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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}

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    For the vast majority of the integers your dirichlet convolution is somewhat close to $\phi(n)/n \cdot n \log n$. The only (possible) exceptions are those integers with very many prime factors, say more than $(\log n)^{1 - \varepsilon}$.2012-11-30

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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, $n \sum_{d \mid n} \frac{\mu(d)}{d} \log \frac{n}{d}=\phi(n)( \ln(n)+\sum_{p\mid n}\frac{\ln(p)}{p-1})$