For $a, b$ relatively prime, I would like to express $$\sum_{n \equiv a\mod b} \frac{\mu(n)}{n^s}$$ in term of Dirichlet L function.
Not sure how to handle this kind of question. Do not find any Theorems in my books state something useful about it either.
Any help please ?
We use character orthogonality to write
\[\sum_{\substack{n = 1 \\ n \equiv a \pmod{q}}}^{\infty} \frac{\mu(n)}{n^s} = \frac{1}{\varphi(q)} \sum_{\chi \pmod{q}} \overline{\chi}(a) \sum_{n = 1}^{\infty} \frac{\mu(n) \chi(n)}{n^s}.\] Then via multiplicativity, we find that \[\sum_{n = 1}^{\infty} \frac{\mu(n) \chi(n)}{n^s} = \prod_p \left(1 - \frac{\chi(p)}{p^s}\right) = \frac{1}{L(s,\chi)}.\] So \[\sum_{\substack{n = 1 \\ n \equiv a \pmod{q}}}^{\infty} \frac{\mu(n)}{n^s} = \frac{1}{\varphi(q)} \sum_{\chi \pmod{q}} \frac{\overline{\chi}(a)}{L(s,\chi)}.\] This is initially valid for $\Re(s) > 1$ (because then both sides converge absolutely), but this then gives a meromorphic extension to all of $\mathbb{C}$.