Okay, so take this post with a Grain of salt. I wanted to look deeper into this question. Because my suspicion is that we can say that sometimes, these are natural analytic continuations. I apologize if this is dumb. I'm just spit balling 
I'd like to add my own two cents on:
\[
f(x) = \sum_{n=0}^\infty \frac{x^n}{1+x^n} \frac{1}{2^n}\\
\]
First of all, the word I used to use for these functions is a function that maps \(\mathbb{C} \to \mathbb{C}\) upto a measure zero set in \(\mathbb{C}\). In this case, the measure zero set is the unit circle \(U\). So that \(f : \mathbb{C}/U \to \mathbb{C}\). There is absolutely zero way to analytically continue these functions, or refer to them as analytic continuations of each other. By standard literature; they are two functions which are holomorphic on disjointed domains of \(\mathbb{C}\).
But yes, ONE is more natural; obviously the one which is just plugging in the number. This is a fallacy that Cauchy called the Generality of Algebra. And it was a stark criticism of Euler. Who upon which, Euler would use the Generality of Algebra to get correct results. But he also got incorrect results.
But this isn't what I want to talk about. I want to use Cauchy's philosophy on "filtering out" when you can use the Generality of Algebra, and when you can't...
Let's take \(|x| > 1\), and let's define a contour \(C\) which is the parameterization of \(|x| = 2\).
\[
F_k = \int_C \frac{f(x)}{x^k}\,dx\\
\]
For \( k \in \mathbb{Z}\). Then this object expands as:
\[
\sum_{n=0}^\infty \int_C \frac{x^n}{1+x^n}\frac{dx}{x^k} \frac{1}{2^n}
\]
So let's evaluate:
\[
a_{kn} = \int_C \frac{x^n}{1+x^n}\frac{dx}{x^k}\\
\]
And:
\[
a_{kn} = 2\pi i b_{kn} + 2\pi i\sum_{q^n = -1} \frac{q^n}{\prod_{p\neq q\,\,\,\,p^n = -1} (q-p)} \frac{1}{q^k}
\]
The product value is actually a simple equation. Despite looking funky. We write this as:
\[
1+x^n = \prod_{p^n = -1} (x-p)\\
\]
The derivative at this for \(q^n = -1\), is simply:
\[
nq^{n-1} = \prod_{q \neq p\,\,\,p^n = -1} (p-q)\\
\]
Where, the first term of the residues of \(b_{kn}\) is just:
\[
\frac{x^n}{1+x^n} = \sum_{k=0}^\infty b_{kn} x^k\\
\]
And for \(k < 0\) we know that \(b_{kn}\) is \(0\) because the Laurent series has no negative terms. So when we expand this we get that:
\[
F_k = \sum_{n=0}^\infty \left(2\pi i b_{kn} + 2\pi i\sum_{q^n = -1} \frac{n}{q^{k-1}}\right)\frac{1}{2^n}\\
\]
Now let's sum this across all \(k\). Let's ignore the convergence of this (also, I may have screwed up a negative sign here and there). But regardless, it looks something like this:
\[
\sum_{k=-\infty}^\infty \sum_{n=0}^\infty a_{kn}= \int_C f(x) \sum_{k=-\infty}^{\infty} x^k\,dx\\
\]
Now, by "the generality of algebra", as I mentioned before:
\[
\begin{align}
\sum_{k=-\infty}^{\infty} x^k &= \sum_{k=-\infty}^{-1}x^k + \sum_{k=0}^\infty x^k\\
&= \frac{1}{x}\frac{1}{1-\frac{1}{x}} + \frac{1}{1-x}\\
&= 0\\
\end{align}
\]
So we've shown that:
\[
\sum_{k\in \mathbb{Z}} \sum_{n=0}^\infty a_{kn} = 0\\
\]
......... right?
NO!
This may seem a little handwavey, because I used the generality of Algebra.
And you'd be write to reject that! Instead, check the math "under the integral"...
You can check the sum:
\[
\sum_{k=-\infty}^\infty \sum_{n=0}^\infty \left(2\pi i b_{kn} + 2\pi i n\sum_{q^n = -1}\frac{1}{q^{k-1}}\right) \frac{1}{2^n} = 1\\
\]
This is some Gaussian bullshit. But the second summand is reducible into a bunch of Fourier sums which sum to zero. You can see this more obviously by writing:
\[
\sum_{q^n = -1}\frac{1}{q^{k-1}} = \sum_{j=1}^n e^{-\pi i (k-1) \frac{j}{n}} = \phi_n(k)\\
\]
This produces a sum of the \(n\)'th roots of \(1\) or the \(n\)'th roots of \(-1\). But, either way \(\phi_n(k) = 0\) for all \(n\) and \(k\).
