I am trying to simplify the function: $$\sum_{n \equiv a(\bmod q)} \frac{\phi(n)-\mu(n)}{n^{s+1}}=\frac{1}{\phi\left(q^{\prime}\right)} \sum_\chi \bar{\chi}\left(a^{\prime}\right) \sum_{n^{\prime}=1}^{\infty} \chi\left(n^{\prime}\right) \frac{\phi\left(b n^{\prime}\right)-\mu\left(b n^{\prime}\right)}{\left(b n^{\prime}\right)^{s+1}}$$
where $b=(a,q), n'=n/b, a'=a/b, q'=q/b$. to the function: $$ \begin{aligned} & \sum_{n \equiv a(\bmod q)} \frac{\phi(n)-\mu(n)}{n^{s+1}} \\ & \quad=\frac{1}{b^{s+1} \phi\left(q^{\prime}\right)} \sum_\chi \frac{\bar{\chi}\left(a^{\prime}\right)(\phi(b) L(s, \chi)-\mu(b))}{L(s+1, \chi) \prod_{p \mid b}\left(1-\chi(p) p^{-s-1}\right)} \end{aligned} $$
But there is some problem due to the product of $b$ and $n'$ since they are not coprime necessartily. what's more, I am confused why the $\prod_{p|b}(1-\chi(p)p_{-s-1})$ comes out because I know how to prove the identity: $$ \sum_n \frac{\chi(n)(\phi(n)-\mu(n))}{n^{s+1}}=\frac{L(s, \chi)-1}{L(s+1, \chi)} $$ Do you know how to deal with it? Thank you for advance !
