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?
[Math] Riemann Zeta Function and Analytic Continuation
complex-analysisriemann-zeta
Related Solutions
The contour is not closed, but the value of the integral is independent (for small radius) of the method you might use to close off the contour near $Re(z)=-\infty$ and take a limit.
Near $-\infty$ means modifying the contour only in a region $Re(z) < a$ for large enough negative $a$. In this problem any $a<0$ will leave unchanged the set of poles and residues inside the contour, and as $a \to -\infty$ any changes in the value of the integral due to the contour modification crossing the branch cut, converge to $0$.
The un-modified contour can be considered as a closed integration path on a compactification of the left half-plane by adding a point at real part $-\infty$, which is justified when integrating a power of $z$ that is suppressed exponentially (in $Re(z)$) when approaching the added point.
Please allow me to elaborate a little on Riemann's and Jeb's elegant exposition.
So imagine the contour $\gamma$ as coming just "above" the real axis from $+\infty$ to $0$ and then swinging counterclockwise around the origin, and then going back just "below" the real axis back to $+\infty$.
As Riemann suggests, we can write $(-x)^{s-1} = e^{(s-1)\log(-x)}$. But what should $\log(-x)$ be equal to for positive $x$?
As Jeb suggested, $-1 = e^{i\pi} = e^{-i\pi}$, so we can write $\log(-x)$ as $\log(e^{i\pi}x) = i\pi+\log{x}$, or as $\log(e^{-i\pi}x) = -i\pi+\log{x}$. However, we must make choices that preserve the continuity of $\log(-x)$ as $x$ moves along the contour in the complex plane. Also, as Riemann specifies, we want $\log(-x)$ to be real when $x$ is negative.
For the piece of the contour from $+\infty$ to $0$, we must choose $$\log(-x) = -i\pi+\log{x}.$$
On the way back from $0$ to $+\infty$, in order to preserve continuity we must have $$\log(-x) = i\pi+\log{x}.$$
Thus, the integral for the part of the contour from $+\infty$ to $0$ is equal to $$\int_{+\infty}^0 \frac{e^{-(s-1)i\pi}e^{(s-1)\log{x}}}{e^x-1}\,dx = e^{-\pi si} \int_0^{+\infty} \frac{e^{(s-1)\log{x}}}{e^x-1}\,dx.$$ And the integral for the part of the contour from $0$ to $+\infty$ is equal to $$\int_0^{+\infty} \frac{e^{(s-1)i\pi}e^{(s-1)\log{x}}}{e^x-1}\,dx = -e^{\pi si} \int_0^{+\infty} \frac{e^{(s-1)\log{x}}}{e^x-1}\,dx.$$
Now add those two pieces together to get Riemann's result.
Best Answer
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.
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.