And the sum:
\[
\sum_{k=-\infty}^\infty \sum_{n=0}^\infty b_{kn} = f(1) = 1\\
\]
So "Generality of Algebra" predicted \(0\) initially--remembering the "under the integral action" reminded us it should be \(1\). Which makes sense in a Dirac sense, the sum:
\[
\int_C ...\sum_{k\in\mathbb{Z}}x^k = \int_C \delta\\
\]
But ignore this, it's unnecessary........
Now it gets beautiful!
Let's try to reconstruct \(f(x)\) for \(|x| < 1\) solely using \(f(x)\) for \(|x| > 1\). Well.... All we need to do is write:
\[
f(x) = \sum_{k=0}^\infty 2 \pi i F_k x^k\\
\]
And this sum converges to the original sum. Because the residues just disappear!!!!!!!!!!!!!!!!!!!!!!
So, sure there is a wall of singularities! But, the residues of the poles on the wall of the singularities tend to zero as you add them up!!!!!!!!!!!!!!!!!!!!!!!!!
This avoids the "Generality of algebra" fallacy, which just rearranges sums. Because for all we know \(f(x)\) for \(|x| > 1\), we may have had a different function \(g(x)\) for \(|x| < 1\) which is actually better than \(f(x)\) for \(|x| < 1\), and is the "not really analytic continuation". But instead, we can take contours \(C_R\) for \(|x| = R\) and \(R \in (0,1) \cup (1,\infty)\). And we get that:
\[
f(x) = \sum_{k=0}^\infty 2 \pi i F_k x^k = \sum_{k=0}^\infty x^k\sum_{n=0}^\infty \frac{b_{kn}}{2^n}\\
\]
And:
\[
F_k = \int_{C_R} \frac{f(x)}{x^k}\,dx\\
\]
Because even when we hit the wall of singularities; the sum of the residues is \(0\)
Hope this helps. If I'm being an idiot again, be sure to tell me! I'm still not entirely certain what you are trying to do. So calling me an idiot only helps me learn
But to me, this relates deeply to the differential nature of a function. Because functions \(f\) which take \(\mathbb{C} \to \mathbb{C}\) almost everywhere, tend to be solutions to first order differential equations. I never had a word for it. I would just write:
\[
f: \mathbb{C}/\mathcal{D} \to \mathbb{C}\\
\]
And
\[
\int_{\mathcal{D}} f(x) dA = 0\\
\]
Where this is a double integral with the standard Lebesgue area measure \(dA\)....
I'm happy to be an idiot though, lmao
Sincere Regards, James
I'd like to add my own two cents on:
\[
f(x) = \sum_{n=0}^\infty \frac{x^n}{1+x^n} \frac{1}{2^n}\\
\]
First of all, the word I used to use for these functions is a function that maps \(\mathbb{C} \to \mathbb{C}\) upto a measure zero set in \(\mathbb{C}\). In this case, the measure zero set is the unit circle \(U\). So that \(f : \mathbb{C}/U \to \mathbb{C}\). There is absolutely zero way to analytically continue these functions, or refer to them as analytic continuations of each other. By standard literature; they are two functions which are holomorphic on disjointed domains of \(\mathbb{C}\).
But yes, ONE is more natural; obviously the one which is just plugging in the number. This is a fallacy that Cauchy called the Generality of Algebra. And it was a stark criticism of Euler. Who upon which, Euler would use the Generality of Algebra to get correct results. But he also got incorrect results.
But this isn't what I want to talk about. I want to use Cauchy's philosophy on "filtering out" when you can use the Generality of Algebra, and when you can't...
Let's take \(|x| > 1\), and let's define a contour \(C\) which is the parameterization of \(|x| = 2\).
