Divergent Series and Analytical Continuation (LONG post)

[Caleb]

Okay, I'll stop blowing up this thread after this last identification.

Let:

\[
L(z) : \mathbb{C}/ U \to \mathbb{C}\\
\]

Where \(U\) is the unit circle. Let's assume that:

\[
\int_{|\zeta| = R} L(\zeta)\frac{d\zeta}{\zeta^{k-1}} = 0\\
\]

For all \(R > 1\), and for all \(k > 0\). Then call the sum:

\[
F(z) = -\sum_{k\in \mathbb{Z}} z^{-k}\int_{|\zeta| = R} L(\zeta)\frac{d\zeta}{\zeta^{k-1}}\\
\]

This just reduces to:

\[
F(z) = -A - \sum_{k=1}^\infty \frac{1}{2\pi i}\int_{|\zeta| = R} L(\zeta)z^k \zeta^{k-1}\,d\zeta\\
\]

Which is just:

\[
F(z) = -A + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\
\]

The conjecture by Caleb; is that:

\[
F(z) = L(z) + C\\
\]

Where the constant \(C\) is easily discoverable. Where it can be written more compactly as:

\[
L(z) = L(0) + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\
\]

For all \(|z| < 1\). This actually creates a reflection formula between \(L(1/z)\) and \(L(z)\). Where we can simply evaluate this integral as:

\[
\frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\ = L(1/z) + A\\
\]

Where:

\[
A = \frac{1}{2\pi i} \int_{|z| = R} \frac{L(\zeta)}{\zeta}\,d\zeta\\
\]

Because:

\[
\frac{L(\zeta)}{\zeta - \frac{1}{z}} = L(\zeta) \sum_{k=1}^\infty z^{1-k}\zeta^{-k}
\]

When \(|z\zeta| > 1\); which is always true if \(|\zeta| = R\) is large enough. And the only non-zero term, is when \(k = 1\).

I HAVE CONFIRMED THIS NUMERICALLY FOR ALL GENERALIZED LAMBERT FUNCTIONS!!!!!!!

I think this might be confirmation of a much much deeper identity; but we have to prove Caleb's conjecture first. Which is that:

\[
L(z) = L(0) + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta-\frac{1}{z}}\,d\zeta\\
\]

For all \(1 < \left|\frac{1}{z}\right| < R\), \(\frac{1}{R} < |z| < 1\).

This would intimately relate both functions \(L(z)\) and \(L(1/z)\) that I can't even believe is possible. This is like striking straight up gold!

BEAUTIFUL INTUITION, CALEB! This is a reflection formula for the books! And I think it'll go a long way in doing what you want to do! But do it without needing to fall back on Fourier Analysis/ Ramanujan shit!

Also, for you modular nerds! If \(L(0) = 0\), then we have shown that:

\[
L(1/z) = L(z) + A\\
\]

If we also have that \(A =0\)... then \(L(1/z) = L(z)\)! So if we assume that \(L(\infty) =0\) and \(L(0) = 0\); then these functions are carbon copies.

Which is the first step in identifying a modular relationship. The next would be a periodic condition... But I can imagine this as "analytically continuing modular functions"

YOU ARE ONTO SOMETHING HERE CALEB!!!!!!!!!!!!!!!!

Sincere regards, SUPER PUMPED, James

I apoligise for the late response, I should have mentioned this earlier- but I belive we have only stubled upon a trivial property  . In particular, its not that \( L(z) \) and \( L(1/z) \) are related in some interesting way-- they are actually just the same evaulation. In paritcular, notice that
\[ f(x) = \sum \frac{x^n}{1+x^n} a_n \]
Using the geometric series formula for \( |x|<1 \) gives
\[ f(x) = \sum_n a_n  \sum_k x^{n(k+1)}\]
Now take a look at 
\[f(1/x) = \sum \frac{(\frac{1}{x})^n}{1+(\frac{1}{x})^n} a_n = \sum a_n \frac{1}{x^n +1} \]
But now, for \(|x|<1\), we have that 
\[\sum a_n \frac{1}{x^n +1} = \sum_n a_n \sum_{k=0}^\infty (x)^{nk}\]
But now the only missing term is \( k=0\), so 
\[f(1/x) - f(x) = \sum_n a_n \sum_{k=0}^0 = \sum_n a_n \]
So, unforunately I don't think anything interseting is actually going on here, at least for this relationship.