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Let $\pi(x)$ be the number of primes not exceeding $x$ and $Li(x) = $. Define $Li(x)=\lim_{\epsilon \rightarrow 0^{+}}\Big(\int_{0}^{1-\epsilon} + \int_{1+\epsilon}^{x}\Big) \frac{dt}{\log t} \mathrm{d}t$. Consider the prime zeta function

$$\sum_{p} p^{-s} = \sum_{m=1}^{\infty} \frac{\mu(m)}{m}\log \zeta(ms)$$ for $\Re(s)=\sigma>1$, where $\mu$ and $\zeta$ denote the Mobius and Riemann zeta functions, respectively.

Applying partial summation to the left-hand side sum over primes $p$ together with the identity $\int_{1}^{\infty} s Li(x)x^{-s-1} \mathrm{d}x=-\log(s-1)$ for $\sigma>1$ yields

$$s\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x-\log((s-1)\zeta(s))=\sum_{m=2}^{\infty} \frac{\mu(m)}{m}\log \zeta(ms)$$ for $\sigma>1$.

The integral on the left-hand side shall be referred to as $F (s)$ forthwith. Denote by $\Theta$ the supremum of the real parts of the zeros of $\zeta$, and suppose $\Theta>1/2$.

We know that $|π(x) − Li(x)| \ll x ^{\Theta} \log x$ and $\Theta$ is the abscissa of convergence of $F (s)$ (Theorems 1.3 and 15.2 of Montgomery-Vaughan). Thus the domain of the above equation (hereafter referred to as $(1)$) extends by analytic continuation to the half-plane $H = \lbrace s : σ > Θ \rbrace,$ thus both sides of $(1)$ must exhibit similar behavior in the neighborhood of $s=\Theta$. This fact shall be crucial for the rest of the argument.

Since $|\mu(m)\log \zeta(ms)| \ll 2^{-m\sigma}$ for all $m\geq 2$ and $\sigma>1/2$, letting $s\rightarrow \Theta^+$ results in the right-hand side of $(1)$ being convergent hence the corresponding left-hand side must also be convergent at $s=\Theta$. Hence the domain of $(1)$ on the real axis must include all $s\geq \Theta$. Thus both sides of $(1)$ must behave similarly as $s\rightarrow (\Theta-\epsilon)^+$ where $\epsilon$ is an arbitrarily small ositive number. Because $(s-1)\zeta(s)\neq 0$ for all real $s>0$ and the right-hand side of $(1)$ is absolutely convergent whenever $\sigma>1/2$, letting $s\rightarrow (\Theta-\epsilon)^+$ where $\epsilon$ is an arbitrarily small positive number, reveals that $F(\Theta-\epsilon)$ must be convergent. But this is a contradiction, since the abscissa of convergence of $F(s)$ is $\Theta$, as noted earlier. Thus we conclude that our supposition must be false and the Riemann hypothesis follows.

Notice that the right-hand side of the above equation converges whenever $σ > 1/2$ since $|μ(m) \log ζ(ms)| \ll 2^{ −mσ}$ for all $m ≥ 2$ and $σ > 1/2.$ Thus we arrive at $Θ ≤ 1/2$, which proves the Riemann hypothesis ?

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What seems to be an issue is that you have only defined $F(s)$ for $\sigma > \Theta$. Further down you claim that the convergence of $F(\Theta - \epsilon)$ follows from the convergence of the RHS at $\Theta - \epsilon$. This is not clear. For example, the expression $$\int_0^\infty e^{-t} t^{s} \frac{dt}{t} - \frac 1s$$ defines a holomorphic function on $\{ \sigma > 0 \}$ which has an analytic continuation to $\{ \sigma > -1 \}$. But the integral does not converge for $\sigma \leq 0$.

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    $\begingroup$ I did not claim from nowhere that $F(s)$ converges at $s=\Theta-\epsilon$. I deduced this from the fact that both sides of the equation must exhibit similar behavior as $s$ approaches $\Theta-\epsilon$ since the equality holds at $s=\Theta$. $\endgroup$ – user697626 Sep 9 at 11:21
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    $\begingroup$ The crux of the argument is: $G, H$ are continuous AND the equality $G(s)=H(s)$ holds for all $s\geq a$, then $G(s)$ and $H(s)$ must exhibit similar behavior as $s \rightarrow a^{+}$. $\endgroup$ – user697626 Sep 9 at 11:25
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    $\begingroup$ @Puncturesd desk, please recheck the argument. I clearly deduced that $(1)$ holds for all $s\geq \Theta$. Will be back after 30 minutes. $\endgroup$ – user697626 Sep 9 at 11:31
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    $\begingroup$ I think have explained enough for you to realize that your answer is not valid. May you please delete it ? $\endgroup$ – user697626 Sep 9 at 14:22
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    $\begingroup$ @user697626 Please change your attitude. punctured dusk is perfectly right : the function $$G(s)=s\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x-\log((s-1)\zeta(s))$$ extends analytically to $\Re(s) > 1/2$ it doesn't mean the integral converges nor that the expression on the RHS has to be well-defined. The RH is about the abscissa of convergence of $\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x$, it is not about the domain of analyticity of $G(s)$. If you take simple examples (as punctured dusk did) you'll see how obvious it is. $\endgroup$ – reuns Sep 10 at 1:56
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I voted for this post to be closed as is opinion based and the OP is very rude and uses word games to hang on any little thing one doesn't spell super explicitly in answers or comments, while also dissembling about the authorship of the argument (once it is a friend, another time it is the OP), but since it seems not to yet be closed and the OP persists in his belief, I will explain shortly why this version of the argument is wrong, capitalizing stuff just to make sure OP's word games are harder to do.

