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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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    Well, according to the answers in https://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. –  Oct 01 '19 at 07:05
  • In particular, does anyone have a reference for the formula of the abscissa of convergence of the above integral that was claimed in https://mathoverflow.net/a/337899/146617 ? –  Oct 01 '19 at 07:10
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    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 – Yemon Choi Oct 14 '19 at 20:33

1 Answers1

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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 $\Re s \le 1$

Conrad
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  • It seems you're badly misunderstanding something here. The gist of the matter lies in the theory of analytic functions. Note that if $f(s)=g(s)$ for $\Re(s)>a$ and both $f(s), g(s)$ are analytic in the larger half-plane $Re(s)>b$, then the domain of $f(s)=g(s)$ extends by analytic continuation to $\Re(s)>b$. –  Oct 01 '19 at 12:33
  • that's true but that doesn't imply that the original definition of $f$ still works; $\zeta(i+s)-\frac{1}{s+i-1}$ extends to an entire function but that doesn;t imply that the original Dirichlet series of $\zeta(i+s)$ converges for $\Re s \le 1$ – Conrad Oct 01 '19 at 12:35
  • $\zeta(i+s)$ is not analytic for $\Re(s)\leq 1$. Unlike our case, the original defintion of both the LHS and the RHS are analytic for $\Re(s)>1/2$ –  Oct 01 '19 at 12:37
  • $\int_{1}^{\infty}(\pi(x)-Li(x))x^{-s-1} \mathrm{d}x$ is not analytic for $\Re s \le 1$ either (or at least we do not know where it is - RZ and weaker version of it are equivalent to various domains of analyticity for it); in your example you have $\int_{1}^{\infty}(\pi(x)-Li(x))x^{-s-1} \mathrm{d}x$ - $f(s)$ where $f(s)=\log((s-1)\zeta)$ anayltic there, in my example I have $\zeta(i+s)-f(s)$] too – Conrad Oct 01 '19 at 12:38
  • That's not what i said at all. Check carefully. Rather, i said $G(s)=s\int_{2}^{\infty}(\pi(x)-Li(x))x^{-s-1} \mathrm{d}x-\log((s-1)\zeta(s))$ is analytic for $\Re(s)>1/2$. This was deduced after showing that $G(s)$ is identical to $s\int_{2}^{\infty}(\pi(x)-\Pi(x))x^{-s-1} \mathrm{d}x$. And please try to make your presentation a bit clearer. –  Oct 01 '19 at 12:42
  • $G(s)==s\int_{2}^{\infty}(\pi(x)-Li(x))x^{-s-1} \mathrm{d}x-\log((s-1)\zeta(s))$ is anayltic for $\Re s > .5$ true but again, that doesn;t mean anything as $\zeta(i+s)-\frac{1}{i+s-1}$ is analytic for $\Re s > .5$ too – Conrad Oct 01 '19 at 12:45
  • There are crucial details that you're leaving out here ! We have $G(s)=\sum_{m=2}^{\infty} \frac{\mu(m)}{m}\log \zeta(ms)$. Notice that the RHS is trivially analytic and absolutely convergent for $\Re(s)>1/2$. Thus both sides are analytic and convergent for $\Re(s)>1/2$, implying that the domain of the identity indeed extends by analytic continuation to $\Re(s)>1/2$. –  Oct 01 '19 at 12:48
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    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 $G(s)$ – Conrad Oct 01 '19 at 12:50
  • Can you answer this question: Does the domain of $G(s)=\sum_{m=2}^{\infty} \frac{\mu(m)}{m}\log \zeta(ms)$ extend to $\Re(s)>1/2$ ? –  Oct 01 '19 at 12:55
  • yes that is true – Conrad Oct 01 '19 at 12:56
  • Perfect, so if that identity holds for $\Re(s)>1/2$, particularly, real $s>1/2$ and we know that the RHS sum converges for $s>1/2$, doesn't that enatil that the LHS converges for real $s>1/2$ ? Let's do things in stages. –  Oct 01 '19 at 13:00
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    no it doesn't imply that – Conrad Oct 01 '19 at 13:01
