Does the Riemann hypothesis predict an upper bound for
$$\left|f(x)-\left(\operatorname{li}(x)-\frac{x}{\log x} \right)\right|,\quad x\ge 2\tag{1}$$
where
$$f(x)=\sum\limits_{n=2}^x \frac{\Lambda(n)}{\log^2(n)}\tag{2}$$
and $\Lambda(n)$ is the von Mangoldt function?
The Riemann hypothesis is equivalent to the following statement: $$f(x)=\mathrm{li(x)}-\frac{x}{\log x}+O(\sqrt{x}),\qquad x\geq 2.$$ Note that $$\mathrm{li(x)}=\mathrm{li(2)}+\frac{x}{\log x}-\frac{2}{\log 2}+\int_2^x\frac{dt}{\log^2 t},$$ hence the claim is that the Riemann hypothesis is equivalent to $$f(x)=\int_2^x\frac{dt}{\log^2 t}+O(\sqrt{x}),\qquad x\geq 2.\tag{$\ast$}$$
1. The Riemann hypothesis implies $(\ast)$ as follows: \begin{align*} f(x)&=\int_{2-}^x\frac{d\psi(t)}{\log^2 t}\\ &=\int_2^x\frac{dt}{\log^2 t}+\int_{2-}^x\frac{d(\psi(t)-t)}{\log^2 t}\\ &=\int_2^x\frac{dt}{\log^2 t}+\left[\frac{\psi(t)-t}{\log^2 t}\right]_{2-}^x+2\int_2^x\frac{\psi(t)-t}{t\log^3 t}\,dt. \end{align*} Here $\psi(t)-t=O(\sqrt{t}\log^2 t)$ by the Riemann hypothesis, hence we conclude $(\ast)$.
2. Let us denote by $g(x)$ the integral in $(\ast)$. Then $(\ast)$ implies RH as follows: \begin{align*} \psi(x)&=\int_{2-}^x(\log^2 t)\,df(t)\\ &=\int_2^x(\log^2 t)\,dg(t)+\int_{2-}^x(\log^2 t)\,d(f(t)-g(t))\\ &=x-2+\left[(\log^2 t)\,(f(t)-g(t))\right]_{2-}^x-2\int_2^x\frac{\log t}{t}(f(t)-g(t))\,dt. \end{align*} Here $f(t)-g(t)=O(\sqrt{t})$ by $(\ast)$, hence we infer that $\psi(x)=x+O(\sqrt{x}\log^2 x)$, which is a form of the Riemann hypothesis.
