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Following On the integral $I_s =\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} dx$, define

$$I_{s}=\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x,$$ where $\pi$ denotes the prime counting function and $Li$ the logarithmic integral.

Does $I_s$ converge at $s=\Theta$, where $\Theta$ is the minimal real number such that $\pi(x)-Li(x) \ll x^{\Theta+ \epsilon}$ for any $\epsilon>0$ ?

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    $\begingroup$ Presumably no, by Landau's theorem? See Section 15.1 of Montgomery/Vaughan's Multiplicative Number Theory I. $\endgroup$
    – Greg Martin
    Commented Sep 13, 2019 at 16:41
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    $\begingroup$ @GregMartin, Landau's lemma is for integrals $\int_{1}^{\infty} f(x) x^{-s - 1} \, dx$ for which $f(x)$ is of constant sign for all sufficiently large $x$. Theorem 15.2, however, shows that $\pi(x) - \mathrm{Li}(x) = \Omega_{\pm}(x^{\Theta - \varepsilon})$. $\endgroup$
    – Peter Humphries
    Commented Sep 13, 2019 at 17:12
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    $\begingroup$ I think that both "yes" and "no" are consistent with our current knowledge. If, say, there is no zero with $\Re(s)\geq\Theta$ and the zeros with $\Re(s)\approx\Theta$ are sparsely located, then I think that $I_\Theta$ can converge. If, say, there is a single zero with $\Re(s)=\Theta$ and there is no zero with $1/2<\Re(s)<\Theta$, then surely $I_\Theta$ diverges. I have not verified these claims, but this was my immediate thought! $\endgroup$
    – GH from MO
    Commented Sep 13, 2019 at 17:13
  • $\begingroup$ If $\Theta=1/2$ or if there is a zero on $\Re(s) =\Theta$ then it diverges, if $\Theta=1$ then it converges. $\endgroup$
    – reuns
    Commented Sep 13, 2019 at 19:37

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