In lecture notes, there was an exercice that I struggle with. Here, $\psi$ denotes the Chebychev function. I assumed that $$\psi(x)-x\in o(x^{1-\varepsilon})$$ for some $0<\varepsilon<1/2$ (which would come from a less strong version of the Riemann Hypothesis, that is some zero-free region of the form $\{\sigma>c\}$). By using summation by parts, I wrote $$\pi(x)=\frac{\psi(x)}{\log(x)}+\int_2^x\frac{\psi(t)}{t\log^2(t)}\text{d}t,$$ which allowed me to prove that $$\pi(x)-\text{Li}(x)\in O(x^{1-\varepsilon}).$$
Now, in the rest of the exercice, I'm asked to prove that $$\pi(x)-\frac{x}{\log(x)}\notin o(x^{1-\delta})$$ for any $\delta>0$ (this is the sense in which $\text{Li}(x)$ is a better approximation to $\pi(x)$ than simply $\frac{x}{\log(x)}$). How to prove this ?
If I assume the converse, that is, $$\pi(x)-\frac{x}{\log(x)}\in o(x^{1-\delta})$$ for some $\delta>0$, then I have no idea which contradiction I'm supposed to get.
I tried the classical inequalities : $$\pi(x)\log(x)\geqslant\psi(x)\geqslant\sum_{x^{1-\eta}\leqslant p\leqslant x}\log(p)=(1-\eta)[\pi(x)+O(x^{1-\eta})]\log(x),$$ but I doubt it's leading me anywhere. Does anyone have any hint/idea ? The exercice doesn't give any besides some trivia.
Based on what you proved, we have $$ \pi (x) - \frac{x}{{\log x}} = \operatorname{Li}(x) + \mathcal{O}(x^{1 - \varepsilon }) - \frac{x}{{\log x}} = \operatorname{Li}(x) - \frac{x}{{\log x}} + \mathcal{O}(x^{1 - \varepsilon }). $$ Now, it can be shown, using integration by parts, that $$ \operatorname{Li}(x) = \frac{x}{{\log x}} + \frac{x}{{\log ^2 x}} + \mathcal{O}\!\left( {\frac{x}{{\log ^3 x}}} \right). $$ Therefore, $$ \pi (x) - \frac{x}{{\log x}} = \frac{x}{{\log ^2 x}} + \mathcal{O}\!\left( {\frac{x}{{\log ^3 x}}} \right) + \mathcal{O}(x^{1 - \varepsilon }) = \frac{x}{{\log ^2 x}} + \mathcal{O}\!\left( {\frac{x}{{\log ^3 x}}} \right) \notin o(x^{1 - \delta }). $$