8

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?

Did
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Elie Bergman
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  • Yes. The result is $-\frac{1}{12}-i \frac{1}{12}.$ – Giuseppe Negro Jun 01 '14 at 14:45
  • I doubt it but thanks for the guess – Elie Bergman Jun 01 '14 at 14:47
  • That's actually little more than a joke. Sorry if it wasn't clear. – Giuseppe Negro Jun 01 '14 at 14:53
  • I suppose this is equivalent to proving a Zeta reflection formula-like identity (see here) for $f(s)$. – Meow Jun 01 '14 at 15:07
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    Which values of $s$ should one consider? For non real integer values of $s$, there is no general definition of $(a+bi)^s$. – Did Jun 01 '14 at 15:59
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    On the other hand, note that $\sum\limits_{(a,b)}|a+bi|^{-s}$ converges absolutely for every complex $s$ such that $\Re s\gt1$. – Did Jun 01 '14 at 16:01
  • A good point on the non integer bit. I would define for general complex z $ z^s=\exp(s\ln(z))$. Where exp and log are define by taylor series and thus single valued. – Elie Bergman Jun 01 '14 at 16:03
  • Perhaps make it explicit in the question that you're looking for the analytic continuation of $f(s)$, not just a special case of the analytic continuation. – Meow Jun 01 '14 at 16:53
  • Nice problem. From a naive evalutation of the $s=-1$ sum it seems like $f(-1) = (1+i)\zeta(0)\zeta(-1) = \frac{1}{24}(1+i)$ would be a likely value. – Winther Jun 01 '14 at 16:59
  • How did you arrive at this? – Elie Bergman Jun 01 '14 at 17:02
  • It was just a loose conjecture. With $f(-1) = \sum_a \sum_b a+ib = (1+i)(\sum_a a^1) (\sum_b b^0) = (1+i)\zeta(0)\zeta(-1)$. btw this is considering the sum over positive $a,b$ only! – Winther Jun 01 '14 at 17:23
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    Except that the logarithm cannot be defined by its Taylor series (and which Taylor series by the way?). – Did Jun 01 '14 at 18:06
  • It can for this problem. We can use log(1-x). It is like saying $1^{1/2}\ne 1$ because it is multivalued. Yes, it $is$, but nonetheless we can force it to be single valued. – Elie Bergman Jun 01 '14 at 18:09
  • In other words, one knows how to define $(a+ib)^s$ when $|a+bi-1|\lt1$. Not many points with integer coordinates in there, I am afraid... :-) – Did Jun 01 '14 at 18:31
  • You would do it in terms of $(a+bi)^s = Re^{si\theta}$ – Elie Bergman Jun 01 '14 at 19:55
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    Then you would have to specify how to determine theta... Right, since you make me repeat the same basic point over and over again and do not wish to listen, I shall stop here. – Did Jun 01 '14 at 21:15

2 Answers2

5

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.

Winther
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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.

Darren J M England
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    You are missing that the sum of a series that is not absolutely convergent can be made to be anything at all, by suitable rearrangement of the terms. The situation is the same as with Grandi's series, $\sum_{i=0}^\infty (-1)^i = 1-1+1-1+\ldots = 0+0+0+\ldots = 0$ and also $= 1-(1-1)-(1-1)-\ldots = 1-0-0-\ldots = 1$. – MJD Jun 01 '14 at 15:28
  • Or similarly consider just the “sum of the integers”. You claim that $\sum_{i\in\Bbb Z} i = 0$, because you have grouped it as $0 + (1 + -1) + (2+-2) + \ldots$. But if you group it instead as $(0+1) + (-1 + 2) + (-2 + 3) + \ldots = 1+1+1+\ldots$ the sum is infinite. – MJD Jun 01 '14 at 15:50
  • Yeh, I see where you are coming from, basically -1-(-1) two minuses become a +, but 1+(+1) still a plus. – Darren J M England Jun 01 '14 at 16:09
  • The other way of looking at it is that 0 can either be considered as a number or not, for example, the Fibonacci sequence can start at 0 but only becomes sequensial with the rules once the first +1. – Darren J M England Jun 01 '14 at 16:21