How to regularize $\sum_{(a,b) \in \mathbb{Z}^2}' \frac{1}{\sqrt{a^2 + b^2}} $?

Question

I would like to evaluate the Dedekind zeta function of $\mathbb{Z}[i]$ at $\frac{1}{2}$. Naively the Dekind zeta function should be

$$ \zeta_{\mathbb{Z}[i]} (s) = \sum_{(a,b) \in \mathbb{Z}^2}' \frac{1}{(a^2 + b^2)^s} $$

Hopefully I have indexed the ideals of $I \subseteq \mathbb{Z}[i]$ correctly. Then to make matter worse, I am going to evluate at $s = \frac{1}{2}$:

$$ \zeta_{\mathbb{Z}[i]} (\frac{1}{2}) = \sum_{(a,b) \in \mathbb{Z}^2}' \frac{1}{\sqrt{a^2 + b^2}} $$

This is a divergent series, but all is not lost. Since if we define $\zeta(s)$ for $\mathrm{Re}(s) > 2$ then we could try to analytic continue to the rest of $\mathbb{C}$. (Does this series converge for $\mathrm{Re}(s) > 1$ ?)

---

The strategy I am seeing, uses Fermat little Theorem to express $\zeta_{\mathbb{Z}[i]}(s)$ in terms of $\zeta(s)$ and $L(s, \chi_4)$. In that case, I'd like to see how to regularize:

\begin{eqnarray*} \zeta(\frac{1}{2}) &"="& 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots \\ L(\frac{1}{2}, \chi_4) &"="& 1 - \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{5}} - \dots \end{eqnarray*}

I think the second one already has a value without quotation marks, but the first one we need to subtract out the infinite part, $2\sqrt{N}$.

Answer 1

$$L(s) = \zeta_{\mathbb{Z}[i]}(s)=\sum_{(a,b)\in\mathbb{Z}^2}'\frac{1}{(a^2+b^2)^s}=\sum_{n\geq 1}\frac{r_2(n)}{n^s}=\sum_{n\geq 1}\frac{4\sum_{d\mid n}\chi_4(d)}{n^s} $$ leads to $$ L(s) = 4\cdot L(\chi_4,s)\cdot\zeta(s) $$ where $\chi_4$ is the non-principal Dirichlet's character $\!\!\pmod{4}$. It follows that the $\zeta$-regularization of $\sum_{(a,b)\in\mathbb{Z}^2}'\frac{1}{(a^2+b^2)^s}$ at $s=\frac{1}{2}$ equals $$ 4\,\zeta\left(\frac{1}{2}\right)\sum_{n\geq 0}\frac{(-1)^n}{\sqrt{2n+1}}\approx -3.90026492. $$ The average value of $r_2(n)$ is $\pi$ by Gauss circle problem, hence $\sum_{(a,b)\in\mathbb{Z}^2}'\frac{1}{(a^2+b^2)^s}$ is convergent for any $s:\text{Re}(s)>1$, by summation by parts.