# Is this function familiar to anyone?

Consider

$$f(z)=\sum_{w\in C}\frac{1}{z-w}$$ Where $C$ is the set of complex integers. What I would like to know is where can I find any information about this function (name perhaps). For instance, I don't even know how the summation would be carried out considering there are so many different ways to sum over the lattice. Does it converge? How fast does it grow? Etc... I think it must be related to elliptic functions because it is doubly periodic, but knowing nothing else of EF, I am sortof stuck. Any thoughts would be great.

• Why do you think that function is defined? – Thomas Andrews Jun 1 '14 at 14:36
• Well it seems like the simplest function periodic in two directions. I'm hoping it exists (converges, that is) because it is related to how many integer lattice points lie within a circle of certain radius (Gauss' circle problem), and is just generally interesting IMO. – Elie Bergman Jun 1 '14 at 14:38
• It doesn't work, unfortunately. The basic examples of doubly-periodic functions involve squaring the $\frac{1}{z-w}$ because otherwise there is no limit for this sum. See en.wikipedia.org/wiki/Weierstrass's_elliptic_functions – Thomas Andrews Jun 1 '14 at 14:42

The following is taken pretty much verbatim from Apostol. In general, consider the function $f_a:\mathbb{C}-\Lambda\to \mathbb{C}$ (with $\Lambda =\{n+im|(n,m) \in \mathbb{Z}^2\}$) $$f_a(z)=\sum_{(m,n )\in \Lambda}\frac{1}{(z-(n+im))^a}.$$ Consider the unit square with centre $C$ that $z$ is in. The minimum and maximum distance from $z$ to the vertices of the cell are $R,r>0$ respectively. Then consider the square with the same centre with side-length $3$. It has $4 \cdot 2$ points, all of which are at a distance between $2\cdot R$ and $2\cdot r$ from $z$. Repeating this procedure to the all squares with centre $C$ and edd side length, we have
$$\sum_{k \ge1}\frac{4k}{(kr)^a}\le \sum_{(m,n) \in \Lambda}\left|\frac{1}{(z-(n+im))^a}\right|\le \sum_{k \ge 1}\frac{4k}{(kR)^a}$$ $$\frac{4}{r^a}\zeta(a-1)\le \sum_{(m,n) \in \Lambda}\left|\frac{1}{(z-(n+im))^a}\right|\le \frac{4}{R^a}\zeta(a-1).$$
Thus $f_a(z)$ is convergent iff $\zeta(a-1)$ is convergent (i.e. $a>2$).
• It's foreseeable that you could analytically continue $f_a(z)$ for all complex $a \ne 2$. – Meow Jun 1 '14 at 16:37