A probably wrong proof of the Riemann Hypothesis, but where is the mistake?

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 ?

Solution 1:

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...

Solution 2:

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 $P(s)=\sum_p p^{-s}$

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

and

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

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.