I would like to show that $$ \sum_{p\leq x} \frac{1}{p^{1+2/\log x}}\left(\frac{\log\left(x/p\right)}{\log(x)}\right)^2=\log\log x +\mathcal{O}(1) $$

What I have tried

Since we know that $$ \sum_{p\leq x}\frac{1}{p}=\log\log x+\mathcal{O}(1) $$ (see this post) I thought I could use Abel's summation to prove the above estimate. Abel's summation says that given a sequence of real numbers $(a_n)_{n=0}^{\infty}$, we can define the partial sum $$ A(t)=\sum_{n\leq t}a_n $$ and, if $\phi$ is a continously differentiable function on $[1,x]$, we have $$ \sum_{1\leq n\leq x}a_n\phi(n)=A(x)\phi(x)-\int_1^x A(u)\phi^{\prime}(u)\,du $$ In our case I would take $$ a_n = \begin{cases} 1/p &\text{if } n=p \text{ is prime}, \\ 0 &\text{otherwise}, \end{cases} $$ and $$ \phi(t)=\frac{1}{t^{2/\log x}}\left(\frac{\log \left(x/t\right)}{\log x}\right)^2 $$ but in this way I have $$ \phi(x)=0 $$ and therefore I am only left with the integral from Abel's summation formula which looks pretty ugly to me $$ \sum_{p\leq x} \frac{1}{p^{1+2/\log x}}\left(\frac{\log\left(x/p\right)}{\log(x)}\right)^2=\int_1^x \left(\sum_{p\leq u}\frac{1}{p}\right)\frac{2\log\left(x/u\right)\left(\log(x/u)+\log x\right)}{u^{2/\log x+1}(\log x)^3}\,du $$ Is there any another way I could proceed? Do you have any hint? Thank you for your help!


Since $$ \left(\frac{\log\left(x/p\right)}{\log(x)}\right)^2\leq 1 $$ and $$ \frac{1}{p^{2/\log x}}\leq 1 $$ I clearly have that $$ \sum_{p\leq x} \frac{1}{p^{1+2/\log x}}\left(\frac{\log\left(x/p\right)}{\log(x)}\right)^2\leq \sum_{p\leq x}\frac{1}{p}=\log\log x+\mathcal{O}(1) $$ hence I only need to prove that I similar lower bound holds too.

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    $\begingroup$ Do you have any good reason to look at $\sum_{p\leq x} \frac{1}{p^{1+2/\log x}}\left(\frac{\log\left(x/p\right)}{\log(x)}\right)^2$ ? $\endgroup$ – reuns Jun 10 at 11:53
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    $\begingroup$ I am (still) going through the paper "Sharp conditional bounds for the moments of the Riemann Zeta-Function" by A. Harper. In the last section of the paper he shows that, even if we are no able to rigorously get the right order of the constant in the asymptotic of the 2k-th moment, we can recover order of the conjectured factor $f(k)$ if we treat Dirichlet polynomials a Gaussian random variable. In this case $$\sum_{p\leq x} \frac{1}{p^{1+2/\log x}}\left(\frac{\log\left(x/p\right)}{\log(x)}\right)^2$$ would be the variance of the Dirichlet polynomial under consideration. $\endgroup$ – asd Jun 10 at 12:03

Here is a general strategy. You are interested in $$\sum_{p \leq x} \frac{1}{p^{1 + \frac{2}{\log x}}} - \frac{2}{\log x} \sum_{p \leq x} \frac{\log p}{p^{1 + \frac{2}{\log x}}} + \frac{1}{(\log x)^2} \sum_{p \leq x} \frac{(\log p)^2}{p^{1 + \frac{2}{\log x}}}.$$ We use partial summation, noting that $$\sum_{p \leq x} \frac{1}{p} = \log \log x + b + O\left(\frac{1}{\log x}\right), \qquad \sum_{p \leq x} \frac{\log p}{p} = \log x + O(1), \qquad \sum_{p \leq x} \frac{(\log p)^2}{p} = \frac{(\log x)^2}{2} + O(1).$$ Here $b$ is some explicit constant.

The first terms via partial summation are $$e^{-2} \log \log x + b e^{-2}, \qquad -2e^{-2}, \qquad \frac{e^{-2}}{2}.$$ The error is $O(1/\log x)$ for the first two and $O(1/(\log x)^2)$ for the third.

