There has been a paper doing rounds on Facebook for the past several days, claiming a proof of the Riemann hypothesis. I feel sure that the argument is flawed, but can't see where exactly. It goes as follows:

Let $\pi(x)$ be the number of primes not exceeding $x$ and $Li(x) = \int_{1}^{x} \frac{dt}{\log t}$. 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 $\sigma>1$, where $\Theta\leq 1$ denotes the supremum of the real parts of the zeros of $\zeta$. The integral on the left-hand side shall be referred to as $F (s)$ forthwith.

We know that $|π(x) − Li(x)| \ll x ^{\Theta} \log x$ and $\Theta$ is the abscissa of convergence of $F (s)$ (Theorem 1.3 of Montgomery-Vaughan). Thus the domain of the above equation extends by analytic continuation to the half-plane $H = \lbrace s : σ > Θ \rbrace.$

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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    $\begingroup$ Hint: if the RH would be solvable in 2 paragraphs it would already have been done so. $\endgroup$ – Klangen Aug 27 '19 at 10:32
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    $\begingroup$ I’m not an expert in analytic number theory, but $F(s)$ can have a meromorphic continuation well beyond its convergence abscissa ($\zeta$ is an example), invalidating the very last deduction. $\endgroup$ – Mindlack Aug 27 '19 at 10:33
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    $\begingroup$ @Klangen: while it is a sound rule of thumb, it does little to actually find the fault in the “proof”. $\endgroup$ – Mindlack Aug 27 '19 at 10:35
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    $\begingroup$ @Klangen, that was also my heuristic argument against the proof. But i would be much more satisfied if i could find a mathematical reason against it. $\endgroup$ – user697626 Aug 27 '19 at 10:35
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    $\begingroup$ Maybe someone could let us know where to find this (attempted) proof of the Riemann Hypothesis and/or the author's name? $\endgroup$ – Gerry Myerson Aug 27 '19 at 13:41

While @reuns clearly showed the fallacy, the following is a simple explanation without getting bogged down in details that one can spin around in circles as we saw so often in this purported "proofs" here or on MO

The "proof" has the logical structure: RH is equivalent to the analyticity of $A(s), \Re s > \frac{1}{2}$ which is equivalent to the analyticity of $B(s), \Re s > \frac{1}{2}$.

We show that $A(s)-B(s)$ extends analyticaly to $\Re s > \frac{1}{2}$ hence we conclude RH (hence) that both $A(s), B(s)$ extend analyticaly to $\Re s > \frac{1}{2}$.

I think that this shows clearly the fallacy of the proof since for example $f(s)-f(s)$, where $f$ is any analytic functions on some domain, extends analytically to the whole plane...

  • $\begingroup$ That's not the structure of the ''proof'' at all. At no point does the author claim that $s\int_{1}^{\infty}(\pi(x)-Li(x))x^{-s-1} \mathrm{d}x$ and $\log((s-1)\zeta(s))$ are analytic because the RHS is analytic. $\endgroup$ – user697626 Aug 27 '19 at 14:29
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    $\begingroup$ no, it claimed RH which is EQUIVALENT to each term on LHS analytic, so while it did not make the explicit claim, it made it implicitly by claiming RH $\endgroup$ – Conrad Aug 27 '19 at 14:33
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    $\begingroup$ what is the last line then: "Thus we arrive at $\Theta$ ≤1/2" (this as noted is equivalent to RHS) so this is a passive-aggressive way to claim RH while questioning the claim... So yes the above "proof" claims RH, so sorry but your comment is disingenuous $\endgroup$ – Conrad Aug 27 '19 at 14:41
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    $\begingroup$ the proof is still wrong because we still have the fact that if two functions have a given domain convergence, their difference may have a much higher domain of convergence $\endgroup$ – Conrad Sep 9 '19 at 13:43
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    $\begingroup$ very good for you - the argument is still wrong, but I am waiting for it to be published and then we can talk further $\endgroup$ – Conrad Sep 9 '19 at 14:31

If we care of $Li(x)=1_{x>2}\int_2^x\frac{dt}{\log t}$ it is because its Mellin transform is $$L(s)= \int_2^\infty Li(x)x^{-s-1}dx= \int_2^\infty \frac1{\log x}\frac{x^{-s}}{s}dx=\frac1s (2Li(2)-\int_2^s\int_2^\infty x^{-z}dxdz)$$$$=\frac1s(2Li(2)-\int_2^s \frac{2^{1-z}}{z-1}dz)=\frac{F(s)}{s}-\frac{\log (s-1)}{s}$$

where $F(s)=2Li(2)-\int_2^s \frac{2^{1-z}-1}{z-1}dz$ is entire. With $P(s)=\sum_p p^{-s}$

$$\frac{P(s)}{s}+\frac{\log(s-1)}{s}=\int_0^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x+\frac{F(s)}{s} $$


$$s\int_0^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x-\log\zeta(s)-\log(s-1)=-F(s) -\sum_{p^k,k\ge 2}\frac{p^{-sk}}{k}$$

where $\color{red}{the\ RHS}$ converges and is analytic for $\Re(s) >1/2$, it doesn't mean the integral on the LHS converges for $\Re(s)>1/2$ which is what the RH is about.

So this doesn't tell anything of the Riemann hypothesis.

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    $\begingroup$ I was wondering when reuns would show up :) $\endgroup$ – Klangen Aug 27 '19 at 11:53
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    $\begingroup$ Where is the flaw now ? $\endgroup$ – Peter Aug 27 '19 at 12:00
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    $\begingroup$ Thanks reuns. But how does this answer the asked question ? Can you explain further ? $\endgroup$ – user697626 Aug 27 '19 at 12:03
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    $\begingroup$ @Reuns it seems what you wrote is not very different from what the claimer wrote. The only addition you made is your last statement ''this doesn't tell anything of the RH''. What we want to understand is why does it have nothing to do with the RH ? $\endgroup$ – user697626 Aug 27 '19 at 12:23
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    $\begingroup$ The RH is equivalent to the analyticity of $\log \zeta(s)+\log (s-1),P(s)+\log (s-1)$ for $\Re(s)>1/2$ and from the Tauberian theorems to the convergence of their corresponding Mellin transforms integrals $\endgroup$ – reuns Aug 27 '19 at 13:05

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