Question: If $p$ and $q$ are prime numbers how do I write $\sum\limits_{p \leq q \text{ prime}}\log (p-1)$ using Chebyshev's function(s) ?
I would like to think $\sum\limits_{p \leq q \text{ prime}}\log (p-1)=\vartheta(q-1)$ but disturbingly I lack faith in that thought.
Which is correct (the lack of faith I mean), because $$\sum\limits_{p \leq q \text{ & prime}}\log{(p-1)}= \sum\limits_{p \leq q \text{ & prime}} \left(\log{p}+\log{\left(1-\frac{1}{p}\right)}\right)=\\ \sum\limits_{p \leq q-1 \text{& prime}} \log{p}+\log{q}+\sum\limits_{p \leq q \text{ & prime}}\log{\left(1-\frac{1}{p}\right)}=\\ \vartheta(q-1)+\log{q}+\sum\limits_{p \leq q \text{ & prime}}\log{\left(1-\frac{1}{p}\right)}=\\ \vartheta(q-1)+\log{\left(q\cdot\prod\limits_{p \leq q \text{ & prime}}\left(1-\frac{1}{p}\right)\right)}=\\ \vartheta(q-1)+\log{\left(\frac{q}{\log{q}}\cdot \log{q}\prod\limits_{p \leq q \text{ & prime}}\left(1-\frac{1}{p}\right)\right)} = ...$$
which, according to Mertens' third theorem
$$... = \vartheta(q-1)+\log{\left(\frac{q}{\log{q}}\right)} + \log{\left( \log{q}\prod\limits_{p \leq q \text{ & prime}}\left(1-\frac{1}{p}\right)\right)} \sim \\ \vartheta(q-1)+\log{\left(\frac{q}{\log{q}}\right)} -\gamma$$ for large prime $q$'s.