How to prove $\sum_{d\mid q}\frac{\mu(d)\log d}{d}=-\frac{\phi(q)}{q}\sum_{p\mid q}\frac{\log p}{p-1}$?

Question

Prove that $$\sum_{d\mid q}\frac{\mu(d)\log d}{d}=-\frac{\phi(q)}{q}\sum_{p\mid q}\frac{\log p}{p-1},$$ where $\mu$ is Möbius function, $\phi$ is Euler's totient function, and $q$ is a positive integer.

I can get \begin{align} \sum_{d\mid q} \frac{\mu(d)\log d}{d}& = \sum_{d\mid q}\frac{\mu(d)}{d}\sum_{p\mid d}\log p \\ & = \sum_{p\mid q} \log p \sum_{\substack{d\mid q \\ p\mid d}} \frac{\mu(d)}{d} = \sum_{p\mid q} \log p \sum_{\substack{d \\ p\mid d \mid q}} \frac{\mu(d)}{d}, \end{align} Let $d=pr$, then $\mu(d)=\mu(p)\mu(r)=-\mu(r)$, $$ \sum_{p\mid q} \log p \sum_{\substack{d \\ p\mid d \mid q}} \frac{\mu(d)}{d}= - \sum_{p\mid q} \frac{\log p}{p} \sum_{\substack{r\mid q \\ p \nmid r}} \frac{\mu(r)}{r}.$$ But I don't know why $$- \sum_{p\mid q} \frac{\log p}{p} \sum_{\substack{r\mid q \\ p \nmid r}} \frac{\mu(r)}{r}=-\frac{\phi(q)}{q} \sum_{p\mid q} \frac{\log p}{p-1}?$$

Can you help me?

Answer 1

Let me write $n$ instead of $q$. We have \begin{align} \sum_{d|n}\frac{\mu(d)\log(d)}d &=\sum_{d|n}\frac{\mu(d)}d\sum_{p|d}\log(p)\\ &=\sum_{p|n}\log(p)\sum_{p|d|n}\frac{\mu(d)}d\\ &=\frac 1n\sum_{p|n}\log(p)\sum_{p|d|n}\mu(d)\frac nd \end{align} Write $n=p^em$ with $p\nmid m$. Then $\varphi(n)=p^{e-1}(p-1)\varphi(m)$ and \begin{align} \sum_{p|d|n}\mu(d)\frac nd &=\sum_{d\mid m}\sum_{i=1}^e\mu(p^id)\frac{p^em}{p^id}\\ &=\sum_{d\mid m}\mu(pd)\frac{p^em}{pd}\\ &=-\sum_{d\mid m}\mu(d)\frac{p^em}{pd}\\ &=-p^{e-1}\varphi(m)\\ &=-\frac{\varphi(n)}{p-1} \end{align}

Answer 2

I find a paper "On some identities in multiplicative number theory", Olivier Bordellès and Benoit Cloitre, arXiv:1804.05332v2 https://arxiv.org/abs/1804.05332v2

Using Dirichlet convolution \begin{eqnarray*} - \frac{\varphi(n)}{n} \sum_{p \mid n} \frac{\log p}{p-1} &=& - \frac{1}{n} \sum_{p \mid n} \varphi \left( \frac{n}{p} \right) \log p \\ &=& - \frac{1}{n} \left( \Lambda \ast \varphi \right) (n) \\ &=& - \frac{1}{n} \left( - \mu \log \ast \mathbf{1} \ast \mu \ast \mathrm{id} \right) (n) \\ &=& \frac{1}{n} \left( \mu \log \ast \mathrm{id} \right) (n) \\ &=& \sum_{d \mid n} \frac{\mu(d) \log d}{d}. \end{eqnarray*}