I am trying to understand whether some non trivial explicit estimate is known about the following $$\sum_{n \leq x} \mu(n) \chi(n)$$ as a function of $x$.
Here $\chi$ is a Dirichlet character modulo $q$ and $q > (\log x)^2$ (for example).
I am trying to understand whether some non trivial explicit estimate is known about the following $$\sum_{n \leq x} \mu(n) \chi(n)$$ as a function of $x$.
Here $\chi$ is a Dirichlet character modulo $q$ and $q > (\log x)^2$ (for example).
Estimating this sum is much the same as estimating the error term in the prime number theorem for arithmetic progressions. See Exercises 7-8 in Section 11.3 of Montgomery-Vaughan: Multiplicative number theory I (Cambridge University Press, 2006). (Your sum is denoted by $M(X,\chi)$ in this book, as introduced by (11.39) there.)
In particular, there is a constant $c>0$ such that for any $A>0$ we have $$ \sum_{n \leq X} \mu(n) \chi(n)\ll_A x\exp(-c\sqrt{\log x})$$ as long as $q\leq(\log x)^A$ .
Explicit estimates for the Moebius function are much harder than for the prime counting function. Ramare (From explicit estimates for the primes to explicit estimates for the Moebius function, Acta Arithmetica 157, (2013), 365-379) has shown how to translate estimates for primes to estimates for the Moebius function, as far as I know this approach is superior to a direct estimate using complex integration. It should be possible without too much work to change Sections 4, 5, 6 of that paper to the character case, as there is a lot of literature on various sums involving primes in arithmetic progressions (see the work of Rosser, Schoenfeld, McCurley, Rumely, Ramare). I would be worried about Lemma 9.1, but this is not necessary for your question.
There is an equivalent of the Riemann explicit formula for those :
If all the zeros of $L(s,\chi) = \sum_{n=1}^\infty \chi(n) n^{-s}$ are simple,
then $\displaystyle\frac{1}{L(s,\chi)} = \sum_{n=1}^\infty \mu(n)\chi(n) n^{-s}= s\int_1^\infty (\sum_{n < x}\mu(n) \chi(n))x^{-s-1}dx$ leads to $$\sum_{n < x} \mu(n) \chi(n) = \frac{1}{2i\pi}\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{1}{L(s,\chi)}\frac{x^s}{s}ds = \frac{1}{L(0,\chi)} + \sum_\beta \frac{x^{\beta}}{\beta L'(\beta,\chi)}$$ by Mellin inverse transform and the residue theorem, where $\sigma > 1$,
$\beta$ are the zeros (trivial and non-trivial) of $L(s,\chi)$,
and $\sum_\beta$ is convergent only when grouping the terms correctly.
(note if some zeros are of order $k$, it is not so different except you'll get some additional terms $x^\beta (\ln x)^{m}, m < k$)