I have the following question: Let $F(s)$ be the Dirichlet series associated to $$f(n)= \sum_{d\mid n} \frac{\log(d)}{d} $$. My answer has to depend on the zeta function. (i.e simplify F(s) so that we can see its relation to the zeta function) Here is what I have so far:
$$ F(s)= \sum_{n=1}^{infinity} \sum_{d\mid n} \frac{\log(d)}{d} \frac{1}{n^s}$$ $$=\prod_{P} (1+ \frac{f(P)}{P^s} + \frac{f(P^2)}{P^{2s}}+....+)$$ $$ = 1+ \frac{\log P}{P *P^s}+ \frac{\log(P)}{P P^{2s}}+ \frac{\log(P^2)}{P^2 P^{2s}}+.... $$
I am stuck. I don't know how to write the sum in function of $$\zeta(s)= \prod_{P} (1+ 1/P^s + 1/P^{2s}....)$$
thank you!
Let $D(f, s)$ denote the Dirichlet series over $f$. We known by a standard convolution identity that $\log = \Lambda \ast 1$, that $D(\Lambda, s) = -\zeta^{\prime}(s) / \zeta(s)$, and that $D(h \ast g, s) = D(h, s) \cdot D(g, s)$ for any arithmetic functions $h$ and $g$. So your Dirichlet series is given by the product $$D(f, s+1) = D(\log \ast \operatorname{Id}_{-1}) = -\frac{\zeta^{\prime}(s)}{\zeta(s)} \cdot \zeta(s) \cdot \zeta(s-1),$$ which implies that $D(f, s) = -\zeta^{\prime}(s+1) \zeta(s)$.