I would like to prove that

Assume RH. Let $T$ large, $2\leq x \leq T^2$ and $T\leq t \leq 2T$, then $$ \log|\zeta(1/2+it))|\leq \Re\left(\sum_{p\leq x}\frac{1}{p^{1/2+1/\log x+it}}\frac{\log(x/p)}{\log x}+\sum_{\min\{\sqrt{x},\log T\}}\frac{(1/2)}{p^{1+2it}}\right)+\frac{\log T}{\log x}+\mathcal{O}(1) $$

which is Proposition 1 from the paper "Sharp Conditional Bound for Moments of the Riemann Zeta Function" by Adam Harper.

This is a modification of the Proposition and Lemma 2 from the paper "Moments of the Riemman Zeta" by Kannan Soundararajan.

What I am told to do is to take $\lambda=1$ in Soundararajan's Proposition. In this way we get $$ \log|\zeta(1/2+it)|\leq \Re\left(\sum_{n\leq x}\frac{\Lambda(n)}{n^{1/2+1\log x+it}}\frac{\log(x/n)}{\log x}\right)+\frac{\log T}{\log x}+\mathcal{O}\left(\frac{1}{\log x}\right) $$ So what it remains to do is to estimate the contribution to the sum given by the prime powers. Harper says that we have $$ \Re\left(\sum_{p\leq \sqrt{x}}\frac{(1/2)}{p^{1+2/\log x+2it}}\frac{\log (x/p^2)}{\log x}\right)=\Re\left(\sum_{p\leq \sqrt{x}}\frac{(1/2)}{p^{1+2it}}\right)+\mathcal{O}(1)=\Re\left(\sum_{p\leq\min\{\sqrt{x},\log T\}}\frac{(1/2)}{p^{1+2it}}\right)+\mathcal{O}(1) $$ which is justified, "assuming the Riemann hypothesis, by standard explicit formula arguments". Can anybody tell me where can I found a detail proof of the above equalities or how to prove it? (I do not know what "standart explicit fomula arguments" stands for). The same arguments are used by Soundararajan in his proof of Lemma 2 (which I could not follow either). I found a detail proof of that Lemma 2 in the paper "Upper bounds for the Moments of $\zeta^{\prime}(\rho)$" by Micah Milinovich (it is the proof of Lemma 3.1.).

Thank you for your help!

As suggested by @2734364041 I can use the Mellin inversion to obtain $$ \sum_{n\leq z}\frac{\Lambda(n)}{n^{2it}} = \frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}-\frac{\zeta'}{\zeta}(s+2it)\frac{z^s}{s}ds=\frac{z^{1-2it}}{1-2it} - \sum_{\rho}\frac{z^{\rho-2it}}{\rho-2it} + O(1), $$ (where in the last equality we have shifted the line of integration to the left, till infinity, and the contribution of the residues of the trivial zeros gives us $\mathcal{O}(1)$). We now use the reasoning of page 111 from Multiplicative Number Theory - Davenport to bound the contribution of the sum over the zeros. In particular under RH we have $$ |z^{\rho-2it}|=z^{1/2} $$ We then use the bound $\sum_{|\gamma|\leq T}\frac{1}{\gamma}\ll(\log T)^2$ from Davenport to get $$ \sum_{|\gamma|\leq T}\frac{1}{\rho-2it}\ll(\log T)^2 $$ as well and thus $$ \sum_{|\gamma|\leq T}\frac{z^{\rho-2it}}{\rho-2it}\ll z^{1/2}(\log T)^2 $$ Using now that $T\leq t\leq 2T$ we have $$ \frac{z^{1-2it}}{1-2it}\ll \frac{z}{T} $$ hence $$ \sum_{n\leq z}\frac{\Lambda(n)}{n^{2it}}\ll \frac{z}{T}+z^{1/2}(\log T)^2 $$ We want to use the above estimate for $z\leq \sqrt{x}\leq T$ thus $z/T\ll 1$ and we can rewrite the above estimate more simply as $$ \sum_{n\leq z}\frac{\Lambda(n)}{n^{2it}}\ll z^{1/2}(\log T)^2. $$ Suppose we are able to show that the contribution of prime powers $p^k$ with $k\geq 2$ is negligible, i.e. of the order of $z^{1/2}(\log T)^2$ (how should I show it? Should I use that $\psi(z)-\theta(z)\ll\sqrt{z}(\log z)^2$ using the estimates for $\psi(z)$ and $\theta(z)$ under RH?) then we would get $$ \sum_{p\leq z}\frac{\log p}{p^{2it}}\ll z^{1/2}(\log T)^2 $$ By applying the partial summation we get $$ \sum_{(\log T)^{10}<p\leq \sqrt{x}}\frac{1}{p^{1+2it}}=\sum_{(\log T)^{10}<p\leq \sqrt{x}}\frac{\log p}{p^{2it}}\frac{1}{p\log p}\leq \left(\sum_{p\leq \sqrt{x}}\frac{\log p}{p^{2it}}\right)\frac{1}{x^{1/2}\log x^{1/2}}+\int_{(\log T)^{10}}^{\sqrt{x}}\frac{1}{u^2\log u}\left(\sum_{p\leq u}\frac{\log p}{p^{2it}}\right)du $$ Inserting now the above estimate we get $$ \sum_{(\log T)^{10}<p\leq \sqrt{x}}\frac{1}{p^{1+2it}}\ll x^{1/4}(\log T)^2\frac{1}{x^{1/2}\log x}+\int_{(\log T)^{10}}^{\sqrt{x}}\frac{1}{u^2\log u}u^{1/2}(\log T)^2du $$ Using that $(\log T)^{10}\leq \sqrt{x}$ (otherwise the sum is empty) we have that $(\log T)^2\leq x^{1/10}$ and therefore the first term of the RHS can be bounded by $$ x^{1/4}x^{1/10}\frac{1}{x^{1/2}\log x}=\frac{1}{x^{3/20}\log x}=\mathcal{O}(1) $$ The second term of the RHS is instead $$ \int_{(\log T)^{10}}^{\sqrt{x}}\frac{1}{u^2\log u}u^{1/2}(\log T)^2 du\leq (\log T)^2\int_{(\log T)^{10}}^{\infty}\frac{1}{u^{3/2}}du\ll (\log T)^2 \left(\frac{1}{(\log T)^{10}}\right)^{1/2}=\mathcal{O}(1) $$ hence showing the claim. Is everything right or did I make some mistake? and how do I show the fact that the contribution to $\sum_{p\leq z}\frac{\Lambda(n)}{n^{2it}}$ from the primes powers (with exponent $\geq 2$) is negligible?

