It is known that $$\sum\_{p\leq x} \frac{\log p}{p}=\log x+c.$$


Are any tight bounds on
$$\sum\_{p\leq x} \frac{\log \log p}{p}$$ known?


I haven't managed to find anything in the literature. Trying to approximate via $p\_k\approx k \log k$ doesn't give a tractable integral.



estimate sum of $\log \log p/p$
Let $S(t):=\sum\_{p\leq t}\frac{\log p}{p}$. By Mertens' theorem, $S(t)=\log t+T(t)$, where $T(t)$ is bounded. It follows that
$$ \sum\_{p\leq x}\frac{\log\log p}{p}=\int\_{2-}^x\frac{\log\log t}{\log t}dS(t)=\int\_{2}^x\frac{\log\log t}{\log t}\cdot\frac{dt}{t}+\int\_{2-}^x\frac{\log\log t}{\log t}dT(t).$$
On the right hand side, the first term equals $\frac{1}{2}(\log\log x)^2+c\_1$, where $c\_1$ is a constant. The second term equals, via integration by parts and with further constants $c\_j$,
\begin{align} 
\int\_{2-}^x\frac{\log\log t}{\log t}dT(t) &= \left[\frac{\log\log t}{\log t}T(t)\right]\_{2-}^x-\int\_{2}^x\left(\frac{\log\log t}{\log t}\right)'T(t)\,dt\\
&=c\_2+O\left(\frac{\log\log x}{\log x}\right)+c\_3+O\left(\frac{\log\log x}{\log x}\right)\\
&=c\_4+O\left(\frac{\log\log x}{\log x}\right),
\end{align}
because $\left(\frac{\log\log t}{\log t}\right)'$ is negative for $t>e^e$. In the end,
$$ \sum\_{p\leq x}\frac{\log\log p}{p} = \frac{1}{2}(\log\log x)^2+c\_5+O\left(\frac{\log\log x}{\log x}\right). $$


