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$S = \sum_{k=1}^{\infty} \frac{\cos(\theta\log(k))}{k^a}$

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...

1 Answers 1

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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.

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    Thanks, I'll check out that link J.M.2010-10-18