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NOTE: I've edited the question one last time, to be much simpler, in the hopes of getting more responses.

SETUP: Let $p_n$ denote the $n$th prime, let $p_x = p_{\lceil x \rceil}$ for all $x > 0$, let $\pi(x)$ denote the prime counting function, and let $\operatorname{li}(x) = \int_0^x \frac{dt}{\log t}$ denote the logarithmic integral function.

QUESTION: Does there exist a $t \in \mathbb{R}$ such that $ \pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) \neq O(\sqrt{x} (\log x)^t)$?

MOTIVATION:

The function $\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ seems to approximate $\pi(x)$ much better than $\operatorname{li}(x)$ does, at least for small $x$. Note that $$\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) = E(x)+H(x),$$ where $E(x) = \pi(x)-\operatorname{li}(x)$ and $H(x) = \frac{\operatorname{li}(x)}{p_{\operatorname{li}(x)}}(p_{\operatorname{li}(x)}-x)\sim \frac{1}{\log(x)}(p_{\operatorname{li}(x)}-x)$, and one has $E(x)\geq 0$ if and only if $H(x) \leq 0$ (so all "interference" in $E(x)+H(x)$ is destructive). Let $\Theta$ denote the supremum of the zeros of the Riemann zeta function. Note that both $E(x)$ and $H(x)$ are $O(x^\Theta \log x)$, and neither is $O(\sqrt{x}/\log x)$.

Note that $x \geq p_y$ if and only if $\pi(x) \geq y$, for all $x, y > 0$. (This defines a monotone Galois connection from $\mathbb{R}_{>0}$ to itself, which expresses an adjoint relationship in the corresponding poset category). Thus $x \geq p_{\operatorname{li}(y)}$ if and only if $\pi(x) \geq \operatorname{li}(y)$. The quantity $p_{\operatorname{li(x)}}\pi(x) - x \operatorname{li}(x)$ is thus natural to consider---discrete on one side, continuous on the other, and involving "both sides" of the duality, not just one. If the answer to the QUESTION above is NO, then $p_{\operatorname{li}(x)}\pi(x) - x \operatorname{li}(x)$ is smaller than $x(\pi(x)-\operatorname{li}(x))$ in order of growth, which would imply that the product $p_{\operatorname{li}(x)}\pi(x)$ of the right-left adjoint pair is better approximated by $x \operatorname{li(x)}$ than either adjoint is by $x$ and $\operatorname{li(x)}$, respectively.

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    $\begingroup$ You ask two questions, and they are not the same. In the first question you ask if $\pi(x) -\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ is genuinely smaller than $\pi(x)-\operatorname{li}(x)$. In the second question, you ask if these two differences have the "same order of growth". It is also not clear to me what you mean by "same order of growth". You say this means "being $o$ of the same", but it is not clear what you mean by "same" here. I suggest that you ask only one question, and you ask it precisely (using a different terminology than "same order of growth"). $\endgroup$
    – GH from MO
    Commented Nov 13, 2022 at 11:32
  • $\begingroup$ It's long to explain precisely what I mean by "same order of growth", but I've narrowed the question's scope to make it more precise. $\endgroup$
    – Jesse Elliott
    Commented Nov 14, 2022 at 2:32
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    $\begingroup$ what is the purpose of this? the beauty of $\operatorname{li}(x)$ is that is a smooth function that is easily computable for high values etc so gives a good estimate of an irregular quantity that is quite difficult to analyze; introducing another difficult-to-analyse function like $p_{\operatorname{li}(x)}$ seems to me kind of useless $\endgroup$
    – Conrad
    Commented Nov 14, 2022 at 15:40
  • $\begingroup$ Because there is a duality here, and it's quite beautiful. $x \geq p_y$ if and only if $\pi(x) \geq y$, for all $x, y > 0$, and thus $x \geq p_{\operatorname{li(x)}}$ if and only if $\pi(x) \geq \operatorname{li}(x)$. The quantity $p_{\operatorname{li(x)}}\pi(x) - x \operatorname{li}(x)$ is thus natural to consider---discrete on one side, continuous on the other, and involving "both sides" of the duality, not just one. If the difference were smaller than that of $x(\pi(x)-\operatorname{li}(x))$, then that would be interesting. It doesn't have to be computationally useful to be interesting. $\endgroup$
    – Jesse Elliott
    Commented Nov 14, 2022 at 22:58

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