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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.

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  • $\begingroup$ I know $\log \log x$ is a lower bound. $\endgroup$
    – kodlu
    Dec 22, 2015 at 21:51

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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, $$ \sum_{p\leq x}\frac{\log\log p}{p} = \frac{1}{2}(\log\log x)^2+c_5+O\left(\frac{\log\log x}{\log x}\right). $$

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  • $\begingroup$ Interesting, thanks for providing an alternate technique, which gives a more accurate estimate. $\endgroup$
    – kodlu
    Dec 22, 2015 at 22:45
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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 $\frac12 h^2(x)+O(h(x))$.

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