# What is the dirichlet series of $f(n)=\sum_{d | n}(\log d) / d$ function? [closed]

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.

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