Return to Answer

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 $\theta<1$ such that $|\pi(x)-Li(x)|\ll x^{\theta}$, the claim follows.

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.

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 $\theta<1$ such that $|\pi(x)-Li(x)|\ll x^{\theta}$, the claim follows.

But out of curiosity, why are you interested in this convergence ? Did you receive a certain paper that i also received a few days ago ?

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 $\theta<1$ such that $|\pi(x)-Li(x)|\ll x^{\theta}$, the claim follows.

Claim: 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 $\theta<1$ such that $\pi(x)-Li(x)\ll x^{\theta}$, the claim follows.

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 $\theta<1$ such that $|\pi(x)-Li(x)|\ll x^{\theta}$, the claim follows.

But out of curiosity, why are you interested in this convergence ? Did you receive a certain paper that i also received a few days ago ?