For the remaining range $[\log T, (\log T)^{10}]$ we have $$ \sum_{\log T<p\leq (\log T)^{10}}\frac{1}{p^{1+2it}}\ll\sum_{\log T<p\leq (\log T)^{10}}\frac{1}{p}=\log\log(\log T)^{10}-\log\log \log T+\mathcal{O}(1)=\log 10+\mathcal{O}(1)=\mathcal{O}(1). $$

  • $\begingroup$ Looking at $\psi_1(x)=\int_1^x \psi(y)dy, \sum_\rho \frac{x^\rho}{\rho(\rho+1)}, \frac{1}{s+1} \frac{\zeta'(s)}{\zeta(s)}, \sum_\rho \frac{1}{\rho+1} \frac{1}{s-\rho}$ is easier because everything converges absolutely,the density of zeros implies error terms when summing over the zeros up to $T$ and replacing $\int_1^\infty$ by $\int_1^N$ in the Mellin transform of $\psi_1(x)$ and its sum over zeros approximation $\endgroup$ – reuns May 16 '19 at 20:25
  • $\begingroup$ I'm sorry, I cannot follow you. What should I do exactly? $\endgroup$ – asd May 16 '19 at 20:49

Observe via Mellin inversion that

$\sum_{n\leq x}n^{-2it}\Lambda(n) = \frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}-\frac{\zeta'}{\zeta}(s+2it)\frac{x^s}{s}ds=\frac{x^{1-2it}}{1-2it} - \sum_{\rho}\frac{x^{\rho-2it}}{\rho-2it} + O(1),$

where $\rho=\beta+i\gamma$ ranges over the zeros of $\zeta(s)$. On RH, $\beta=1/2$ always. Estimate the sum over zeros as in Chapters 17 and 18 in Davenport. The contribution to the LHS arising from $n = p^k$ with $k\geq 2$ is negligible, on the order of $\sqrt{x}$ ignoring logs. To properly estimate the sum over zeros, you will need to truncate the contour integral and estimate the remainder separately; again, see Chapters 17 and 18 of Davenport.

This is what "standard explicit formula arguments" means (express sum as a contour integral, shift to the left, estimate sum over zeros). This is also the same route that Micah Milinovich takes in his lemma (though he estimates the integral directly rather than expanding it into a sum over zeros). The desired result should follow from partial summation, once you separately consider the ranges $[1,\log T]$, $[\log T,(\log T)^{10}]$, and $[(\log T)^{10},\sqrt{x}]$ as described by Harper in the comments following his Proposition 1.

| cite | improve this answer | |
  • $\begingroup$ thanks for replying! What I get using your hints is that $$ \sum_{n\leq z} \frac{\Lambda(n)}{n^{2it}}\ll \frac{z}{T}+z^{1/2}(\log T)^2+\frac{z}{T}(\log zT)^2. $$ Since we have $z\leq x\leq \sqrt{T}$, the previous bound is $$\ll \frac{z}{T}+z^{1/2}(\log zT)^2 $$Is it right? Now I should use this bound, together with the partial summation to estimate $$\sum_{(\log T)^2\leq p\leq \sqrt{x}} \frac{1}{p^{1+2it}} $$ isn't it? $\endgroup$ – asd May 26 '19 at 12:02
  • $\begingroup$ when using the partial summation (en.wikipedia.org/wiki/Abel%27s_summation_formula) don't I need also a lower bound for $$\sum_{n\leq z}\frac{\Lambda(n)}{n^{2it}} $$ $\endgroup$ – asd May 26 '19 at 12:43
  • $\begingroup$ I have edited my question and tried to use your help to prove the claim. Does it make sense? How should I show that the contribution to $\sum_{p\leq z}\frac{\Lambda(n)}{n^{2it}}$ from the prime power $p^k$ with $k\geq 2$ is negligible? $\endgroup$ – asd May 26 '19 at 16:36
  • 1
    $\begingroup$ @asd It is straightforward to show that $|\psi(x)-\theta(x)|\ll (\log x)\pi(\sqrt{x}) + (\log x)\sum_{3\leq k\leq (\log x)/(\log 2)}\pi(x^{1/k})$. Now invoke Chebyshev's bound. $\endgroup$ – 2734364041 May 29 '19 at 9:18
  • $\begingroup$ I thought I could also use that under RH we have $$ \psi(x)=x+\mathcal{O}(x^{1/2}(\log x)^2) $$ and $$ \theta(x)=x+\mathcal{O}(x^{1/2}(\log x)^2) $$ (see chapter 13 of Montgomery-Vaughan) and therefore $$ \psi(x)-\theta(x)\ll x^{1/2}(\log x)^2. $$ $\endgroup$ – asd May 29 '19 at 10:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.