Riemann Hypothesis: Proving a relation between $\psi(x)$ and $\pi(x)$

Question

I am trying to prove the following

If $\pi(x) = \operatorname{Li}(x) + O(x^{\frac{1}{2}}\log(x))$ then $\psi(x) = x + O(x^{\frac{1}{2}} \log^2(x))$

I have tried using $\psi(x) = \theta(x) + O(x^{\frac{1}{2}} \log^2(x))$ and then bounding $\theta(x)$ but could not reach anywhere.

I have tried this problem for many hours now.

The definitions are as follows:

• $\operatorname{Li}(x) = \int_2^x \frac{1}{\ln t} dt$

• $\pi(x) = \sum_{p \leq x} 1$ with $p$ prime

• $\theta(x) = \sum_{p \leq x} \ln p$ with $p$ prime

• $\psi(x) = \sum_{p^k \leq x, k \geq 1} \ln p$ with $p$ prime

Answer 1

You can obtain the result by writing $\theta(x)$ as a Riemann-Stieltjes integral and integration by parts.

\begin{align} \theta(x) &= \int_{[2,x]} \ln t \, d\pi(t)\\ &= \pi(x)\ln x - \pi(2^-)\ln 2 - \int_2^x \pi(t)\,d\ln t\\ &= \pi(x)\ln x - \int_2^x \frac{\pi(t)}{t}\,dt. \end{align}

Now replace $\pi(y)$ by $\operatorname{Li}(y) + O(\sqrt{y}\ln y)$ and estimate the errors. To evaluate

$$\int_2^x \frac{\operatorname{Li}(t)}{t}\,dt,$$

change the order of integration.

Answer 2

The Riemann hypothesis is equivalent to the statement that $\pi(x)=Li(x)+O(\sqrt{x}\log(x))$. This implies that $\psi(x)=x+O(\sqrt{x}\log(x)^2)$. For a reference see the article "Schoenfeld, Lowell (1976), "Sharper bounds for the Chebyshev functions θ(x) and ψ(x). II", Mathematics of Computation 30". In fact, Schoenfeld proved that RH implies that $$ |\psi(x)-x|<\frac{1}{8\pi}\sqrt{x}\log(x)^2 $$ for all $x\ge 73.2$.