Dirichlet Convolution

Question

I was wondering how to simplify the expression: $id \ast (\mu\phi)$, where $\ast$ denotes Dirichlet convolution, $\mu$ is the Mobius function, $\phi$ is the Euler's totient function and $id$ is the identity function. Juxtaposition represents pointwise multiplication.

Answer 1

$$\left(\text{id}*(\mu\varphi)\right)(n)=\sum_{d\mid n}\frac{n}{d}\mu(d)\varphi(d)$$ is a multiplicative function and $$\left(\text{id}*(\mu\varphi)\right)(p^k)=p^k\sum_{d\mid p^k}\frac{\mu(d)}{d}\varphi(d)=p^k\left(1-\frac{p-1}{p}\right)=p^{k-1}$$ hence we simply have $$\left(\text{id}*(\mu\varphi)\right)(n)=\frac{n}{\text{rad}(n)} $$ where $\text{rad}(n)=\prod_{p\mid n}p$.