# Can $\pi(x)-\operatorname{Li}(x)=O(\sqrt x\log(x))$ be proven without using $\psi(x)$?

Assuming the truth of the Riemann hypothesis, $\pi(x)-\operatorname{Li}(x)=O(\sqrt x\log(x))$.

Apparently the proof that $\psi(x)$ approaches $1$ for sufficiently large $x$ proves the error term. However, is there a more "intuitive" way to understand the error bound? (I do admit to not understanding the $\psi(x)$ function, although that could probably be solved with some more diligent research.) If there isn't I'd appreciate a link to an explanation or a mention of information you think would be easy for someone to miss.

Your phrasing is a bit odd. We don't currently even know that the error is $\sqrt x \log x$, so we have no idea about whether or not we can show it using any technique.
I assume that by $\psi(x)$ you mean the Chebyshev function $$\psi(x) = \sum_{p^k \leq x} \log p.$$ If you don't mean that, then my answer here is meaningless to you. As you can quickly see, it's never the case that $\psi(x) \to 1$. However, the prime number theorem is equivalent to showing that $\dfrac{\psi(x)}{x} \to 1$, which is almost what you were talking about. This has little to do with the error term though.
The link between the Riemann zeta function $\zeta(s)$ and the prime number theorem is through $\psi(x)$. In particular, $\psi(x)$ is the sum of the first $x$ coefficients of $\dfrac{\zeta'(s)}{\zeta(s)}$ (when this is written as a Dirichlet series). So analytic data about $\zeta(s)$ gives a good understanding of $\psi(x)$.
The error term $O(\sqrt x \log x)$ is equivalent to showing that $\lvert \psi(x) - x \rvert = O(\sqrt x)$, which is equivalent to there being no $0$ of $\zeta(s)$ in the right half of the critical strip.