Kronecker Zeta function

Question

If we define the Kronecker symbol K(a,n) as at Wikipedia, can we define

$$\zeta_K(s,a) = \sum_n \dfrac{K(a,n)}{n^s}$$? If so, what does it equal?

Answer 1

Although not a complete answer, I will list some observations.

First of all, the series is certainly defined for $\text{Re}(s)>1$, since the absolute value of $\left(\frac an\right)$ is bounded ($\le1$).

Now, since $K(a,n)$, by definition, is completely multiplicative with respect to $n$, we have the following Euler product:

$$\zeta_K(s,a)=\prod_{p:\text{prime}}\left(1-\left(\frac ap\right)p^{-s}\right)^{-1}.$$

Moreover, in the case $a$ is of the form $2^k(4m+1)$, because of quadratic reciprocity we have

$$\zeta_K(s,a)=\sum_{n=1}^\infty\left(\frac na\right)n^{-s},$$

where clearly $\left(\frac na\right)$ is a Dirichlet character modulo $a$. Thus, we have $\zeta_K(s,a)=L(s,\chi)$ with $\chi(n)=\left(\frac n a\right)$.