# What is the qualitative relationship between an analytic function and its analytic continuation on the complement of the originally defined domain?

Definition of analytic continuation which I am familiar with is that if $$f_1$$ is an analytic function in a connected open set $$\Gamma_1$$ then $$(f_2, \Gamma_2)$$ is an analytic continuation of $$(f_1, \Gamma_1)$$ if $$\Gamma_1\cap \Gamma_2\neq \varnothing$$, $$f_1\equiv f_2$$ on $$\Gamma_1\cap \Gamma_2$$ and $$f_2$$ is analytic on $$\Gamma_2$$.

Suppose that $$(f_1,\Gamma_1)$$ is an analytic function with an appropriate domain and $$(f_2, \Gamma_2)$$ is an analytic continuation for $$f_2$$. My question is that what if any qualitative relationship there is between the original analytic function $$f_1$$ and its analytic continuation $$f_2$$ on $$\Gamma_1^c\cap \Gamma_2$$? My question stems from trying to understand what if any is the relationship between the Riemann zeta function's defintion $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$$ on $$\{z \in \mathbb{C}\mid \mathfrak{Re}(z) > 1\}$$ and its analytic continuation which gives $$\zeta(-1) = \frac{-1}{12}$$.

It is possible that an acceptable answer to this question is "next to nothing" as the set we are investigating is not part of the domain of definition of the original function. I am asking this question because my level of maturity with complex analysis/analytic continuation isn't that high, so I cannot/don't know how to appreciate short overviews given in standard text books.

The first thing is that your definition of analytic continuation is not quite right. To clarify what is happening, let $$f_1, f_2$$ be given analytic functions on connected open sets $$\Gamma_1,\Gamma_2$$, respectively. Then $$D := \Gamma_1\cap\Gamma_2$$ is the domain where both functions are defined and analytic. By the identity theorem, if $$f_1 \equiv f_2$$ on a set of points $$S \subset D$$ that have at least one accumulation point, then $$f_1 \equiv f_2$$ on their shared domain $$D$$. But in general we don't say that $$f_1, f_2$$ are analytic continuations of each other, but rather they are both analytic continuations of $$f$$ on $$D$$ (where $$f$$ is either $$f_1$$ or $$f_2$$ since they agree on $$D$$). For example, we could construct using the complex logarithm function three domains $$\Gamma_1,\Gamma_2,\Gamma_3$$ that altogether encircle the origin and overlap in three places so that $$f_1 \equiv f_2$$ and $$f_1 \equiv f_3$$ on their respective overlaps but $$f_2 \not\equiv f_3$$ on their overlap (to be a bit more concrete, take $$\Gamma_1$$ to be the open right half-plane Re $$z > 0$$ and $$\Gamma_2,\Gamma_3$$ the entire complex plane minus the positive and negative imaginary axis, respectively). If we used your definition then analytic continuation would no longer be unique, an undesirable property. The case generally considered is the one where $$\Gamma_1 \subset \Gamma_2$$, in which case $$f_2$$ on $$\Gamma_2$$ is an analytic continuation of $$f_1$$ on $$\Gamma_1$$. This will be unique by the identity theorem.
Now with that being said, what you are basically asking for is if we can infer anything about what $$f_2$$ looks like on $$\Gamma_2\setminus \Gamma_1$$ based on $$f_1$$. I'm not sure what type of qualitative answer you're looking for, but to illustrate why it's not so straightfoward you can look to see the previous example; on the left half-plane Re $$z < 0$$ the value differs between the $$(f_2,\Gamma_2)$$ and $$(f_3,\Gamma_3)$$ analytic continuations. So we can't expect values of an analytic continuation outside the original domain to be well-defined in the absence of a given larger set on which the function is to be extended. So in general there is not much we can say unfortunately.
$$\zeta(s) = 2^s \pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$
where $$\Gamma(s) = (s-1)!$$ extends the factorial function. If say we wanted to calculate $$\zeta(-1)$$, which we couldn't do using the series formula defined on the right half-plane, we could plug into the functional equation to get $$\zeta(-1) = 2^{-1}\pi^{-2}\sin\left(\frac{-\pi}{2}\right)\Gamma(2)\zeta(2) = \frac{-1}{2\pi^2}\zeta(2)$$ where we used $$\Gamma(2) = 1! = 1$$. Now $$2$$ is in our original domain, and it is well-known that $$\zeta(2) = \frac{\pi^2}{6}$$. So from this we get $$\zeta(-1) = -\frac{1}{12}$$; clearly we could keep going in this manner, but the point is that the values after moving to the left are related to those at their starting point by this nice functional equation.