TITLE: How to find the sum of this cos series
QUESTION [5 upvotes]: $$S = \sum_{k=1}^{\infty} \frac{\cos(\theta\log(k))}{k^a}$$
How do I go about finding the value of S, given that $\theta \to \infty$ and $0 < a < 1$.

Any special techniques that might be
  helpful in calculating this sum?

EDIT: 
Just to give some background,
I was actually trying to figure out
$$\sum_{k=1}^{\infty} \frac{\cos(\theta\log(k))}{k^a} - \sum_{k=1}^{\infty} \frac{\cos(\theta\log(k + 0.5))}{(k+0.5)^a}$$
Since that expression was a bit complicated, I decided to write the common version...

REPLY [4 votes]: Your value is basically the value of the Riemann zeta function:
$$
S = \sum_{k=1}^{\infty} \frac{Re (e^{i \theta \log(k)})}{k^a} = Re( \sum_{k=1}^{\infty} k^{i \theta - a} ) = Re( \zeta ( a - i \theta ))
$$
You want to evaluate this on the critical strip $0 < a < 1$.
The good news is that there is an enormous amount of literatue on the Riemann zeta. The bad news is that this function is nasty on the critical strip.