Prove that $\displaystyle{\sum\limits_{p\le x\\ p\text{ prime}} \log p= x+O\left(\frac{x}{\log^2 x}\right)}$
Here I'm using Prime number theorem i.e. $\pi(x)=x/\log x+E(x)$ where $E(x)=o(x/\log x)$
Now $\displaystyle{\sum\limits_{p\le x\\ p\text{ prime}} \log p}$
$\displaystyle{=\int\limits_{2-\delta}^x}\log t\ d\pi(t)$
$\displaystyle{=\pi(t)\log t|_{2-\delta}^x-\int\limits_{2-\delta}^x}\frac{\pi(t)}{t}\ dt$
$\displaystyle{=\pi(x)\log x-\int\limits_2^x \frac{\pi(t)}{t}\ dt}$ (taking $\delta\to 0$)
$=x+o(x)+O\left(\int\limits_2^x \frac{1}{\log t}\ dt\right)$
Now I'm able to show that $O\left(\int\limits_2^x \frac{1}{\log t}\ dt\right)=o(x)$ (Just dividing the integral into two intervals $[0,\sqrt{x}]$ and $[\sqrt{x},x]$.
So finally I'm getting $\displaystyle{\sum\limits_{p\le x\\ p\text{ prime}} \log p=x+o(x)}$.
But it's not the form we are asked to prove. Can anyone help me in this regard? Thanks for your help in advance.
It is impossible to use OP's version of PNT to prove the desired result.
In fact, the remainder term looks like this:
$$ E(x)=\pi(x)-{x\over\log x}\asymp{x\over\log^2x} $$
By partial summation, we have
$$ \vartheta(x)=\sum_{p\le x}\log p=\pi(x)\log x-\int_2^x{\pi(t)\over t}\mathrm dt $$
However, because
$$ \int_2^x{\pi(t)\over t}\mathrm dt\gg\int_2^x{\mathrm dt\over\log t}\gg{x\over\log x} $$
Therefore even with the best possible $E(x)$, OP's method can only deduce
$$ \vartheta(x)=x+\mathcal O\left(x\over\log x\right) $$
On the other hand, if you use a more superior version of prime number theorem:
$$ \pi(x)=\int_2^x{\mathrm dt\over\log t}+\mathcal O\left(xe^{-c\sqrt{\log x}}\right) $$
where $c$ is an effectively computable positive constant (this is due to de la Vallée Poussin in 1898), then it is possible to improve the remainder term of $\vartheta(x)$ considerably:
\begin{aligned} \vartheta(x) &=\int_{2^-}^x\log t\mathrm d\pi(t)=\int_{2^-}^x\mathrm dt+\int_2^x\log t\mathrm d\left\{\mathcal O\left(te^{-c\sqrt{\log t}}\right)\right\} \\ &=x+\mathcal O\left\{x(\log x)e^{-c\sqrt{\log x}}\right\}+\mathcal O\left\{\int_2^xe^{-c\sqrt{\log t}}\mathrm dt\right\} \end{aligned}
To estimate the remaining integral, we can introduce a square root factor:
$$ \int_2^xe^{-c\sqrt{\log t}}\mathrm dt\le x^{1/2}e^{-c\sqrt{\log x}}\int_2^xt^{-1/2}\mathrm dt\ll xe^{-c\sqrt{\log x}} $$
Now, we can pick any $c'\in(0,c)$ to obtain the following asymptotic formula:
$$ \vartheta(x)=x+\mathcal O\left(xe^{-c'\sqrt{\log x}}\right) $$
This error term is a much sharper result than the $\mathcal O(x\log^{-2}x)$ stated by the OP.