# How to prove that $\vartheta(x)=x+O(\frac x{\log(x)})$

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

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

• It would be better if you put the definition of $\vartheta$ in the beginning of the question. – Kemono Chen Nov 11 '18 at 13:56

## 1 Answer

Hint: by the Abel summation formula $$\theta\left(x\right)=\pi\left(x\right)\log\left(x\right)-\int_{2}^{x}\frac{\pi\left(t\right)}{t}dt.$$

• 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. – user604594 Nov 11 '18 at 17:18
• @user604594 Except the equivalent summation by parts, there is no other method – reuns Nov 11 '18 at 19:12