Define $\pi(x)$ to be the prime counting function and Li(x) the logarithmic integral. Let $I_s$ be defined as above.

Is $I_s$ known to be convergent for any real number $s<1$ ?


2 Answers 2


I claim that no such $s$ is known to exist. Indeed, define $\sigma_c$ to be the abscissa of convergence of $I$. Then

$$\sigma_c = \limsup_{x\rightarrow \infty} \frac{\log|\pi(x)-Li(x)|}{\log x}.$$ Since we do not know of any absolute $\theta<1$ such that $|\pi(x)-Li(x)|\ll x^{\theta}$, the claim follows.


The following answer is essentially the same as the one given by ChenClass, but it provides more detail. I claim that currently it is unknown whether $I_s$ converges for some real number $s_0<1$.

Assume that $I_s$ converges for some real number $s_0<1$. Adapting the proof of Theorem 1.1 in "Montgomery-Vaughan: Multiplicative number theory I", we see that $I_s$ converges locally uniformly in the half-plane $H=\{s:\Re(s)>s_0\}$. In particular, by Morera's theorem, $I_s$ is analytic in $H$. Using the explicit formulae on Page 465 of the same book, we infer that $\log(\zeta(s)(s-1))$ has an analytic continuation to $H$. Equivalently, $\zeta(s)$ has no zero on $H$. This is currently unknown, so at the moment we cannot prove that $I_s$ converges for some real number $s_0<1$.

On the other hand, the Riemann Hypothesis implies that $I_s$ converges in the half-plane $\{s:\Re(s)>1/2\}$, so at the moment we cannot disprove either that $I_s$ converges for some real number $s_0<1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.