TITLE: What is the sum of all complex integers?
QUESTION [8 upvotes]: In line with 
$$\zeta(-1)=-1/12$$
Could we, by considering 
$$f(s)=\sum_{a,b\in\mathbb Z,\;(a,b)\neq(0,0)}\frac{1}{(a+bi)^{s}}$$
Evaluate the sum of all complex integers?

REPLY [5 votes]: You might want to restrict to positive integers only since...
$$f(s) = \sum_{a>0,b>0} \frac{1}{(a+ib)^s} + \sum_{a>0,b<0} \frac{1}{(a+ib)^s} + \sum_{a<0,b>0} \frac{1}{(a+ib)^s} \\+ \sum_{a<0,b<0} \frac{1}{(a+ib)^s} + \sum_{a=0,b\in Z}\frac{1}{(ib)^s} + \sum_{a\in Z,b=0} \frac{1}{a^s}$$
giving
$$f(s) = (1 + (-1)^s)\left[\left(\sum_{a>0,b>0} \frac{1}{(a+ib)^s} + \frac{1}{(a-ib)^s}\right) + \zeta(s)(1+i^s)\right]$$
So $f(-1)$ (meaning the analytical continuation) will be $0$ (unless the analytical continuation of $\sum_{a>0,b>0} \frac{1}{(a+ib)^s} + ...$ has a simple pole at $s=-1$). 
The same happens if one consideres the sum of all integers:
$$g(s) = \sum_{n\in Z, n\not= 0} \frac{1}{n^s} = (1 + (-1)^s)\sum_{n=1} \frac{1}{n^s} = \zeta(s)(1 + (-1)^s)$$ 
so the 'sum of all integers' are zero.