# Why should the series representation of the zeta function know about its analytic continuation?

In physics, when we calculate the renormalized sum of $$S=\sum_{n=1}^\infty n$$, we usually use an exponential regulator and instead first calculate

$$S_\epsilon = \sum_{n=1}^\infty ne^{-\epsilon n} = \frac{e^\epsilon}{(e^\epsilon -1)^2}.$$

Now, if we expand this around small $$\epsilon>0$$ we obtain that

\begin{align*} S_\epsilon &= \frac{1}{\epsilon^2} - \frac{1}{12} + {\cal O}(\epsilon^2)\\ &=\zeta(-1) + \frac{1}{\epsilon^2} + {\cal O}(\epsilon^2). \end{align*}

So that when we take $$\epsilon \to 0$$ (in which limit $$S_\epsilon \to S)$$, we obtain that

$$S \to \zeta(-1) + \text{divergent piece.}$$

Now, I know that this "decomposition", if you will, into a convergent and divergent part is independent of the sum regulator, and hence always yields the same result. Furthermore, the domain of validity of the series representation of the $$\zeta$$ function is only when $$\mathbb{R}(s) >0$$. Therefore, we ought to expect it to be of no use to us when naively applying it in regions where $$\mathbb{R}(s)<1$$.

However, it is useful. Namely, its renormalized sum gives us the correct value for the analytic continuation of the $$\zeta$$ function. Furthermore, since the analytic continuation of a complex function is unique I feel there must be a connection between these two things.

Question: How does the series representation "know" about its analytic continuation? That is, what exactly is the connection between the naive summation and its analytic continuation?

• To see what is really happening you should look at $\Gamma(s) \zeta(s) = \int_0^\infty t^{s-1} f(t)dt$ where $f(t) = \sum_{n=1}^\infty e^{-nt}$. That $g(t)=t f(t)$ is $C^\infty$ at $t=0$ implies that $\int_0^1 t^{s-1} (f(t)- t^{-1}\sum_{k=0}^K \frac{g^{(k)}(0)}{k!} t^k)dt$ is analytic for $\Re(s) > -K$, whence the poles of $\Gamma(s) \zeta(s)$ are at the negative integers, of order $1$ and residue $\frac{g^{(k-1)}(0)}{(k-1)!}$ ie. $\zeta(-k) = (-1)^{k} g^{(k-1)}(0)$ – reuns Jan 10 at 6:01