equivalence between Chebyshev estimation and pi estimation in PNT I searched, but though many posts are close, none of them are dublicates.
Our version of PNT states that there is some $c$ s.t.
$$\psi(x)=x+O(x\exp(-c\sqrt{\ln x}))$$
We have to prove equivalence with existence of some c s.t.
$$\pi(x)=li(x)+O(x\exp(-c\sqrt{\ln x}))$$
where li(x) is the logarithmic integral from 2 to x. 
I know s solution that uses Stiltjes integration, but i seek a solution that does not touch stiltjes integration directly, as we have not really used any integrals w.r.t. "d g(t) " . We proved directly in class that 
$$\sum_{n\leq x}a_n f(n)=f(x)A(x)-\int_1^x A(t)f'(t)dt.$$
for $f\in C_1 , A(t)=\sum_{t\leq t} a_n$.
What i have done until now:
I have proved that $\psi(x)=\theta(x)+O(x^{\frac{1}{2}}ln \; x)$. where $\theta(x):=\sum_{p\leq x}\ln p$. The O term is small comparing to the one in the theorem, so i conclude that 
$$\theta(x)=x+O(x\exp(-c\sqrt{\ln x}))$$
So my idea was simply to try get something out of the integration rule with $A=\theta$ and $f=\frac{1}{\ln}$
$$\pi(x)=\frac{\theta(x)}{\ln x}+\int_{2^{-}}^x \frac{\theta (t)}{t\ln^2 t} dt.$$
It smells right, because the above equaity was stated as a previous exercise, but i cannot move on from here, even though it was stated as an easy exercise.
I would be very greatful if you do would not state the solution, but instead write a small hint to get me going from here.
 A: Together with a friend, I derived an answer. 
From the Integral expression written in the question, combined with the O result we get 
$$\pi(x)=\frac{x}{\ln x}+\int_2^x \frac{t}{t\ln^{2}t}dt+O\left(x\exp(-c\sqrt{\ln x})+\int_2^x\frac{t\exp(-c\sqrt{\ln t})}{t\ln^{2} t}dt\right).$$
The thing i missed in my calculations, was that the non-O expressions, can be collected to $Li(x)$ using partial integration "backwards". 
$$=\int_2^x \frac{1}{\ln t} dt+O\left(x\exp(-c\sqrt{\ln x})+\int_2^x\frac{t\exp(-c\sqrt{\ln t})}{t\ln^{2} t}dt\right).$$
Hence we are done if we can show that  $\int_2^x\frac{t\exp(-c\sqrt{\ln t})}{t\ln^{2} t}dt=O(x\exp(-K\sqrt{\ln x}).$ atleast for some $K>0$.
To show it, we start substituting $u=\ln x$ 
$$\int_2^x\frac{t\exp(-c\sqrt{\ln t})}{t\ln^{2} t}dt=\int_{\ln 2}^{\ln x}e^u\exp(-c\sqrt{u}) du.$$
For c small enough the integrand is increasing on the interval, hence
$$\leq (\ln x-\ln 2) x\exp(-c\sqrt{ln x})=O(x(\ln x)\exp(-c\sqrt{\ln x})).$$
A quick consideration shows that this is $O(x\exp(-K\sqrt{\ln x}))$ for any $K<c$.
