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I`m reading Melvin B. Nathanson's "Elementary Methods in Number Theory" and I can't think of a way of deducing Selberg's formula (9.3) from the prime number theorem. This is one of the tasks left for the reader at the end of chapter 9.3 (The Elementary Proof).

$$(9.3) \qquad \qquad \vartheta(x) \log x + \sum_{p \leq x}(\log p) \vartheta(x/p) = 2x\log x +O(x)$$

Simply replacing $\vartheta(x)$ by $x + o(x)$ results in error terms of order $o(x\log x)$ which are strictly worse than error terms of order $O(x)$.

I feel like I'm missing something rather simple, but I've been wasting just a little too much time on that exercise. Thanks in advance!

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The $O(x)$ term can be refined by completely elementary means that are not at all complicated. Filip Saidak (On the prime number lemma of Selberg, Math. Scand. 103 (2008), no. 1, 5–10) proved a much clearer connection between Selberg's result and the prime number theorem. He stated this for $\psi(x) = \sum_{n\leq x}\Lambda(n)$, but this holds with trivial modifications for $\vartheta(x)=\sum_{p\leq x}\log p$.

If $\vartheta(x):=\sum_{n\leq x}\Lambda(n) = x + E(x)$, then

$\vartheta(x)\log x + \sum_{p\leq x} (\log p)\vartheta(x/p)=2x\log x -(2\gamma+1)x+O(E(x)(\log x)^2)$,

where $\gamma$ is the Euler-Mascheroni constant. So if all that you know is that $E(x) = o(x)$, then you cannot recover Selberg's result. Selberg's breakthrough that $-(2\gamma+1)x+O(E(x)(\log x)^2) = O(x)$ appears to be equivalent to the stronger estimate $E(x) \ll x/(\log x)^2$.

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  • $\begingroup$ Thanks for clarifying this! $\endgroup$
    – GH from MO
    Commented Nov 3, 2020 at 1:23
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    $\begingroup$ For anyone who cannot access the paper, the result (Theorem 2.2) states that if $\psi(x)=x+E(x)$, then $\psi(x)\log x +\int_{1}^{x} \psi(x/t) d\psi(t) = 2x\log x -(2\gamma+1)x+O(E(x)(\log x)^2)$. The sparsity of primes to powers greater than 2 would then give the result I posted above. $\endgroup$
    – 2734364041
    Commented Nov 4, 2020 at 13:03
  • $\begingroup$ Thank you both for answering. I was busy during the last two days, but now I have the time to study that paper throughout the next two days, as I luckily have acces to it. $\endgroup$
    – Juu
    Commented Nov 5, 2020 at 14:36
  • $\begingroup$ However: I don't quite understand how to read $\int_{1}^{x} \psi(x/t) \,d\psi(t)$. Is $d \psi(t)$ to be read as a counting measure, where each integer k not less than 1 is given the weight $\psi(k)$? $\endgroup$
    – Juu
    Commented Nov 5, 2020 at 14:45
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    $\begingroup$ @Juu This is an example of a Stieltjes integral. Roughly speaking, you can integrate with respect to a weight $\mathrm{d}\alpha(t)$ for any monotonically increasing function $\alpha(t)$. The usual Riemann integral is recovered by choosing $\alpha(t)=t$. Suitable references are contained in Saidak's paper, since you say you have access to it. $\endgroup$
    – 2734364041
    Commented Nov 5, 2020 at 16:03
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Edit: Answer to GH from MO by author.

Edit2: Changed a sign.

Let $$l(n)=\left\{\begin{array}{11} \log p, & if\ n = p\ prime \\ 0, & else. \end{array}\right.$$ Then $$\sum_{pq\leq x}(\log p)(\log q)=\sum_{n\leq x}(l\ast l)(n)$$ So far, so good. However, using Dirichlet's hyperbola identity, I only obtain $$\sum_{n\leq x}(l\ast l)(n)=2\sum_{p\leq x^\frac{1}{2}}(\log p)\vartheta(\frac{x}{p})-\vartheta(x^\frac{1}{2})^2 $$, where obviously $\vartheta(x^\frac{1}{2})^2=O(x)$.

Thus I am back at my original problem, as I would have to show $$\sum_{p\leq x^\frac{1}{2}}(\log p)\vartheta(\frac{x}{p})=\frac{1}{2}x\log x+O(x)$$

Where do I fail to see the shortpath? Thanks for answering anyways, appreciate it!

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