TITLE: A Dirichlet Convolution involving $\mu(n)$ and $\log n$
QUESTION [8 upvotes]: The following arithmetic identity holds:
\begin{align}
\Lambda(n) = \sum_{d \mid n} \mu(d) \log \frac{n}{d}
\end{align}
where $\mu(n)$ is the Moebius function and $\Lambda(n)$ is the von Mangoldt function. Does the following related Dirichlet convolution simplify to known (or simpler) functions?
\begin{align}
n \sum_{d \mid n} \frac{\mu(d)}{d} \log \frac{n}{d}
\end{align}

REPLY [4 votes]: Yes,$$n \sum_{d \mid n} \frac{\mu(d)}{d} \log \frac{n}{d}=n \sum_{d \mid n} \frac{\mu(d)}{d} (\log(n)-\ln(d))$$
$$= \sum_{d \mid n} \frac{\mu(d)}{d} n\log(n)-n\frac{\mu(d)}{d}\ln(d)=n\ln(n) \sum_{d \mid n} \frac{\mu(d)}{d} -n\sum_{d\mid n}\frac{\mu(d)}{d}\ln(d)$$
$$=\ln(n)\phi(n)-n\sum_{d\mid n}\frac{\mu(d)}{d}\ln(d)=\ln(n)\phi(n)+\phi(n)\sum_{p\mid n}\frac{\ln(p)}{p-1}$$
Thus,
$$n \sum_{d \mid n} \frac{\mu(d)}{d} \log \frac{n}{d}=\phi(n)( \ln(n)+\sum_{p\mid n}\frac{\ln(p)}{p-1})$$