Let $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$$ be the Riemann zeta function. The fact that we can analytically extend this to all of $\mathbb{C}$ and can find a zero free region to the left of the line $Re(s)=1$ shows that $$\pi(x) := |\{p\leq x: p \mbox{ prime }\}| \sim \frac{x}{\log(x)}.$$ Moreover, improving the zero free region improves the error term on $\pi(x)$ with the Riemann hypothesis giving us the best possible error term.

However, there are elementary proofs for this fact about $\pi(x)$ which does not rely on using $\zeta(s)$. My question is, just using the knowledge that $$\pi(x) = \frac{x}{\log(x)} + ET$$ for some error term I call $ET$, can you show that $\zeta(s)$ can be analytically continued with a zero free region to the left of the line $Re(s)=1$, where this region depends on $ET$.

For example, if you assume that $ET = x^{1-\epsilon}$ for some $\epsilon>0$, then can you show that $\zeta(s)$ can be analytically continued to the region $Re(s) > 1-\epsilon$ and that $\zeta(s)$ has no zeros in this region?

Any solution or reference would be greatly appreciated.

  • We have those Mellin transforms for $Re(s) > 1$ : $$G(s) = \int_1^\infty (x-1) x^{-s-1}dx = \frac{1}{s-1}-\frac{1}{s}, \qquad F(s) = \int_1^\infty \frac{1-x}{\ln x} x^{-s-1}dx$$

    We see that $$F'(s) = G(s) \implies F(s) = \ln(s-1)- \ln(s) + C$$

  • Then use the Euler product, again for $Re(s) > 1$ :$$\zeta(s) = \prod_p \frac{1}{1-p^{-s}} \implies \ln \zeta(s) = \sum_p \sum_{k \ge 1} \frac{p^{-sk}}{k} = s \int_1^\infty J(x) x^{-s-1}dx$$ where $J(x) = \sum_{p^k \le x} \frac{1}{k}$ (see Abel summation formula, some sort of integration by parts).

  • Finally, it is easy to see that $J(x) = \pi(x) + \mathcal{O}(x^{1/2})$ so that $$\pi(x)- \frac{x}{\ln x} = \mathcal{O}(x^{\sigma}) \implies J(x)+\frac{1-x}{\ln x} = \mathcal{O}(x^{\sigma}) $$ $$ \implies \ln \zeta(s)+s\ln(s-1)-s \ln(s)+sC = s \int_1^\infty \left(J(x)+\frac{1-x}{\ln x}\right) x^{-s-1}dx$$

    converges absolutely and hence is holomorphic for $Re(s) > \sigma \implies \zeta(s)$ has no zero on $Re(s) > \sigma$.

  • $\begingroup$ the converse is much more complicated : showing that $\zeta(s)$ has no zero on $Re(s) > \sigma \implies \pi(x) - \frac{x}{\ln x} = \mathcal{O}(x^{\sigma+\epsilon})$, for this you can look at Titchmarsh's book page 60 to 66 and the last chapter (consequence of the RH). $\endgroup$ – reuns Jul 18 '16 at 14:51

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.