# Zeta like function summing over Gaussian integers in the first quadrant

Let $$x$$ be a real number and let

$$f(x)=\sum_{ z = re^{\theta i} \in\mathbb{Z}[i] \\ r \le x \\ 0\le \theta \le \pi/2}\frac{1}{z^s}$$

Is it possible to compute in (terms of the $$\zeta$$ function perhaps) $$\xi (s)=\lim_{x\to\infty} f_s(x)$$

It looks just by toying around that $$\xi(2)$$ is divergent. Does $$\xi$$ converge for larger $$s$$?

Here's some empirical evidence that $$\xi(2)$$ diverges.

Let $$\tau(n) : \mathbb{N} \to \mathbb{N}[i]$$ be a pairing function and let $$\tau(n)=x_n+y_ni$$.

We are looking for $$\frac{1}{\tau(n)^2}=\frac{1}{(x_n+y_ni)^2}= \frac{(x_n-y_ni)^2}{(x_n^2+y_n^2)^2}= \frac{x_n^2-2x_ny_ni-y_n^2}{(x_n^2+y_n^2)^2}=\frac{x_n^2-y_n^2}{(x_n^2+y_n^2)^2}+i\frac{2x_ny_n}{{(x_n^2+y_n^2)^2}}$$

We have then $$\sum_{ }\frac{1}{z^2} =\sum_{n=1 \\ x+yi=\tau(n)}^\infty\frac{x^2-y^2}{(x^2+y^2)^2}+2i\sum_{n=1 \\ x+yi=\tau(n)}^\infty\frac{xy}{(x^2+y^2)^2}$$ By numerical methods it seems to me that the sum diverges.

• I like to use $\mathbb{N}[i]$ for the Gaussian integers with real and imaginary parts non-negative. – Mason Dec 4 '18 at 0:35
• No to your first question. Do you know Hecke L-functions ? For example $L(s,\psi^n)$ with $\psi(a+ib) = \frac{a+ib}{a-ib}$, for $n$ even it is a Hecke character depending only on the embeddings $\mathbb{Z}[i] \to \mathbb{C}$ and not on the quotient rings $\mathbb{Z}[i]/(n)$. We need to sum over all the $L(s,\psi^n),n \in \mathbb{Z}$ to obtain the completely multiplicative function $f(a+ib) = i^m$ if $arg(i^{-m} (a+ib)) \in [0,\pi/2)$ – reuns Dec 4 '18 at 0:58
• I don't know about Hecke L-functions. But I guess I'll do some reading – Mason Dec 4 '18 at 1:02
• Oops of course $f$ isn't multiplicative. But summing over all the $L(s,\psi^n)$ make it appear. – reuns Dec 4 '18 at 1:07

Yes, $$\xi(2)$$ diverges. Note that for $$z$$ in this quadrant, $$-\text{Im}(1/z^2) \ge 0$$. So $$-\text{Im} \sum_{z \in \mathbb N[i]: 0 < |z| < r} 1/z^2$$ can be approximated by $$\int_1^r d\rho \int_0^{\pi/2} d\theta\; r \frac{\cos(\theta)}{r^2} \sim {\text {const}} \cdot \log(r)$$ (essentially the sum is a two-dimensional Riemann sum for the integral).
• I, of course, appreciate this answer and will see whether I can use it to answer the question that I proposed but I should just comment that I am not accepting as the answer because it doesn't address my question which was: Does $\xi$ converge for $s>2$ – Mason Dec 8 '18 at 16:17