I have come across the following exercise from Hildebrand's notes on analytic number theory (unfortunately, the link on the author's webpage has expired).
Let $0<\alpha<1$ be fixed. Show that if $$ \theta(x)=x+O\left(x \exp \left(-c(\log x)^\alpha\right)\right) \quad(x \geq 2) $$ with some positive constant $c$, then $$ \pi(x)=\operatorname{Li}(x)+O\left(x \exp \left(-c^{\prime}(\log x)^\alpha\right)\right) \quad(x \geq 2) $$ with some (other) positive constant $c^{\prime}$ (but the same value of the exponent $\alpha$).
I tried to apply partial summation: $$ \pi(x)=\sum_{p \leq x} 1 =\sum_{p \leq x}\log p\cdot\frac{1}{\log p} =\frac{\theta(x)}{\log x}+\int_2^x\frac{\theta(t)}{t(\log t)^2}dt. $$ However the second (error) term is too big: $$ \int_2^x\frac{1}{(\log t)^2}dt\ll\frac{x}{(\log x)^2}, $$ and I'm not sure if one can bound it by $O\left(x \exp \left(-c^{\prime}(\log x)^\alpha\right)\right)$. Also, I don't see how the logarithmic integral may appear in the partial summation.
Any help or hint on this problem would be appreciated. Thanks in advance.
Since
$$ \pi(x)-\operatorname{Li}(x)=\int_{2^-}^x{\mathrm d[\theta(u)-u]\over\log u}={\theta(x)-x\over\log x}+\int_2^x{\theta(u)-u\over u(\log u)^2}\mathrm du, $$
it suffices to show that the remaining integral on the right hand side is $O(xe^{-c'(\log x)^\alpha})$. Plugging in the bound $\theta(x)-x=O(xe^{-c(\log x)^\alpha})$ gives
\begin{aligned} \int_2^x{\theta(u)-u\over u(\log u)^2}\mathrm du &\ll\int_2^x{e^{-c(\log u)^\alpha}\over(\log u)^2}\mathrm du\ll\int_2^xe^{-c(\log u)^\alpha}\mathrm du \\ &\le(\sqrt x-2)e^{-c(\log2)^\alpha}+\int_{\sqrt x}^xe^{-c(\log u)^\alpha}\mathrm du \\ &\le(\sqrt x-2)e^{-c(\log2)^\alpha}+(x-\sqrt x)e^{-c(\log\sqrt x)^\alpha} \\ &\ll xe^{-c'(\log x)^\alpha} \end{aligned}
with $c'=2^{-\alpha}c$.