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I'm trying to prove that $$\sum_{n\le x}(\psi(\frac xn)-\vartheta(\frac xn))\Lambda (n)=O (x) $$

After some calculations, l arrived to $$O (\sqrt{x}\log x\sum_{m=1}^\infty x^{\frac 1m} (\frac{1}{\sqrt{2}})^m)$$

So l need to evaluate that sum. Could anyone help me, please?

Remark: This question is related to problem 4.22 in Apostol's book. The problem is requested to prove that Selberg's asymptotic formula is equivalent to

$$\vartheta (x)\log x+\sum_{p\le x}\vartheta (\frac xp)\log p=2x\log x+O (x) $$

My idea was to first show that

$$\vartheta (x)\log x+\sum_{n\le x}\vartheta (\frac xn)\Lambda (n)=2x\log x+O (x) $$

Thus, i've needed to show that

$$\sum_{n\le x}(\psi(\frac xn)-\vartheta(\frac xn))\Lambda (n)=O (x) $$

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    $\begingroup$ What are $\psi,\theta, \Lambda$? $\endgroup$ – Jimmy R. Jun 1 '19 at 8:15
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    $\begingroup$ I'm really curious where you've got this $1/\sqrt{2}$ in the exponent, let alone its power... $\endgroup$ – Wojowu Jun 1 '19 at 8:20
  • $\begingroup$ @Wojowu I've edited that. $\endgroup$ – user678879 Jun 1 '19 at 12:08
  • $\begingroup$ The sum is at least as big as its first term, which is $\Omega(x)$, so the entire thing cannot be in $O(x)$ $\endgroup$ – Edward H Jun 1 '19 at 13:57
  • $\begingroup$ @EdwardH. $\psi(x)-\vartheta(x)$ is $O(\sqrt{x})$. Even then, $\Omega(x)$ doesn't exclude $O(x)$ $\endgroup$ – Wojowu Jun 1 '19 at 16:03
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$$\sum_{n\le x}(\psi(\frac xn)-\vartheta(\frac xn))\Lambda (n)=O(\sum_{n \le x}(\frac xn)^{1/2} (\psi(n+1)-\psi(n)))\\ = O(\psi(x)+\sum_{n \le x}((\frac xn)^{1/2}-(\frac x{n+1})^{1/2}) \psi(n)) $$

$$ = O(x+\sum_{n \le x}((\frac xn)^{1/2}-(\frac x{n+1})^{1/2}) n)= O(x+\sum_{n \le x} (\frac xn)^{1/2}) = O(x)$$

where I used $\psi(x)-\psi(x/2) = O(\log {x\choose x/2} ) = O(x)$,

$\psi(x) = O(x)$, $\psi(x)-\vartheta(x) = O(x^{1/2})$ and partial summation.

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  • $\begingroup$ Thanks. I've provd the relation $\psi (x)-\vartheta (x) =O (\sqrt x) $ using prime number theorem like below $$\psi (x)-\vartheta (x) =\sum_{m=2}^{\log_2x}\sum_{p\le x^{1/m}}\log p=O (\sum_{p\le\sqrt x}\log p)=O (\frac {\sqrt x}{\log\sqrt x}\log\sqrt x)=O (\sqrt x)$$ Does there exist a proof without use of PNT ? $\endgroup$ – user678879 Jun 6 '19 at 12:38
  • $\begingroup$ The proof is $\psi(x)-\psi(x/2) = O(\log {x\choose x/2} ) = O(x)$ $\endgroup$ – reuns Jun 6 '19 at 15:43
  • $\begingroup$ I looked to chapter 4 in Apostol's book several times but couldn't find any relation between $\psi (x)-\psi ( x/2) $ and $\psi (x)-\vartheta (x) $. Could you please say me how they are related? Where does the $\log {x\choose x/2} $ come from? Should l see another book? $\endgroup$ – user678879 Jun 10 '19 at 5:42
  • $\begingroup$ The proof should be $\psi (x)=\vartheta (x)+\vartheta (\sqrt x)+\vartheta (\sqrt [3] x )+\cdots $. $\endgroup$ – user678879 Jun 11 '19 at 9:18

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