ask to me if is better that I should to use an asymptotic for the second Chebyshev function $$\psi(t)$$, further how I know that the first asymptotic identity is better or worse than the second that I'm theorizing now (I say the second)? Can you discard some of these identities, one as baddest than the other one?

My following question is to deepen in this my disorientation. I did and thought about some tools. I have the cited consequence of Shapiro's theorem, too perhaps I can use Hölder's inequality for $$\sum_{2\leq n\leq x}\frac{\Lambda(n)\operatorname{Li}(n)}{n}$$, where for example $$a_n=\frac{\Lambda(n)}{n}$$ for $$n>1$$, and $$b_n=\operatorname{Li}(n)$$ if $$n\geq 2$$ and $$b_1:=0$$, and take $$p=1+\epsilon$$. Too I imagine and ask to me: could be useful Selberg's asymptotic formula? I say: why no?, if I don't know what I doing, I write for example as an easy deduction that I can relate with a previous identity, since these share summands $$\int_2^x\frac{\psi(t)}{t\log t}dt+\int_{2}^x\frac{\sum_{n\leq t}\Lambda(n)\psi(\frac{t}{n})}{t\log^2 t}dt=2\operatorname{Li}(x)+O\left(\operatorname{Li}_2(x)\right),$$ and it is known that the logarithmic integral $$\operatorname{Li}_2(x)=\int_2^x\frac{dt}{\log^2t}=O\left(\frac{x}{\log^2 x}\right)$$.

Question 2. I did these computations, and ask to you: can you discard (using a mathematical reasoning, from a practical point of view and objective) these usages of for example Hölder, Selberg or Shapiro?

The

Main Question. How do you compute with a clear strategy the asymptotic behaviour as $$x\to\infty$$ of $$\sum_{2\leq n\leq x}\frac{\Lambda(n)}{n}\cdot \operatorname{Li}(n)$$ without expensive methods in analytic number theory?

• I believe that there is a mistake in the computation of the error term in the deduction from Selberg's asymptotic formula, since I take $\int O(something)=O(\int(something))$.
– user243301
Commented Feb 20, 2016 at 7:16
• Thanks for the useful and detailed answer given by the user, below.
– user243301
Commented Feb 21, 2016 at 8:08

In the first application of the Abel's summation you get $$\sum_{n\leq x}\frac{\Lambda\left(n\right)}{n}\textrm{Li}\left(n\right)=\left(\log\left(x\right)+O\left(1\right)\right)\textrm{Li}\left(x\right)-\int_{2}^{x}\left(\log\left(t\right)+O\left(1\right)\right)\left(\textrm{Li}\left(t\right)\right)'dt=$$ and now since $\textrm{Li}\left(n\right)=\frac{n}{\log\left(n\right)}+O\left(\frac{n}{\log^{2}\left(n\right)}\right)$ we have $$=x+O\left(\frac{x}{\log\left(x\right)}\right)-x+2+O\left(\frac{x}{\log\left(x\right)}\right)=O\left(\frac{x}{\log\left(x\right)}\right).$$ For the second application, we can use the asymptotic $$\psi\left(x\right)=x+O\left(\frac{x}{\log^{2}\left(x\right)}\right)$$ and so we can get $$\sum_{n\leq x}\frac{\Lambda\left(n\right)}{n}\textrm{Li}\left(n\right)=\frac{\psi\left(x\right)}{x}\textrm{Li}\left(x\right)-\int_{2}^{x}\frac{\psi\left(t\right)}{t\log\left(t\right)}dt+\int_{2}^{x}\frac{\psi\left(t\right)}{t^{2}}\textrm{Li}\left(t\right)dt=$$ $$=\textrm{Li}\left(x\right)+O\left(\frac{\textrm{Li}\left(x\right)}{\log^{2}\left(x\right)}\right)-\textrm{Li}\left(x\right)+O\left(\int_{2}^{x}\frac{1}{\log^{3}\left(t\right)}dt\right)+\int_{2}^{x}\frac{\textrm{Li}\left(t\right)}{t}dt+O\left(\int_{2}^{x}\frac{\textrm{Li}\left(t\right)}{t\log^{2}\left(t\right)}\right).$$ Now observe that we have, integrating by parts, $$\int_{2}^{x}\frac{\textrm{Li}\left(t\right)}{t}dt=\log\left(x\right)\textrm{Li}\left(x\right)-x+2\tag{1}$$ so if we take again $\textrm{Li}\left(n\right)=\frac{n}{\log\left(n\right)}+O\left(\frac{n}{\log^{2}\left(n\right)}\right)$ we'll get $$\int_{2}^{x}\frac{\textrm{Li}\left(t\right)}{t}dt=O\left(\frac{x}{\log\left(x\right)}\right)$$ which is bigger than the other errors, since $$O\left(\frac{\textrm{Li}\left(x\right)}{\log^{2}\left(x\right)}\right)=O\left(\int_{2}^{x}\frac{1}{\log^{3}\left(t\right)}dt\right)=O\left(\int_{2}^{x}\frac{\textrm{Li}\left(t\right)}{t\log^{2}\left(t\right)}\right)=O\left(\frac{x}{\log^{3}\left(x\right)}\right)\tag{2}$$ so in this case (and only in this, because in the first part we have $O\left(\textrm{Li}\left(x\right)\right)$ and so we are “forced” to $x/\log\left(x\right)$) we can use a better approximation for $\textrm{Li}\left(x\right)$. It is possible to prove, iterating the integration by parts, that $$\textrm{Li}\left(x\right)=\frac{x}{\log\left(x\right)}\sum_{k=0}^{n}\frac{k!}{\log^{k}\left(x\right)}+O_{n}\left(\frac{x}{\log^{n+2}\left(x\right)}\right)$$ so if we take $n=2$ we have $$\textrm{Li}\left(x\right)=\frac{x}{\log\left(x\right)}+\frac{x}{\log^{2}\left(x\right)}+\frac{2x}{\log^{3}\left(x\right)}+O\left(\frac{x}{\log^{4}\left(x\right)}\right)$$ and so if we apply this in $(1)$ we get $$\int_{2}^{x}\frac{\textrm{Li}\left(t\right)}{t}dt=\frac{x}{\log\left(x\right)}+\frac{2x}{\log^{2}\left(x\right)}+O\left(\frac{x}{\log^{3}\left(x\right)}\right)\tag{3}$$ and finally, by $(2)$ and $(3)$ we have $$\sum_{n\leq x}\frac{\Lambda\left(n\right)}{n}\textrm{Li}\left(n\right)=\frac{x}{\log\left(x\right)}+\frac{2x}{\log^{2}\left(x\right)}+O\left(\frac{x}{\log^{3}\left(x\right)}\right)$$ which is a better result.