Let $\pi, \vartheta$ be respectively the prime counting function and the first chebyshev function. As you know, $ \pi(x) \sim x/\log x$, and $\vartheta(x) \sim x$, so that, at first order, seems $\pi(x) \log x \sim \vartheta(x)$. Is it easy to state that $\pi(x) \log x > \vartheta(x)$ (just use definition of $\theta$), but my question is:
Can you find an asymptotics for $\pi(x)\log x - \vartheta(x)$?
I met it while approaching to Combinatorial Interpretation of a Certain Product of Factorials
Using $$\pi\left(x\right)=O\left(\frac{x}{\log\left(x\right)}\right)$$ and, by partial summation$$\theta\left(x\right)=\pi\left(x\right)\log x-\int_{2}^{x}\frac{\pi\left(t\right)}{t}dt$$ you have$$\pi\left(x\right)\log(x)-\theta\left(x\right)=\int_{2}^{x}\frac{\pi\left(t\right)}{t}dt=O\left(\int_{2}^{x}\frac{1}{\log\left(t\right)}dt\right)=O\left(\textrm{li}\left(x\right)\right)=O\left(\frac{x}{\log\left(x\right)}\right)$$ where $\textrm{li}\left(x\right)$ is the logarithmic integral. So$$\pi\left(x\right)\log(x)=\theta\left(x\right)+O\left(\frac{x}{\log\left(x\right)}\right)$$ as $x\rightarrow\infty.$
