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

Question

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?

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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

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

Answer 1

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