In a paper I am reading, I've come across the estimate

$$ 2\pi\sum_{\substack{\sigma<\beta<\sigma_0\\ T<\gamma\leq 2T}}(\beta-\sigma)=\underbrace{\int_T^{2T}\log|\zeta(\sigma+it)|\,dt}_{I_\sigma}-\underbrace{\int_T^{2T}\log|\zeta(\sigma_0+it)|\,dt}_{I_{\sigma_0}}+O(\log T), $$ where $\frac{1}{2}\leq\sigma\leq1$ is fixed, $\sigma_0=4$, and $\rho=\beta+i\gamma$ is a nontrivial zero of Riemann's zeta function.

I'm fine up to this point. Shortly after this, the author makes the statement that the above estimate holds for all complex numbers $a$, so in particular, when $a=0$, it becomes

$$ I_\sigma=O(\log T). $$

I don't understand how to prove this estimate.

Assuming Riemann's hypothesis (RH), the sum on the left-hand side becomes $0$. Thus, we need only to show that $I_{\sigma_0}=O(\log T)$.

A trivial estimate doesn't work because that will only give us $O(T)$ which isn't good enough for my purposes. I've attempted to search for estimation on the logarithm of the zeta function but have failed at successfully finding anything helpful, so I'm asking here.

  • $\begingroup$ What is the title of this article? $\endgroup$ Jun 23 '15 at 11:22
  • $\begingroup$ @MarcoCantarini: It is Kai-Man Tsang's thesis; a link to it is math.sjsu.edu/~goldston/TsangThesis.htm ... The relevant pages are pp.109-110. $\endgroup$
    – Clayton
    Jun 23 '15 at 13:10
  • $\begingroup$ @Clayton : when $\Re(s) \to \infty$, $\zeta(s) \sim 1 + 2^{-s}$, $\ln \zeta(s) \sim 2^{-s}$, that's all ? $\endgroup$
    – reuns
    Jul 18 '15 at 8:23
  • $\begingroup$ @reuns: What you have written is true (obviously), but it doesn't address the problem here; I'm not letting $\Re(s)\to\infty$, but rather staying along a vertical line with $\sigma_0=4$, as Marco Cantarini has shown below. $\endgroup$
    – Clayton
    Jul 18 '15 at 18:55

We can show something stronger. If $\sigma>1 $ we know that holds, by Euler product, $$\log\left(\zeta\left(s\right)\right)=\sum_{p}\sum_{k\geq1}\frac{1}{kp^{ks}} $$ then $$\int_{T}^{2T}\log\left|\zeta\left(4+it\right)\right|dt=\textrm{Re}\left(\sum_{p}\sum_{k\geq1}\frac{1}{kp^{4k}}\int_{T}^{2T}\exp\left(-itk\log\left(p\right)\right)dt\right)\ll\sum_{n\geq1}\frac{1}{n^{4}}\ll1. $$

  • $\begingroup$ I'm sorry; would you mind elaborating a bit more on the integral and explaining why it is $O(1)$? $\endgroup$
    – Clayton
    Jul 18 '15 at 18:56
  • $\begingroup$ @Clayton $$\int_{T}^{2T}e^{-itk\log\left(p\right)}dt=$$ $$=\left.\frac{ie^{-ikt\log\left(p\right)}}{k\log\left(p\right)}\right|_{T}^{2T}$$ $\endgroup$ Jul 19 '15 at 8:06

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.