For a real $x\geq 2$ and when we take $y= [x]$ its integer part, I am trying to study the asymptotic size or growth of $$\sum_{\substack{2\leq n\leq y,\text{n prime}}}n\log\log n,$$ I believe that combining Abel's summation formula with known theorems from analytic number theory perhaps it is possible to prove that it is asymptotically equivalent as $x\to\infty$ with $c\cdot\pi(y)\sigma(y)$, where $\pi(y)$ is the prime-counting function and $\sigma(y)$ is the sum of divisors function. Then

Question. Can you prove or refute $$\sum_{\substack{2\leq n\leq y,\text{n prime}}}n\log\log n\sim c\cdot\pi(y)\sigma(y),$$ for a positive constant $c$? Thanks in advance.

I believe that is true, but I don't know how analyze each summand in my computations.

  • $\begingroup$ I want the vote up for the user who answered this question , thank you. $\endgroup$ – user243301 Feb 12 '16 at 19:34

By partial summation, \[\sum_{p \leq x} p \log \log p = x \log \log x \pi(x) - \int_{2}^{x} \pi(t) \left(\log \log t + \frac{1}{\log t}\right) \, dt. \] Now by the prime number theorem, there exists some constant $c > 0$ such that \[\pi(x) = \mathrm{Li}(x) + O\left(x e^{-c\sqrt{\log x}}\right),\] where \[\mathrm{Li}(x) = \int_{2}^{x} \frac{du}{\log u},\] and so \[\sum_{p \leq x} p \log \log p = x \log \log x \mathrm{Li}(x) - \int_{2}^{x} \int_{2}^{t} \frac{\log \log t}{\log u} \, du \, dt - \int_{2}^{x} \int_{2}^{t} \frac{1}{\log u \log t} \, du \, dt + O\left(x^2 e^{-c'\sqrt{\log x}}\right) \] for some constant $c' > 0$. Now \[\int_{2}^{x} \int_{2}^{t} \frac{\log \log t}{\log u} \, du \, dt = \int_{2}^{x} \frac{1}{\log u} \int_{u}^{x} \log \log t \, dt \, du, \] and via integration by parts, this is \[x \log \log x \mathrm{Li}(x) - \int_{2}^{x} \frac{u \log \log u}{\log u} \, du - \int_{2}^{x} \int_{u}^{x} \frac{1}{\log u \log t} \, dt \, du. \] So we get \[\sum_{p \leq x} p \log \log p = \int_{2}^{x} \frac{u \log \log u}{\log u} \, du + O\left(x^2 e^{-c'\sqrt{\log x}}\right). \] But this has nothing to do with the sum of divisors function! Indeed, the problem is that the sum of divisors function is too random: if $n$ is a prime, then $\sigma(n) = n + 1 \approx n$, while for composite $n$ it can be the case that $\sigma(n) \approx e^{\gamma_0} n \log \log n$.

  • $\begingroup$ I unerstadnd your words about the sum of divisor function, and I've written your computations to study these now, more! Very thnaks much @PeterHumphries $\endgroup$ – user243301 Feb 12 '16 at 7:19

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