This question is a question about nomenclature more than anything. I have shown all the math, but I don't know what to search for for similar results. In such a sense, it is more so a reference request, or a request for someone to point me in the right direction.


We start with the function:


$$
f\_1(z) = \sum\_{n=0}^\infty \frac{z^n}{1+z^n} \frac{1}{2^n}\\
$$


Which is holomorphic for $|z| <1$. And the second function is:


$$
f\_2(z) = \sum\_{n=0}^\infty \frac{z^n}{1+z^n} \frac{1}{2^n}\\
$$


Which is holomorphic for $|z| > 1$.


Both of these functions are expressed by the same summation, but on different domains. Using the usual concept of "Analytic Continuation" the functions $f\_1$ and $f\_2$ are not related. They are analytic on non-intersecting domains. So, to call one a continuation of the other is a fallacy to begin with.


To make matters worse; both functions $f\_1$ and $f\_2$ have a "wall" of singularites along $|z| =1$. Where for all $q^n =-1$ for all $n \in \mathbb{N}$ we have that $f\_1(q) = \infty = f\_2(q)$, which are simple poles. There is no domain $D \subset \mathbb{C}$ which intersects $|z| =1$ where either $f\_1$ or $f\_2$ are holomorphic--because there are a dense amount of singularities on this line. So any standard way of analytic continuation is hopeless.


BUT! And this is a big but. Let us take the contour $C = \{ \zeta \in \mathbb{C}\,|\, |\zeta| = 2\}$. And let's take $|z|<1$. Then:


$$
f\_1(z) = \frac{1}{2\pi i} \int\_C \frac{f\_2(\zeta)}{\zeta-z}\,d\zeta\\
$$


This is a pretty involved result, which basically amounts to. All of the residues of the poles along the "wall" of singularities sum to zero when added up. And all that's left is the function for $|z|<1$. This is one of those perfect moments where everything cancels out and $f\_2(z)$ acts as an analytic continuation of $f\_1$. Despite the fact "analytic continuation" is meaningless.


I like to write this as:


$$
f(z) = \frac{1}{2\pi i} \int\_{|z| = R} \frac{f(\zeta)}{\zeta - z}\,d\zeta\\
$$


For all $|z| < \min(R,1)$ for all $R \neq 1$.


From this, I'd like to ask my question. I'd like to think of $f(z) = f\_1(z) = f\_2(z)$, where $f(z) : \mathbb{C}/\mathcal{U} \to \mathbb{C}$; where $\mathcal{U} = \{z \in \mathbb{C}\,|\, |z| =1\}$. But I know this is kind of nonsense. But it's not entirely nonsense, because Cauchy's integral theorem still works... It's not really an analytic continuation--but it kinda (..?) is.


Any help is greatly appreciated! Any pointing in the right direction is also greatly appreciated!



Analytic continuation for disjoint domains
As I already hinted at in my comment, your formula is based on a miscalculation (the only alternative is that $f\_1$ can actually be continued past $|z|=1$, which even without closer analysis looks highly suspicious since the summands have poles at $n$th roots of $-1$, as you pointed out). Here's a direct quick test that confirms this:


Take $z=0$. We can integrate $\int\_C \frac{f\_2(\zeta)}{\zeta}\, d\zeta$ term by term, so let's look at (for $n\ge 1$)
\begin{align\*}
\int\_{|\zeta|=2 } \frac{\zeta^{n-1}}{\zeta^n+1}\, d\zeta & =2^ni\int\_0^{2\pi}\frac{e^{in t}}{2^ne^{int}+1}\, dt = \frac{2^ni}{n}\int\_0^{2\pi n} \frac{e^{ix}}{2^n e^{ix}+1}\, dx \\
& = i\int\_0^{2\pi} \frac{e^{ix}}{e^{ix}+2^{-n}}\, dx
=\int\_{|w|=1} \frac{dw}{w+2^{-n}} = 2\pi i .
\end{align\*}
I use the standard parametrization $z=Re^{it}$ of the circle here (we can also get this more quickly by right away substituting $u=\zeta^n$). Also, when $n=0$, the integral equals $\pi i$.
Hence
$$
\frac{1}{2\pi i}\int\_{|\zeta|=2} \frac{f\_2(\zeta)}{\zeta}\, d\zeta = \frac{1}{2}+\sum\_{n\ge 1} 2^{-n}= \frac{3}{2} \not= f\_1(0)= \frac{1}{2} .
$$


