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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?

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In the case $a=0$, $b=1$, we have $F(s)=1/\zeta(s)$. –  Gerry Myerson Oct 28 '11 at 6:02

1 Answer 1

up vote 8 down vote accepted

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

$$F(s)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{s}}\left(\frac{1}{\phi(b)}\sum_{\chi\ \text{mod}\ b}\chi(n)\overline{\chi(a)}\right)$$

$$=\frac{1}{\phi(b)}\sum_{\chi\ \text{mod}\ b}\overline{\chi(a)}\sum_{n=1}^{\infty}\frac{\chi(n)\mu(n)}{n^{s}}$$

$$=\frac{1}{\phi(b)}\sum_{\chi\ \text{mod}\ b}\overline{\chi(a)}\frac{1}{L(s,\chi)}.$$

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