Does $O(\log^2(x))$ imply $O(x)$

Question

Does $O(\log^2(x))$ imply $O(x)$

I have to prove the following:

$$\sum\limits_{\substack{n\in\mathbb N\\n\le x}}\Lambda(n)\log(n)=\psi(x)\log(x)+O(x)$$

Applying partial sum I get;

$\displaystyle\sum\limits_{\substack{n\in\mathbb N\\n\le x}}\Lambda(n)\log(n)=\psi(x)\log(x)-\int_1^x\frac1t\left(\sum\limits_{\substack{n\in\mathbb N\\n\le t}}\Lambda(n)\right)dt$

so to show is: $\displaystyle\int_1^x\frac1t\left(\sum\limits_{\substack{n\in\mathbb N\\n\le t}}\Lambda(n)\right)dt=O(x)$

Can I use the fact that; $\sum\limits_{\substack{n\in\mathbb N\\n\le x}}\frac{\Lambda(n)}{n}=\log(x)+O(1)$ (already proved) then I obtain:

$\displaystyle\int_1^x\frac1t\left(\sum\limits_{\substack{n\in\mathbb N\\n\le t}}\Lambda(n)\right)dt\le\int_1^x\sum\limits_{\substack{n\in\mathbb N\\n\le t}}\frac{\Lambda(n)}{n}dt=O(x)+O(\log^2(x))$

Is this correct ?

EDIT OK I integrated wrong, but If I use

$\frac{\psi(t)}{t}\le\frac{4\log(t)}{\sqrt{t}}+\frac{\theta(t)}{t}$, with $\theta(t)=\sum\limits_{p:\text{prime}\le t}\log(p)$

Now $\theta(t)\le(4\log(2))t$ (this and the previous inequality are proved)

then everything should be OK

$\int_1^x\frac{\psi(t)}{t}\le\int_1^x\frac{4\log(t)}{\sqrt{t}}+\frac{(4\log(2))t}{t}dt=O(\sqrt{x}\log(x))+O(x)=O(x)$

Answer 1

If you already proved that $$\sum_{n\leq x}\Lambda\left(n\right)=O\left(x\right) $$ you can conclude directly because $$\int_{1}^{x}\frac{1}{t}\left(\sum_{n\leq t}\Lambda\left(n\right)\right)dt=O\left(\int_{1}^{x}\frac{1}{t}tdt\right)=O\left(x\right) $$ and this complete your proof. If you use your argument $$\sum_{n\leq x}\frac{\Lambda\left(n\right)}{n}=\log\left(x\right)+O\left(1\right) $$ you get $$\int_{1}^{x}\frac{1}{t}\left(\sum_{n\leq t}\Lambda\left(n\right)\right)dt\leq\int_{1}^{x}\sum_{n\leq t}\frac{\Lambda\left(n\right)}{n}dt=\int_{1}^{x}\log\left(t\right)dt+O\left(x\right)=x\log\left(x\right)+O\left(x\right) $$ which is too bigger.

Addendum With the edit, your proof is valid.

Answer 2

$$\lim_{x\to\infty}\frac{\log(x)^n}{x^m}= 0$$ for any $n>0$ and $m>0$

so $O(\log(x)^n)$ is a tighter bound and can be replaced by $O(x^m)$ if it is more convenient. However the inverse will not be true.