Find $ \sum_{ (m,n) \neq (0,0)} \sqrt{m^2 + n^2}$ or $\zeta_{\mathbb{Q}[i]}(-1)$

Question

I would like to compute $\zeta_{\mathbb{Q}[i]}(-1)$ - a Dedekind zeta function. Mimicking the computation for $\zeta(-1)$, we can observe the following diverges:

$$ \frac{1}{4}\sum_{ (m,n) \neq (0,0)} \sqrt{m^2 + n^2} = 1+\sqrt{2}+2 + 2 \sqrt{5} + \sqrt{8} + \dots $$

and I would like to gives infinite divergent sum a finite value along the same line as these answers:

• Why does $1+2+3+\cdots = -\frac{1}{12}$?

In particular there is Abel's theorem which I am going to misuse slightly. If $\sum a_n$ converges then:

$$ \lim_{x \to 1^{-}} \sum a_n x^n = \sum a_n $$

which is a statement about continuity of the infinite series in $x$. Trying to make it work here.

$$ \sum_{(m,n) \neq (0,0)} \sqrt{m^2 + n^2} \;x^{\sqrt{m^2 + n^2}} = \frac{d}{dx}\Bigg[\sum_{(m,n) \neq (0,0)} x^{\sqrt{m^2 + n^2}} \Bigg]$$

This is not so helpful as I now have a Puisieux series (what on earth is $x^\sqrt{2}$ ?) and there is no closed form. What about:

$$ \sum_{(m,n) \neq (0,0)} \sqrt{m^2 + n^2} \;x^{m+n} = \frac{d}{dx}\bigg[\sum_{(m,n) \neq (0,0)} x^{m+n} \bigg]$$ This could converge as long as we have an estimate for the sum (this could be a separate strategy): $$ \sum_{m+n = N} \sqrt{m^2 + n^2} $$ maybe zeta-function regularization is our only option. The Dedekind function does have a Mellin transform

$$ \sum_{(m,n) \neq (0,0)} \sqrt{m^2 + n^2} \;e^{t\sqrt{m^2 + n^2}} = \frac{d}{dt}\Bigg[\sum_{(m,n) \neq (0,0)} e^{t\sqrt{m^2 + n^2}} \Bigg]$$

similar to what I have found. So that zeta regularization and Abel regularization are kind of the same.

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Note As I've written it $\sum \sqrt{m^2 + n^2} = \zeta_{\mathbb{Q}(i)}(-\frac{1}{2})$ which I imagine should not attain any special value :-/

Answer 1

You may find here a derivation of the identity : $$\tag{1}\sum_{(n,k)\neq (0,0)}\frac{1}{\left ( n^2+k^2 \right )^s}=4\,\zeta(s)\,\beta(s),\quad\Re(s)>1$$ with $\beta$ the Dirichlet beta function

so that analytic continuation to the (out of bounds) value $\,s=-\frac 12\,$ should give you the (regularized) series : \begin{align} \tag{2}\frac{1}{4}\sum_{ (m,n) \neq (0,0)} \sqrt{m^2 + n^2} &=\zeta\left(-\frac 12\right)\,\beta\left(-\frac 12\right)\\ &\approx -0.0572060775943\\ \end{align}