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Let, $$A(x)=\sum_{p\leq x}f(p)$$ Where $p$ is a prime number. Under the Prime Number theorem we have that, $$\pi(x)=Li(x)+O\left(\frac{x}{e^{a\sqrt{\ln(x)}}}\right) $$ as $x$ approach infinity. Now, $$A(x)=\sum_{k\leq x}f(k)(\pi(x)-\pi(x-1))=\int_{2}^{x}f(u)d\pi(u)$$ Now how to use the Prime Number theorem to simplify that Integral further?

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    $\begingroup$ If $f$ is reasonable (e.g. differentiable), then you can integrate by parts and try to use the prime number theorem to study the other side. Depending on $f$, this may or may not be sufficient to get a good understanding of $A(x)$. Does that make sense? $\endgroup$
    – davidlowryduda
    Commented Aug 19, 2022 at 14:08
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    $\begingroup$ Did you mean $A(x)=\sum\limits_{k\leq x}f(k)(\pi(k)-\pi(k-1))$? $\endgroup$
    – Steven Clark
    Commented Aug 19, 2022 at 22:55
  • $\begingroup$ From the definition of logarithmic integral, it is natural to believe $\sum_{p\le x}f(p)\approx\int_2^x{f(t)\over\log t}\mathrm dt$ when $f(x)$ is continuously differentiable and satisfies certain other good properties. $\endgroup$
    – TravorLZH
    Commented Aug 24, 2022 at 6:48

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