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 ?