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

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:

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.

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 :-/

• For the record, $x^{\sqrt2}$, when $x>0$, can be seen as $x^k$ for $k\in\mathbb Q$ where we take $k\to\sqrt2$. (mind my informal language) – Simply Beautiful Art Dec 2 '16 at 22:08
• Try consider $f(x)=\sum(m^2+n^2)^x$ and analytically continue it to $x=1/2$, if you are merely interested in some numerical value, this could work. – Simply Beautiful Art Dec 2 '16 at 22:13
• $\zeta_K(-1)$ has nothing to do with the series representation in the half-plane, it's outside of that one's area of convergence. – Adam Hughes Dec 4 '16 at 3:10

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}
• @ RaymondManzoni It is far to be elementary that $\zeta_{\mathbb{Q}(i)}(s) =4 \zeta(s)\beta(s)$. You can use some abstract algebra (splitting of prime ideals in $\mathbb{Z}[i]$) or the theory of modular forms ($\vartheta(q)^4$ is a modular form of weight $2$ which is a 2 dimensional vector space). Did you find a simpler way ? – reuns Dec 3 '16 at 17:42
• @user1952009: I provided an analytic solution, the other ones (in my two last 'here' links) are rather algebraic and their exposition is shorter (Jack D'Aurizio wrote "I believe that the algebraic approach is way easier to follow"). 'Simpler' is a relative concept depending of your background (I have nothing really simple to propose here) but the proposed links may at least help people to decide what is simpler for them. Different approaches may be useful too for more than two dimensions or if a signed coefficient multiplies $n^2$ and so on... – Raymond Manzoni Dec 3 '16 at 19:21
• Just to point out, perhaps this is not helpful. It's fairly easy to show that for any Galois number field $K$ one has that $\zeta_K$ is the $L$-function of the regular representation of $\text{Gal}(K/\mathbb{Q})$. So, if $K/\mathbb{Q}$ is abelian then $\zeta_K$ is a product of the Artin $L$-functions attached to the set of all characters of $\chi:\text{Gal}(K/\mathbb{Q})\to\mathbb{C}^\times$. In particular, for $K=\mathbb{Q}(i)$ this gives you the above decomposition. – Alex Youcis Dec 3 '16 at 23:39