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Consider the prime zeta function, defined for $\Re(s)>1$, by the infinite series

$$\sum_{p} p^{-s} = \sum_{m=1}^{\infty}\frac{\mu(m)}{m}\log \zeta(ms)$$ where $p$ denotes a prime, $\mu$ the Mobius function and $\zeta$ the Riemann zeta function. By partial summation, one finds that

$$s\int_{2}^{\infty} \pi(x)x^{-s-1} \mathrm{d}x =\log \zeta(s) + \sum_{m=2}^{\infty}\frac{\mu(m)}{m}\log \zeta(ms)$$ for $\Re(s)>1$, where $\pi$ is the prime counting function. Let $Li(x)=\int_{2}^{x} \frac{dt}{\log t}$. We know that $s\int_{2}^{x} Li(x)x^{-s-1} \mathrm{d}x=-\log(s-1)+r(s)$, where $r(s)$ is entire. Putting this into the above gives

\begin{equation} s\int_{2}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x -\log ((s-1)\zeta(s)) -r(s)= \sum_{m=2}^{\infty}\frac{\mu(m)}{m}\log \zeta(ms) \\ \end{equation} for $\Re(s)>1$. Note that $\log \zeta(s)=s\int_{2}^{\infty}\Pi(x)x^{-s-1} dx$ hence $\log((s-1)\zeta(s))=s\int_{2}^{\infty}(\Pi(x)-Li(x))x^{-s-1} dx$ for $\Re(s)>1$ where $\Pi$ is the prime power counting function. Putting this into the above identity, we find that the identity can be written as

$$s\int_{2}^{\infty}(\pi(x)-\Pi(x))x^{-s-1} \mathrm{d}x-r(s)=\sum_{m=2}^{\infty}\frac{\mu(m)}{m}\log \zeta(ms)$$

for $\Re(s)>1$. Notice that both sides of the preceding identity are analytic and convergent for $\Re(s)>1/2$ since $\Pi(x)-\pi(x)=O(x^{1/2}), r(s)$ is entire and $\mu(m)\log \zeta(ms) \ll 2^{-m\Re(s)}$ as $m\rightarrow \infty$ for $\Re(s)>1/2$. By analytic continuation, this implies that the preceding identity also holds for $\Re(s)>1/2$.

Because $(s-1)\zeta(s)>0$ for every real $s>0$, it follows from this idenity that $\int_{2}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x$ converges on the real axis for $s>1/2$ ?

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closed as off-topic by Yemon Choi, user44191, David Handelman, LeechLattice, LSpice Oct 18 at 19:36

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    $\begingroup$ Well, according to the answers in mathoverflow.net/q/337897/146617, the convergence of $\int_{1}^{\infty}(\pi(x)-Li(x))x^{-s-1} \mathrm{d}x$ on the real axis for $s>1/2$ is equivalent to the Riemann hypothesis. However, those answers appear not to have sufficient detail for me. ChenClass pulls of some formula for the abscissa of convergence of the integral out of nowhere, and GHfromMO's answer relies on adapting the proof of Theorem 1.1 of Montgomery-Vaughan (MV) However, Theorem 1.1 of MV was proved for infinite series, not integrals. $\endgroup$ – user146617 Oct 1 at 7:05
  • $\begingroup$ In particular, does anyone have a reference for the formula of the abscissa of convergence of the above integral that was claimed in mathoverflow.net/a/337899/146617 ? $\endgroup$ – user146617 Oct 1 at 7:10
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    $\begingroup$ I'm voting to close this question because circumstantial evidence suggests this was yet another disingenuous attempt by TIK to promote his attempts to prove RH, combined with hostile responses to other users who try to indicate gaps in the logic $\endgroup$ – Yemon Choi Oct 14 at 20:33
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the answer to the original question is obviously we do not know (as noted the question is equivalent to RH), while the reasoning above doesn't work for the same reason that, for example, the fact that $\zeta(i+s)-\frac{1}{s+i-1}$ extends to an entire function from the original definition on $\Re s >1$ and obviously $\frac{1}{s+i-1}$ doesn't have any singularity for $s$ real, doesn't imply that the Dirichlet series (which can be expressed as an integral if one wishes) of $\zeta(i+s)$ converges beyond $\Re s >1$.

The reason in both cases that we cannot conclude anything definite from the extension to a bigger half-plane plus lack of singularities on the real line (while we have singularities somewhere else on the original abscissa of convergence) is that the coefficients of the series (or the integrand $\pi(x)-Li(x)$) are not eventually positive or negative - in my contrived example they are complex, in the example in the post the integrand is real but change sign indefinitely.

Edit later - just to note that we have for $\Re s >1$ the identity:

$\sum{\frac{1}{n^{i+s}}}-\frac{1}{i+s-1}=(s+i)\int_{1}^{\infty}\frac{[x]-x+.5}{x^{s+i+1}}dx + .5$ where RHS converges uniformly for $\Re s > \sigma >0$ so extends to the half plane $\Re s >0$, and also $\frac{1}{i+s-1}$ has no singularity for $s$ real, so we have the precise setting of the post but here we do know that $\sum{\frac{1}{n^{i+s}}}$ actually doesn't converge for any $\Re s \le 1$

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    $\begingroup$ nobody disputes that $G(s)$ extends to $\Re s > .5$ but that implies nothing about the convergence of the integral part from the LHS of $G(s)$ $\endgroup$ – Conrad Oct 1 at 12:50
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    $\begingroup$ no it doesn't imply that $\endgroup$ – Conrad Oct 1 at 13:01
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    $\begingroup$ $\zeta{(i+s)}-\frac{1}{i+s-1}$ extends to $\Re s > .5$; $\zeta({i+s})-\frac{1}{i+s-1}$ has no singularity for real $s$. It doesn;t follow that $\sum{\frac{1}{n^{i+s}}}$ converges real $s > .5$ and actually it is not true for any $\Re s \le 1$ $\endgroup$ – Conrad Oct 1 at 13:05
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    $\begingroup$ @QUB: would you please calm down? MO is not FB. $\endgroup$ – Mark Sapir Oct 1 at 14:03
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    $\begingroup$ this is the 4th, 5th time (that I know of) this RH "proof" (or variants thereof with curlies and obfuscations) has been presented here or on MSE under various user names (though similar offensive attitude as the last comment - I am referring to the one regarding upvotes just in case it gets deleted - which is pretty much same as on MSE shows); also these posts tend to change when various objections are raised so I tend to screenshot the original just in case and locking them can be useful $\endgroup$ – Conrad Oct 1 at 15:52