What is the sum of all complex integers? 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?
You might want to restrict to positive integers only since...

$$f(s) = \sum_{a&gt;0,b&gt;0} \frac{1}{(a+ib)^s} + \sum_{a&gt;0,b&lt;0} \frac{1}{(a+ib)^s} + \sum_{a&lt;0,b&gt;0} \frac{1}{(a+ib)^s} \\+ \sum_{a&lt;0,b&lt;0} \frac{1}{(a+ib)^s} + \sum_{a=0,b\in Z}\frac{1}{(ib)^s} + \sum_{a\in Z,b=0} \frac{1}{a^s}$$

giving

$$f(s) = (1 + (-1)^s)\left[\left(\sum_{a&gt;0,b&gt;0} \frac{1}{(a+ib)^s} + \frac{1}{(a-ib)^s}\right) + \zeta(s)(1+i^s)\right]$$

So $f(-1)$ (meaning the analytical continuation) will be $0$ (unless the analytical continuation of $\sum_{a&gt;0,b&gt;0} \frac{1}{(a+ib)^s} + ...$ 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.
I'm not sure how to go about proving it in any form of notation, but if we consider the real number axis from -x to +x, and the imaginary axis from -iy to +iy, then all the complex integers will be plotted as iy against x.  Since there are no limits to the size of either real or imaginary parts and given that they may only be integers.  I would have thought that the sum of all the complex integers would be zero.  Mainly because the sum of all imaginary numbers from -iy to +iy is zero, and all the real numbers from -x to +x is zero.  Am I missing something?

I see what you are saying, so 1+(1-1-1)+(1-1-1)...=-∞
But 1-(1-(1-1-1))-(1-(1-1-1))...=+∞

I'm not sure that eventually if all the combinations of number series is included in the argument a situation would arise where the net result is asymmetric.

Thank you Winther, I don't know if everyone will agree with you, but I do. But I'm beggining to see now that with complex numbers the problem may well be different.