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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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Brick
    Commented Oct 11, 2019 at 17:15
  • $\begingroup$ 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! $\endgroup$
    – Brick
    Commented Oct 11, 2019 at 17:17
  • $\begingroup$ 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? $\endgroup$
    – Jair Taylor
    Commented Oct 11, 2019 at 17:34
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    $\begingroup$ 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 $\hat{\zeta}$. $\endgroup$
    – Jair Taylor
    Commented Oct 11, 2019 at 17:52
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    $\begingroup$ 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. $\endgroup$
    – Jair Taylor
    Commented Oct 11, 2019 at 17:56

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