For the second terms, we note that $$\frac{d}{dt} \frac{1}{t^{\frac{2}{\log x}}} = -\frac{2}{\log x} \frac{1}{t} \exp\left(-\frac{2 \log t}{\log x}\right),$$ and so we must evaluate $$\frac{2}{\log x} \int_{2}^{x} \log \log t \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}, \qquad \frac{2b}{\log x} \int_{2}^{x} \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}, \qquad -\frac{4}{(\log x)^2} \int_{2}^{x} \log t \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}, \qquad \frac{1}{(\log x)^3} \int_{2}^{x} (\log t)^2 \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}.$$ (There are also error terms; the method below shows that they are $O(\log \log x/\log x)$.) We make the change of variabes $t \mapsto e^t$, so that these become $$\frac{2}{\log x} \int_{\log 2}^{\log x} \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt, \qquad \frac{2b}{\log x} \int_{\log 2}^{\log x} \exp\left(-\frac{2 t}{\log x}\right) \, dt, \qquad -\frac{4}{(\log x)^2} \int_{\log 2}^{\log x} t \exp\left(-\frac{2 t}{\log x}\right) \, dt, \qquad \frac{1}{(\log x)^3} \int_{\log 2}^{\log x} t^2 \exp\left(-\frac{2 t}{\log x}\right) \, dt.$$ The latter three can be explicitly evaluated (directly for the second, via integration by parts for the last two); they give $$-b e^{-2} - b \exp\left(-\frac{2 \log 2}{\log x}\right), \qquad 3e^{-2} - \exp\left(-\frac{2 \log 2}{\log x}\right) \left(\frac{2 \log 2}{\log x} + 1\right), \qquad -\frac{5}{4} e^{-2} + \frac{1}{4} \exp\left(-\frac{2 \log 2}{\log x}\right) \left(\frac{2 (\log 2)^2}{(\log x)^2} + \frac{2 \log 2}{\log x} + 1\right).$$ Note that $\exp(-2\log 2/\log x) = 1 + O(1/\log x)$. For the former, we integrate by parts once, getting $$-e^{-2} \log \log x + \log \log 2 \exp\left(-\frac{2 \log 2}{\log x}\right) + \int_{\log 2}^{\log x} \frac{1}{t} \exp\left(-\frac{2 t}{\log x}\right) \, dt.$$ For this second term, we make the change of variables $t \mapsto \frac{\log x}{2} t$, so that this becomes $$\int_{\frac{2 \log 2}{\log x}}^{2} \frac{e^{-t}}{t} \, dt = E_1\left(\frac{2 \log 2}{\log x}\right) - E_1(2),$$ where $E_1(z)$ is the exponential integral. Since $E_1(z) = -\log z - \gamma_0 + O(z)$ about $z = 0$, we get additional terms $$\log \log x - \log 2 - \log \log 2 - \gamma_0 - E_1(2) + O\left(\frac{1}{\log x}\right).$$

So now we combine everything and find that the desired asymptotic is $$\log \log x - b + \frac{e^{-2}}{4} + \frac{1}{4} - \log 2 - \gamma_0 - E_1(2) + O\left(\frac{\log \log x}{\log x}\right).$$

Alternatively, you can use your approach to get the expression $$\frac{4}{\log x} \int_{2}^{x} \log \log t \exp\left(-\frac{2 \log t}{\log x}\right) \, \frac{dt}{t} - \frac{6}{(\log x)^2} \int_{2}^{x} \log t \log \log t \exp\left(-\frac{2 \log t}{\log x}\right) \, \frac{dt}{t} + \frac{2}{(\log x)^3} \int_{2}^{x} (\log t)^2 \log \log t \exp\left(-\frac{2 \log t}{\log x}\right) \, \frac{dt}{t}.$$ (Note that you seem to have calculated the derivative incorrectly.) We again make the change of variables $t \mapsto e^t$, yielding $$\frac{4}{\log x} \int_{\log 2}^{\log x} \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt - \frac{6}{(\log x)^2} \int_{\log 2}^{\log x} t \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt + \frac{2}{(\log x)^3} \int_{\log 2}^{\log x} t^2 \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt.$$ Now integrate by parts repeatedly, integrating the exponential and differentiating everything else.

  • $\begingroup$ thank you very much! I was trying to use the hint you'd given me in the comment to the other post and didn't realize you had already given a thorough answer. I will try go through it now! $\endgroup$ – asd Jun 27 at 18:40
  • $\begingroup$ finally able to work over all these integrals! thank you again, I didn't know how to work with the exponential integral when I first tried to use Abel's summation! $\endgroup$ – asd Jun 28 at 8:51

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