estimate sum of $\log \log p/p$ 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.
 A: 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,
$$ \sum_{p\leq x}\frac{\log\log p}{p} = \frac{1}{2}(\log\log x)^2+c_5+O\left(\frac{\log\log x}{\log x}\right). $$
A: UPD. GH's approach is, of course, better, the reason is that it uses a stronger asymptotical estimate: asymptotics $\sum_{p\leqslant x} \log p/p=\log x+O(1)$ implies $\sum_{p\leqslant x} 1/p=\log\log x+O(1)$, but not viceversa.
We use $$F(x):=\sum_{p\leqslant x} \frac1{p}=\log\log x+O(1),$$
see here about more precise statement.
Then apply Abel transform, for $h(p)=\log\log p$:
$$
\sum_{p\leqslant x}\frac{h(p)}{p}=\sum_{n\leqslant x} h(n)(F(n)-F(n-1))=h([x])F([x])-\sum_{n\leqslant x-1} F(n)(h(n+1)-h(n))
$$
First guy $h([x])F([x])$ equals $h^2(x)+O(h(x))$. As for the subtracted term, at first we replace each $F(n)$ to $h(n)+O(1)$, then sum of errors is 
$$
O\left(\sum_{n\leqslant x-1} (h(n+1)-h(n))\right)=O(h(x)).
$$ 
Next, for $h(n)(h(n+1)-h(n))$, it equals 
$$
h(n)(h(n+1)-h(n))=\frac{(h(n+1))^2-(h(n))^2}2-
\frac{(h(n+1)-h(n))^2}{2}.
$$
First terms cancel telescopically and give $\frac{h^2(x)}{2}+O(1)$, the second terms are really small and give just $O(1)$. 
To summarize, the answer is $\frac12 h^2(x)+O(h(x))$.
