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](https://en.wikipedia.org/wiki/Von_Mangoldt_function)?



Does the Riemann hypothesis predict a bound for this prime-counting 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.


