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My opinion is ;

We may use id(d)=d arithmetic function and log*id dirichlet convolution in the question. i thought that ; when we multiply and divide n with $(\log d) / d$ we obtain $F(S)=\sum_{n=1}^{\infty} \frac{1}{n^{s+1}} \cdot\left(\log*id\right)$ so

$F(S)=D(log,s+1).D(id,s+1)$

So we get

$F(S)=-\zeta^{\prime}(s+1). \zeta(s)$

i am curious that my solution is right or not.You may write your solutions and different ideas. thanks for your helps.

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We can verify this by direct calculation. By definition, $$F(s)=\sum_{n=1}^\infty\frac{f(n)}{n^s}=\sum_{n=1}^\infty\frac{1}{n^s}\sum_{d\mid n}\frac{\log d}{d}.$$ We rearrange the double sum so that $d$ comes first, and then write $n=dm$ to separate variables. We get \begin{align*} F(s)&=\sum_{d=1}^\infty\frac{\log d}{d}\sum_{m=1}^\infty\frac{1}{(dm)^s}=\left(\sum_{d=1}^\infty\frac{\log d}{d^{s+1}}\right)\left(\sum_{m=1}^\infty\frac{1}{m^s}\right). \end{align*} On the right hand side, the first factor is $-\zeta'(s+1)$, while the second factor is $\zeta(s)$, so $$F(s)=-\zeta'(s+1)\zeta(s).$$

P.S. Your question is not of research level, but I felt like answering it.

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    $\begingroup$ it's nice solution , i understand better it now. thanks:) $\endgroup$ – queen028 Apr 29 at 0:18
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you may look to this answer by mds in SE

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 $D(f, s) = -\zeta^{\prime}(s+1) \zeta(s)$.

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  • $\begingroup$ thanks for solution. its different perspective. $\endgroup$ – queen028 Apr 29 at 7:41

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