Assume $\gcd(a,b)=1$, how to express $F(s)=\sum_{n \equiv a \pmod b} \frac{\mu(n)}{n^s}$ in the half plane where $\Re(s)>1$ in terms of Dirichlet L-function?
Express Dirichlet series in terms of Dirichlet L-function
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0In the case $a=0$, $b=1$, we have $F(s)=1/\zeta(s)$. – 2011-10-28
1 Answers
Recall the orthogonality relations between the dirichlet characters modulo $b$ :
$\frac{1}{\phi(b)}\sum_{\chi\ \text{mod}\ b}\chi(n)=\left\{ \begin{array}{cc} 1 & \text{if }n\equiv1\text{ mod }b\\ 0 & \text{otherwise} \end{array}\right\}.$
Consequently we may rewrite the indicator function for the arithmetic progression $n\equiv a \text{ mod } b$ as
$\frac{1}{\phi(b)}\sum_{\chi\ \text{mod}\ b}\chi(n)\overline{\chi(a)}=\left\{ \begin{array}{cc} 1 & \text{if }n\equiv a\text{ mod }b\\ 0 & \text{otherwise} \end{array}\right\}.$
Hence your sum becomes
$F(s)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{s}}\left(\frac{1}{\phi(b)}\sum_{\chi\ \text{mod}\ b}\chi(n)\overline{\chi(a)}\right)$
$=\frac{1}{\phi(b)}\sum_{\chi\ \text{mod}\ b}\overline{\chi(a)}\sum_{n=1}^{\infty}\frac{\chi(n)\mu(n)}{n^{s}}$
$=\frac{1}{\phi(b)}\sum_{\chi\ \text{mod}\ b}\overline{\chi(a)}\frac{1}{L(s,\chi)}.$