It is well known¹ that

\[ \sum_{p\leq x}\ln p = x + O\biggl(\frac{x}{e^{c\sqrt{\ln x}}}\biggr) \]

where $c>0$ is a constant and the summation runs over the primes. In fact, under the Riemann hypothesis, one even has

\[ \sum_{p\leq x}\ln p = x + O(x^{1/2+\epsilon}) \]

for any $\epsilon>0$. Since $e^{c\sqrt{\ln x}}$ grows slower than any power of $x$, the second statement gives a better approximation.

A related question, but one I wasn’t familiar with, is to give a similar asymptotic result for the summation with $\ln p$ replaced by $\ln\ln p$. In other words, to estimate the quantity

\[ \sum_{p\leq x}\ln\ln p . \]

To do this, we may employ Abel’s summation formula with

\[ a_n := \begin{cases}
1 & \text{if $n$ is prime} \\
0 & \text{otherwise}
\end{cases} \]

and $\phi(n):=\ln\ln n$. Then we have

\[ \sum_{p\leq x}\ln\ln p = \pi(x)\ln\ln x-\int_2^x\frac{\pi(t)}{t\ln t}\mathrm{d}t . \]

By the prime number theorem we have $\pi(t)=t/\ln t+O(t/\ln(t)^2)$, so

\[ \int_2^x\frac{\pi(t)}{t\ln t}\mathrm{d}t = \int_2^x\frac{\mathrm{d}t}{\ln(t)^2}+O\biggl(\int_2^x\frac{\mathrm{d}t}{\ln(t)^3}\biggr) . \]

By Wolfram Alpha we have that

\[ \int_2^x\frac{\mathrm{d}t}{\ln(t)^2} = \DeclareMathOperator{\li}{li}\li(x)-\frac{x}{\ln x} + O(1) = \frac{x}{\ln(x)^2} + O\biggl(\frac{x}{\ln(x)^3}\biggr) , \]

with the latter equality following from the asymptotic expansion of the logarithmic integral.

It remains to estimate the integral $\int_2^x\ln(t)^{-3}\,\mathrm{d}t$. Actually, this is not entirely straightforward, but a trick is to split the integral into two (around $\sqrt{x}$) and then estimate each, as follows:

\[ \int_2^{\sqrt{x}}\frac{\mathrm{d}t}{\ln(t)^3} + \int_{\sqrt{x}}^x\frac{\mathrm{d}t}{\ln(t)^3} \leq \frac{\sqrt{x}-2}{\ln(2)^3} + \frac{x-\sqrt{x}}{\ln(\sqrt{x})^3} = O\biggl(\frac{x}{\ln(x)^3}\biggr) \]

Putting everything together, we find the result

\[ \sum_{p\leq x}\ln\ln p =\pi(x)\ln\ln x-\frac{x}{\ln(x)^2} +O\biggl(\frac{x}{\ln(x)^3}\biggr) . \]