The argument (in this version as it had a bunch of other versions including a deleted post etc) is actually the subtlest of all and it is an INCORRECT interpretation of an old one appearing 100+ years ago in the writings of Littlewood and others who used it to deduce stuff when RH is FALSE ($\Theta > .5$), NOT TO DEDUCE that RH is TRUE.

One can find references in the classic (1934, reissued 1990) book of Ingham, The Distribution of Prime Numbers starting on page 90 (Thm 32), while on page 91, Ingham actually explicitly notes, exact quote as I have the book in front of my eyes:

"It will be noted that the argument depends essentially on the fact that $\zeta(s)$ has no zeros on the positive real axis, a property which is not usually important."

In a nutshell, if one has a function expressed as a Dirichlet integral or series as below:

$F(s)=s\int_1^{\infty}\frac{G(x)}{x^{s+1}}dx$ or (in the usual textbook representation $\frac{F(s)}{s}=\int_1^{\infty}\frac{\frac{G(x)}{x}}{x^{s}}dx$,though in our case the first representation is more useful since the number theoretic $G$ of interest here are functions for which $|G|$ lies between positive powers of $x$, for arbitrarily high $x$) with positive abscissa of convergence $\Theta$, then we know that $F$ IS ANALYTIC if $\Re s > \Theta$, while IF we know that $F$ converges at the REAL VALUE $s=\Theta$ on the abscissa itself, as for example in the case $G_{\pi, \Pi}$ below and $\Theta >.5$, we can deduce ONLY that $G(x)$ CHANGES SIGN INDEFINITELY and NOTHING MORE.

In particular we can use this to deduce that in the Riemann Zeta case, if $\Theta$ (the supremum of the real parts of the RZ zeros) is strictly greater than $\frac{1}{2}$ (hence RH is FALSE) then both $G_{\Pi}(x)=\Pi(x)-Li(x), G_{\pi}(x)=\pi(x)-Li(x)$ must change sign indefinitely, which is the CORRECT conclusion, not that RH is true.

Actually even more is true, namely that both $\Pi(x)-Li(x), \pi(x)-Li(x)$ are $\Omega_{\pm}(x^{\Theta-\epsilon})$, for any $\epsilon >0$.

Note also that since $\Pi(x)-\pi(x)=O(x^{.\frac{1}{2}+\epsilon})$, in the case $\Theta > \frac{1}{2}$ the Dirichlet integrals given by $\Pi-Li$ and $\pi - Li$ which are $\log((s-1)\zeta(s))+h_{\Pi,\pi}(s)$ HAVE the SAME BEHAVIOUR on the line $\Re s = \Theta$ as CONVERGENCE GOES.

Actually it is easy to see that $h_{\Pi}$ (the one for $\Pi$) is an ENTIRE function, while $h_{\pi}$ (the one in the OP argument for $\pi$) has a singularity at $\frac{1}{2}$ confirming this and showing why IF RH is TRUE the proof that $\pi-Li$ changes sign indefinitely is so much harder (there is no convergence now for the $\pi-Li$ integral at $s=.5$ hence we need much subtler arguments to deduce $G_{\pi}$ changes sign indefinitely, while for $\Pi-Li$ we still have convergence there!) than the corresponding proof for $\Pi-Li$ and why it was (and still is 100+ years later) considered such a major achievement of Littlewood.

In the Ingham book above, those subtle arguments start on page 92, while Ingham shows nicely using only prime number theory why we actually need them, and how the direct adaption of the arguments for $\Pi -Li$ fails because some constant depending on the first, ie lowest positive imaginary part, zero of RZ on the critical line is ~$.07$ and the direct adaption requires the constant to be higher than $1$

(here for completeness we restate that $\pi(x)$ is the prime counting function while $\Pi(x)$ is the prime power counting function)

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