I want to re-evaluate the infinite sum used in this paper (https://www.researchgate.net/publication/265112276_FAVORED_CARDINALITIES_OF_SCALES) over just the primes.
They have (presumably the author just looked this up in a table or book, because it's not really derived, and setting $\sigma$ = 0):
$$ \sum_{n=1}^\infty Cos(2\ \pi\ x\ Log\ n ) = Re\ \zeta(2\ \pi\ i\ x) $$
At any rate, does anyone know what this infinite sum evaluates to if n is restricted over just the primes?:
$$ \sum_{n\ \epsilon\ Primes}^\infty Cos(2\ \pi\ x\ Log\ n ) = \ ?$$
As a start, note that $\cos(z) =\Re e^{iz} $.
Therefore
$\begin{array}\\ \sum_{n=1}^\infty Cos(2\ \pi\ x\ Log\ n ) &=\sum_{n=1}^\infty \Re e^{2\ \pi\ i\ x\ Log\ n }\\ &=\Re\sum_{n=1}^\infty e^{2\ \pi\ i\ x\ Log\ n }\\ &=\Re\sum_{n=1}^\infty n^{2\ \pi\ i\ x }\\ &= \Re\ \zeta(2\ \pi\ i\ x)\\ \end{array} $
If you do the same thing for primes, you get
$\begin{array}\\ \sum_{n\in \mathbb{P}} Cos(2\ \pi\ x\ Log\ n ) &=\sum_{n\in \mathbb{P}} \Re e^{2\ \pi\ i\ x\ Log\ n }\\ &=\Re\sum_{n\in \mathbb{P}} e^{2\ \pi\ i\ x\ Log\ n }\\ &=\Re\sum_{n\in \mathbb{P}} n^{2\ \pi\ i\ x }\\ \end{array} $
I'm not sure where to go from here.