Prime Counting: Relationship between Chebyshev's function and the Prime counting function

Question

How do I show that if $\psi(x)=x+O(x^{1/2}\log^2(x))$ then $\pi(x)=\int_2^x \frac{dt}{logt} + O(x^{1/2}\log x)$

Where $\psi(x)$ is Chebyshev's second function and $\pi(x)$ is the prime counting function

Answer 1

Step 1: Note that the prime powers do not contribute very much, and show that that $$\psi(x)=\theta(x)+O\left(\sqrt{x}\log x\right).$$ From this and the assumption that $\psi(x)-x=O(\sqrt{x}\log^2 x)$, it follows that $$\theta(x)-x=O\left(\sqrt{x}\log^2 x\right).$$

Step 2: Use partial summation to prove that $$\pi(x)-\text{li}(x)=\frac{\theta(x)-x}{\log x}+\int_{2}^{x}\frac{\theta(t)-t}{t\log^{2}t}dt+O(1),$$ and deduce the desired result.