Divergent Series and Analytical Continuation (LONG post)

[Caleb]

Okay, I want to add a stronger result to my last result.

Assume \(|x| \neq 1\). Call the contour \(C\) the parameterization of \(|z| =R\) for \(R\neq 1\). Then for all \(|x| < R\) and \(|x|\neq 1\), we have that:

\[
f(x) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z-x}\,dz\\
\]

At this point I have only shown this result for \(|x| < \text{min}(R,1)\). I want to show it for \(|x| < R\).

This is actually pretty simple...

Let's take the summand \(\frac{z^n}{(1+z^n)(z-x)}\)... The value of all the residues of this summand are:

\[
\frac{x^n}{1+x^n} + \frac{1}{n}\sum_{q^n = -1}\frac{q}{q-x}\\
\]

The sum:

\[
\sum_{q^n = -1}\frac{q}{q-x} = \sum_{j = 0}^{n-1} \frac{\sqrt[n]{-1}e^{2\pi i \frac{kj}{n}}}{\sqrt[n]{-1}e^{2\pi i \frac{kj}{n}}-x}\\
\]

So that:

\[
\sum_{j=0}^{n-1} \sum_{m=0}^\infty \sqrt[n]{-1}^{-m} e^{-2\pi i \frac{kjm}{n}}x^m=0\\
\]

Because:

\[
\sum_{j=0}^{n-1} e^{-2\pi i \frac{kjm}{n}} = 0\\
\]

And that, all we end up with is that:

\[
\int_C \frac{z^n}{1+z^n} \frac{dz}{z-x} = \frac{x^n}{1+x^n}\\
\]

So that the wall of singularities disappear even better than I originally thought.

---

My conclusion is that Caleb's function \(f(x)\) is holomorphic for \(\mathbb{C}/U\) where \(U\) is the unit circle. And that for all disks \(|z| = R\) where \(R \neq 1\)-- we have Cauchy's integral Formula:

For all \(|x| < R\) and \(|x| \neq 1\) and \(R \neq 1\)--where \(C\) is the parameterization of \(|z| =R\):

\[
f(x) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z-x}\,dz\\
\]

For all \(|x| < R\)...

Fuck this function is cool as fuck!

This becomes not a question of "analytic continuation"--but whether it satisfies Cauchy's integral formula... This has to do with functional analysis (where advanced interpretations of Cauchy's integral formula exist). I'm super excited by your work, Caleb!

Hmm, I did a little bit of numerical testing and I've found that
\[\frac{1}{2\pi i} \int_C \frac{f(z)}{z-x} dz = -\frac{3}{2}\]
if the contour \(C\) encloses the whole natural boundary. In particular, I think we actually have that 
\[\frac{1}{2\pi i} \int_C \frac{x^n}{x^n+1} \frac{1}{z-x} = 1 \]
As long as \(C\) encloses all the poles of \(x^n+1\).

Let me know if any of these computations are wrong, I did these in a rush and I'll check them later