# Are products of analytic continuations also analytic?

The question of the value, if any depending on which answer you choose, of $$\sum_{n=1}^\infty n$$ has been addressed a few times. At least here Does $\zeta(-1)=-1/12$ or $\zeta(-1) \to -1/12$? and here Why does $1+2+3+\cdots = -\frac{1}{12}$?. I do not want to re-open that question in general, but I have a question about a specific step of one of the approaches (or purported approaches as you may like) to computing the result.

Under the zeta function regularization technique, one ultimately observes that $$\left( 1 - 2^{1-s} \right) \zeta(s) = \eta(s)$$ for the Riemann zeta function $$\zeta$$ and the Dirichlet eta function $$\eta$$. One usually arrives at this result by using the series representations of these two functions and performing manipulations on them that are valid for complex values of $$s$$ where the series representations of $$\zeta$$ and $$\eta$$ converge.

That seems fine as far as it goes, under the assumption that each function is evaluated at a value of $$s$$ where the series converges. The method then continues to assert that the relationship holds for the analytic continuations of $$\zeta$$ and $$\eta$$. That's the step that motivates my question.

Is it generally true that if $$f(s) g(s) = h(s)$$ on an open set $$U$$ that this relationship will continue to hold for their analytic continuations to larger sets? If not generally true, what is the special property of $$\zeta$$ and $$\eta$$ that makes it true for the case outlined above?

My sense is that it's not generally true because of differences in which potential supersets of $$U$$ each individual function has an analytic continuation, but I'm operating well on the fringe of my understanding of this topic.

Yes, it is true, by the identity theorem, as stated by e.g., Wikipedia:

Given functions $$f$$ and $$g$$ holomorphic on a domain $$D$$ (open and connected subset), if $$f = g$$ on some $$S\subseteq D$$, $$S$$ having an accumulation point, then $$f = g$$ on $$D$$.

In particular, $$f(s), g(s), h(s)$$ are analytic functions and $$f(s)g(s) = h(s)$$ on any open set $$U\subseteq D$$ (or indeed, any set $$S$$ that has a limit point), then $$f(s)g(s) = h(s)$$ on the whole set $$D$$.

You may gain some intuition on the identity theorem by expecting that analytic functions behave, to some extent, like high-degree polynomials - which may be expected since they have power series representations. Any two degree-$$n$$ polynomials are identical if they agree on any $$n+1$$ points. Similarly, any two analytic functions are identical if they agree on any infinite set of points - with the important caveats that the set has a limit point, and that the domain they are defined on is connected.

• I'm not clear that this explains it. Say $f(s) = s$, $g(s) = 1/s$ and $h(s) = 1$ on an open set that doesn't include the origin. The analytic continuations of $f$ and $h$ to the whole plane are trivial, but $g$ is not analytic at the origin (and has no continuation there). Perhaps my question was not precisely enough worded, in which case I welcome an edit, but I don't think the identity theorem here connects me to each function, separately, having an analytic continuation at any given point. I see the point about the product though. Commented Oct 11, 2019 at 17:15
• Ah, but yes, I see the top-level question leads you to this answer. I probably should have worded it the other way around. Maybe I'll need to ask a new question to get it in that direction! Commented Oct 11, 2019 at 17:17
• Hmm, I'm not sure what your question is, then. I was answering the question "Is it generally true that if $f(s)g(s) = h(s)$ on an open set $U$ that this relationship will continue to hold for their analytic continuations to larger sets?" I was assuming you meant that $f,g$ and $h$ all had analytic continuations to some domain $D$ containing $U$, and asking whether $fg = h$ continued to hold in $D$. In the case you give, we must take $D = \mathbb{C} \backslash \{0\}$. Are you asking about conditions under which analytic continuations exist? Commented Oct 11, 2019 at 17:34
• Technically speaking, we define a new function $\hat{\zeta}(s) = \eta(s) / (1 - 2^{1-s})$ and then we can evaluate $\hat{\zeta}(-1)$ if $\eta(-1)$ is defined. What the identity theorem does for us is to justify the abuse of notation $\hat{\zeta} = \zeta$, since it shows that any analytic continuation $f$ of $\zeta$ must agree with $\hat{\zeta}$. Commented Oct 11, 2019 at 17:52
• In particular, the analytic continuation $\hat{\zeta}$ does not satisfy $\hat{\zeta}(s) = \sum_{k=1}^\infty k^{-s}$, so it doesn't say anything about the sum of positive integers in the usual sense. Commented Oct 11, 2019 at 17:56