If we assumed R.H. what is the best upper/lower bounds for $ E +\ln \theta(x) - \sum \limits_{p \leq x} \frac{\ln p}{p}$
And if one can also give the best upper/lower bounds for $ E +\ln x + (\frac{\theta(x)}{x}-1)- \sum \limits_{p \leq x} \frac{\ln p}{p}$
Where $E = -\gamma - \sum \limits_{P} \frac{\ln p}{p(p-1)} \approx -1.33258$
Any ref that gives partial answer (only lower bounds, or for one of the inequalities) will be helpful.
Thanks.
Note that $$\sum_{n\leq N}\frac{\Lambda\left(n\right)}{n}-\sum_{p\leq N}\frac{\log\left(p\right)}{p}\leq\sum_{p\geq2}\frac{\log\left(p\right)}{p\left(p-1\right)}$$ and, using the explicit formula $$\sum_{n\leq N}{}^{^{\prime}}\frac{\Lambda\left(n\right)}{n}=\log\left(N\right)-\gamma+\sum_{\rho}\frac{N^{-\rho}}{\rho}-\frac{1}{N}+\frac{1}{2}\log\left(\frac{N+1}{N-1}\right)$$ where the symbol $\prime$ indicates that if $N$ is a prime power then the term $n=N$ must be multiplied by $1/2$, we have $$E-\sum_{p\leq N}\frac{\log\left(p\right)}{p}-\log\left(N\right)\leq-\sum_{\rho}\frac{N^{-\rho}}{\rho}+\frac{1}{N}-\frac{1}{2}\log\left(\frac{N+1}{N-1}\right)$$ and so, if we assume RH, we obtain $$E-\sum_{p\leq N}\frac{\log\left(p\right)}{p}-\log\left(N\right)=O\left(N^{-1/2}\log^{2}\left(N\right)\right).$$ Now remains to observe that, since under RH we have $$\theta\left(N\right)=N+O\left(N^{1/2}\log^{2}\left(N\right)\right),$$ then $$\log\left(\theta\left(N\right)\right)-\log\left(N\right)=\log\left(1+O\left(N^{-1/2}\log^{2}\left(N\right)\right)\right)$$ and so $$E-\sum_{p\leq N}\frac{\log\left(p\right)}{p}+\log\left(\theta\left(N\right)\right)=O\left(N^{-1/2}\log^{2}\left(N\right)\right),$$ as indicated by Greg Martin.