\[
F_k = \int_C \frac{f(x)}{x^k}\,dx\\
\]
For \( k \in \mathbb{Z}\). Then this object expands as:
\[
\sum_{n=0}^\infty \int_C \frac{x^n}{1+x^n}\frac{dx}{x^k} \frac{1}{2^n}
\]
So let's evaluate:
\[
a_{kn} = \int_C \frac{x^n}{1+x^n}\frac{dx}{x^k}\\
\]
And:
\[
a_{kn} = 2\pi i b_{kn} + 2\pi i\sum_{q^n = -1} \frac{q^n}{\prod_{p\neq q\,\,\,\,p^n = -1} (q-p)} \frac{1}{q^k}
\]
The product value is actually a simple equation. Despite looking funky. We write this as:
\[
1+x^n = \prod_{p^n = -1} (x-p)\\
\]
The derivative at this for \(q^n = -1\), is simply:
\[
nq^{n-1} = \prod_{q \neq p\,\,\,p^n = -1} (p-q)\\
\]
Where, the first term of the residues of \(b_{kn}\) is just:
\[
\frac{x^n}{1+x^n} = \sum_{k=0}^\infty b_{kn} x^k\\
\]
And for \(k < 0\) we know that \(b_{kn}\) is \(0\) because the Laurent series has no negative terms. So when we expand this we get that:
\[
F_k = \sum_{n=0}^\infty \left(2\pi i b_{kn} + 2\pi i\sum_{q^n = -1} \frac{n}{q^{k-1}}\right)\frac{1}{2^n}\\
\]
Now let's sum this across all \(k\). Let's ignore the convergence of this (also, I may have screwed up a negative sign here and there). But regardless, it looks something like this:
\[
\sum_{k=-\infty}^\infty \sum_{n=0}^\infty a_{kn}= \int_C f(x) \sum_{k=-\infty}^{\infty} x^k\,dx\\
\]
Now, by "the generality of algebra", as I mentioned before:
\[
\begin{align}
\sum_{k=-\infty}^{\infty} x^k &= \sum_{k=-\infty}^{-1}x^k + \sum_{k=0}^\infty x^k\\
&= \frac{1}{x}\frac{1}{1-\frac{1}{x}} + \frac{1}{1-x}\\
&= 0\\
\end{align}
\]
So we've shown that:
\[
\sum_{k\in \mathbb{Z}} \sum_{n=0}^\infty a_{kn} = 0\\
\]
......... right?
NO!
This may seem a little handwavey, because I used the generality of Algebra.
And you'd be write to reject that! Instead, check the math "under the integral"...
You can check the sum:
\[
\sum_{k=-\infty}^\infty \sum_{n=0}^\infty \left(2\pi i b_{kn} + 2\pi i n\sum_{q^n = -1}\frac{1}{q^{k-1}}\right) \frac{1}{2^n} = 1\\
\]
This is some Gaussian bullshit. But the second summand is reducible into a bunch of Fourier sums which sum to zero. You can see this more obviously by writing:
\[
\sum_{q^n = -1}\frac{1}{q^{k-1}} = \sum_{j=1}^n e^{-\pi i (k-1) \frac{j}{n}} = \phi_n(k)\\
\]
This produces a sum of the \(n\)'th roots of \(1\) or the \(n\)'th roots of \(-1\). But, either way \(\phi_n(k) = 0\) for all \(n\) and \(k\).
And the sum:
\[
\sum_{k=-\infty}^\infty \sum_{n=0}^\infty b_{kn} = f(1) = 1\\
\]
So "Generality of Algebra" predicted \(0\) initially--remembering the "under the integral action" reminded us it should be \(1\). Which makes sense in a Dirac sense, the sum:
\[
\int_C ...\sum_{k\in\mathbb{Z}}x^k = \int_C \delta\\
\]
But ignore this, it's unnecessary........
Now it gets beautiful!
Let's try to reconstruct \(f(x)\) for \(|x| < 1\) solely using \(f(x)\) for \(|x| > 1\). Well.... All we need to do is write:
\[
f(x) = \sum_{k=0}^\infty 2 \pi i F_k x^k\\
\]
And this sum converges to the original sum. Because the residues just disappear!!!!!!!!!!!!!!!!!!!!!!
So, sure there is a wall of singularities! But, the residues of the poles on the wall of the singularities tend to zero as you add them up!!!!!!!!!!!!!!!!!!!!!!!!!
This avoids the "Generality of algebra" fallacy, which just rearranges sums. Because for all we know \(f(x)\) for \(|x| > 1\), we may have had a different function \(g(x)\) for \(|x| < 1\) which is actually better than \(f(x)\) for \(|x| < 1\), and is the "not really analytic continuation". But instead, we can take contours \(C_R\) for \(|x| = R\) and \(R \in (0,1) \cup (1,\infty)\). And we get that:
\[
f(x) = \sum_{k=0}^\infty 2 \pi i F_k x^k = \sum_{k=0}^\infty x^k\sum_{n=0}^\infty \frac{b_{kn}}{2^n}\\
\]
And:
\[
F_k = \int_{C_R} \frac{f(x)}{x^k}\,dx\\
\]
Because even when we hit the wall of singularities; the sum of the residues is \(0\)
Hope this helps. If I'm being an idiot again, be sure to tell me! I'm still not entirely certain what you are trying to do. So calling me an idiot only helps me learn
But to me, this relates deeply to the differential nature of a function. Because functions \(f\) which take \(\mathbb{C} \to \mathbb{C}\) almost everywhere, tend to be solutions to first order differential equations. I never had a word for it. I would just write:\[
f: \mathbb{C}/\mathcal{D} \to \mathbb{C}\\
\]
And
\[
\int_{\mathcal{D}} f(x) dA = 0\\
\]
Where this is a double integral with the standard Lebesgue area measure \(dA\)....
I'm happy to be an idiot though, lmao
Sincere Regards, James
