How to calculate the sum $\sum_{n=1}^\infty n^{-a}(\log n)^b$

Question

Assume, $a>1,b>0$, how to calculate the infinite sum: $$\sum_{n=1}^\infty n^{-a}(\log n)^b$$

I tried with WolframAlpha, I found that this sum converge bu running sum (1/n^a)*(log n)^b, n=1 to infinity, But it does not give a result. How to calculate this sum? It seems does not has a closed form. But how can I write the result with the $\zeta$ function?

Answer 1

$$\underset{N\to\infty}{\text{lim}}\left(\sum_{n=1}^N\log^b(n)\ n^{-s}\right)=(-1)^b\ \zeta^{(b)}(s)\tag{1}$$

where

$$\Re(s)>1\land b\in\mathbb{Z}\land b>0\tag{2}$$

so the formula

$$\zeta^{(b)}(s)=\underset{N\to\infty}{\text{lim}}\left((-1)^b\ \sum_{n=1}^N \log^b(n)\ n^{-s}\right)\tag{3}$$

can be used to evaluate the $b^{th}$-order derivative of $\zeta(s)$ for $\Re(s)>1$.

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The formulas above are based on the term-wise differentiation

$$\frac{\partial\, n^{-s}}{\partial s}=-\log(n)\ n^{-s}\tag{4}$$

which leads to the more general result

$$\frac{\partial^b\, n^{-s}}{\partial s^b}=(-1)^b\, \log^b(n)\ n^{-s}\tag{5}.$$