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Suppose that we have $\pi(x)\log(x)=x+O(\frac x{\log(x)})$. I want to prove that $\vartheta(x)=\sum_\limits{p\le x}\log(p)=x+O(\frac x{\log(x)})$.

I've used the relation $\lim\limits_{x\to\infty}\frac{\pi(x)\log (x)-\vartheta(x)}{x}=0$. So given $\epsilon>0$, there exists some $N>0$ such that for all $x\ge N$

$$-\epsilon x<\pi(x)\log (x)-\vartheta(x)<\epsilon x$$

or

$$-\epsilon x<\pi(x)\log (x)-x+x-\vartheta(x)<\epsilon x$$

Since $\pi(x)\log(x)=x+O(\frac x{\log(x)})$, there exists some $M>0$ such that for all $x\ge 2$,

$$-M\frac x{\log(x)}\le\pi(x)\log(x)-x\le M\frac x{\log(x)}$$ Thus we have

$$-\epsilon x+\vartheta(x)-x\le\pi(x)\log(x)-x\le M\frac x{\log(x)}$$

and

$$-M\frac x{\log(x)}\le\pi(x)\log(x)-x\le\epsilon x+\vartheta(x)-x$$

So taking $\epsilon\to0$, implies that

$$-M\frac x{\log(x)}\le\vartheta(x)-x\le M\frac x{\log(x)}$$ or $\vartheta(x)-x=O(\frac x{\log(x)})$. But this is for all $x\ge N$ and not for $x\ge 2$. So I want to know if there is a way to eliminate this problem? Could anyone give me some suggestion, please?


I've also tried to solve the problem like below

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

But I couldn't prove that

$$\vartheta(x)\ge x+O(\frac x{\log(x)}).$$

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    $\begingroup$ It would be better if you put the definition of $\vartheta$ in the beginning of the question. $\endgroup$ – Kemono Chen Nov 11 '18 at 13:56
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Hint: by the Abel summation formula $$\theta\left(x\right)=\pi\left(x\right)\log\left(x\right)-\int_{2}^{x}\frac{\pi\left(t\right)}{t}dt.$$

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  • $\begingroup$ As you probably know, if I use the Abel summation formula, then it would be the problem 4.19 in Apostol's analytic number theory. But my question is the problem 4.18. So I'm looking for another method. $\endgroup$ – user604594 Nov 11 '18 at 17:18
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    $\begingroup$ @user604594 Except the equivalent summation by parts, there is no other method $\endgroup$ – reuns Nov 11 '18 at 19:12

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