  • Are you kidding me ??? That's an IDENTITY for goodness sake ! Meaning whatever happens to the RHS also happens to the LHS in the domain ! –  Oct 01 '19 at 13:01
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    $\zeta{(i+s)}-\frac{1}{i+s-1}$ 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 $\Re s \le 1$ – Conrad Oct 01 '19 at 13:05
  • That's an IDENTITY for goodness sake ! Meaning whatever happens to the RHS also happens to the LHS in the domain ! – Conrad Oct 01 '19 at 13:08
  • To be honest, i'm getting tired of explaining these trivial details. In your ''counter-example'', you DO NOT have BOTH sides of the identity being analytic for $\Re(s)>1/2$, meaning that you cannot extend the domain f the identity analytically to $\Re(s)>1/2$. I suggest you thoroughly familiarise yourself with the concept of analytic continuation, otherwise i can't continue in this conversation anymore. You can also see this answer https://mathoverflow.net/a/164878/146617 –  Oct 01 '19 at 13:09
  • what do you mean: $\zeta({i+s})-\frac{1}{i+s-1}=(s+i)\int_{1}^{\infty}\frac{[x]-x+.5}{x^{s+i+1}}dx + .5$ is precisely of the form $F(s)=G(s)$ where the RHS converges (uniformly) for any $\Re s > \sigma >0$ – Conrad Oct 01 '19 at 13:10
  • you clearly need to thoroughly familiarise yourself with basic complex analysis. This will be my last comment. In your edit, we indeed have for $\Re(s)>1$ $\sum{\frac{1}{n^{i+s}}}-\frac{1}{i+s-1}=(s+i)\int_{1}^{\infty}\frac{[x]-x+.5}{x^{s+i+1}}dx + .5$ where RHS converges uniformly for $\Re s > \sigma >0$ so extends to the half plane $\Re s >0$. However, the original defintion of the corresponding LHS IS NOT ANALYTIC for $\Re(s)\leq 1$, implying that its domain cannot be extended analytically to $\Re(s)>\sigma_0$, where $\sigma_{0}\leq 1$. –  Oct 01 '19 at 13:32
  • This is EVIDENTLY DIFFERENT from our case whereby BOTH sides of our identity are analytic for $\Re(s)>1/2$. Bye. –  Oct 01 '19 at 13:34
  • We just proved that LHS can be extended; in your case,it is the same and no obfuscation (as there are a few identities tossed around, some trivially anayltic, some not so clearly so) will work – Conrad Oct 01 '19 at 13:35
  • you didn't prove anything here. You really need some basic complex analysis, rather than waste time on these trivially wrong arguments of yours. Bye bye (twice). –  Oct 01 '19 at 13:41
  • neither did you prove anything – Conrad Oct 01 '19 at 13:42
  • Okay, convince yourself whatever you wish, but we will see where it ends. –  Oct 01 '19 at 14:02
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    @QUB: would you please calm down? MO is not FB. –  Oct 01 '19 at 14:03
  • @MarkSapir, it's hard to calm down when someone is arrogantly imposing their elementary misunderstandings. But to avoid unnecessarily stirring up emotions, i will not comment any further. –  Oct 01 '19 at 14:05
  • Oh, this ''answer'' was upvoted again, what a joke. –  Oct 01 '19 at 14:07
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    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 – Conrad Oct 01 '19 at 15:52
  • @Conrad: there have been many more posts, which you might not have seen since they were deleted. FWIW I would recommend not engaging with T. K. Isaac / TIK (for it is he) since there are basic underlying confusions here and a persistent refusal to learn from comments/corrections made by other users – Yemon Choi Oct 14 '19 at 03:15
  • @Yemon - good to know; the only comment I would make is that unless these are taken down immediately he will spam various mathematicians blogs with links bragging about how he proved RH - this is actually how I first saw one of those posts – Conrad Oct 14 '19 at 10:50