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

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

There's nothing controversial about this, but an example by Lubos Motl can be found here.

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?

  • 1
    $\begingroup$ 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. $\zeta(-k) = (-1)^{k} g^{(k-1)}(0)$ $\endgroup$ – reuns Jan 10 at 6:01

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