Riemann Zeta Function and Analytic Continuation

Question

The Riemann Zeta Function is defined as $ \displaystyle \zeta(s) = \sum\limits_{n=1}^{\infty} \frac{1}{n^s}$. It is not absolutely convergent or conditionally convergent for $\text{Re}(s) \leq 1$. Using analytic continuation, one can derive the fact that $\displaystyle \zeta(-s) = -\frac{B_{s+1}}{s+1}$ where $B_{s+1}$ are the Bernoulli numbers. Can one obtain this result without resorting to analytic continuation?

Answer 1

Using Euler--MacLaurin summation, one can obtain the following formula for $\zeta(s)$:

$$ \zeta(s) = \frac{1}{s-1}+\frac{1}{2} + \frac{B_2}{2} s + \cdots \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $$ $$ \cdots + \frac{B_{2k}}{(2k)!}s(s+1)\cdots (s + 2k-2) + \frac{s(s+1)\cdots(s+2k-1)}{(2k)!}f(s), $$ where $f(s)$ is an integral involving $s$ which converges when $\Re(s) > -2k$. (My favourite reference for $\zeta(s)$ is Edwards's book Riemann's zeta function. This particular formula is gotten by setting $N = 1$ in formula (1) on p. 114.)

So this gives a formula for $\zeta(s)$ which is defined when $\Re(s) > -2k$. If you substitute in $s = -2k+1$, you will get (after some rearrangement) that $\zeta(-(2k-1)) = -B_{2k}/2k.$

Of course this is a form of anaytic continuation (as others have noted, it is hard to make sesne of what $\zeta(-(2k-1))$ would mean otherwise). But it perhaps a little different to the standard approach.

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ADDED: An approach which seems quite different to analytic continuation --- at least at first --- is the Abelian regularization approach used by Euler. (Please excuse the anachronism of labelling Euler's method with Abel's name!) This is disussed in some of the answers to this question.

The idea is first to multiply by $(1-2^{-s+1})$, which eliminates the pole at $s=1$, and replaces $\zeta(s)$ by the function $\eta(s):= \sum_{n=1}^{\infty} (-1)^{n-1} n^{-s}$. (Clearly if we can evaluate $\eta(s)$, we can evaluate $\zeta(s)$, simply by divising through by $(1-2^{-s+1})$.) Then, one computes $\eta(-k)$ via the following formula: $$\eta(-k) = \lim_{T \to 1} \sum_{n=1}^{\infty}(-1)^{n-1}n^k T^n.$$ The point is that the series in $T$ converges (when $|T| < 1$) to a rational function of $T$, which we can then evaluate at $T = 1$.

This method can be seen directly to lead to the usual formula in terms of Bernoulli numbers. One can also relate it to the usual description of $\zeta(s)$ (or --- equivalently --- $\eta(s)$) via analytic continuation (by considering the two variable function $\sum_n (-1)^n n^{-s} T^n$), but it is the approach I know which is a priori furthest removed from analytic continuation.

Answer 2

What are complex Bernoulli numbers?

http://en.wikipedia.org/wiki/Bernoulli_number

Bernoulli numbers are only defined for natural numbers.

So you probably mean $s$ being an integer. As KCrad points out in the comments: Euler has proven a functional equation for the Riemann zeta function before Riemann using integrals and only for real numbers, so that is a method.

But if you plugin complex values in the integral, you are back to analytic continuation, so the answer is most likely no.

E.g., do you have a nice heuristic why $\zeta(-1) =1+2+3+\dots = 1/12$?