Let's say we have a function:
\[
E(z) : \mathbb{H} \to \mathbb{C}\\
\]
Where \(\mathbb{H} = \{\Im(z) > 0\}\)...
Let's write the value:
\[
E(-\infty) = E_0 = \lim_{z\to -\infty} E(z)\\
\]
Now let's write the value:
\[
A = \int_{-\infty}^{\infty} E(t) \,d\mu\\
\]
Where \(d\mu\) is the measure induced on a circle mapped to the real line with the weight \(d\zeta/\zeta\).
Let's assume both of these things converge.
\[
E(\overline{z}) = E(z) -E_0 + A\\
\]
We have defined this function on the lower half plane, solely from its value on the upper half plane!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
EDIT!!!
To explain what I mean, let's take:
\[
F(z) = \prod_{n=1}^\infty \left(1-x^n\right)\\
\]
The value \(F(0) = 1\). Now there is a natural boundary at \(|z| = 1\). But this boundary is integrable. We have that:
\[
\int_{|z| =1} F(z)\frac{dz}{z} = 0\\
\]
We also have that:
\[
\int_{|z| =1} F(z)\frac{dz}{z^{k+1}} = 0\\
\]
For all \(k \ge 0\).
Using this Lambert identity we have that:
\[
F(1/z) = F(z) - 1+ 0\\
\]
And not only is this a viable function; it satisfies our uniqueness conditions:
\[
\int_{|z| = R} \frac{F(z)}{z^{k+1}}\,dz = \int_{|z| = 1/R} \left(F(z)-1\right)z^{k-1}\,dz\\
\]
If \(k \ge 1\)... And therefore this is a natural analytic continuation!
This shit is highly non-trivial BRO!
\[
E(z) : \mathbb{H} \to \mathbb{C}\\
\]
Where \(\mathbb{H} = \{\Im(z) > 0\}\)...
Let's write the value:
\[
E(-\infty) = E_0 = \lim_{z\to -\infty} E(z)\\
\]
Now let's write the value:
\[
A = \int_{-\infty}^{\infty} E(t) \,d\mu\\
\]
Where \(d\mu\) is the measure induced on a circle mapped to the real line with the weight \(d\zeta/\zeta\).
Let's assume both of these things converge.
\[
E(\overline{z}) = E(z) -E_0 + A\\
\]
We have defined this function on the lower half plane, solely from its value on the upper half plane!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
EDIT!!!
To explain what I mean, let's take:
\[
F(z) = \prod_{n=1}^\infty \left(1-x^n\right)\\
\]
The value \(F(0) = 1\). Now there is a natural boundary at \(|z| = 1\). But this boundary is integrable. We have that:
\[
\int_{|z| =1} F(z)\frac{dz}{z} = 0\\
\]
We also have that:
\[
\int_{|z| =1} F(z)\frac{dz}{z^{k+1}} = 0\\
\]
For all \(k \ge 0\).
Using this Lambert identity we have that:
\[
F(1/z) = F(z) - 1+ 0\\
\]
And not only is this a viable function; it satisfies our uniqueness conditions:
\[
\int_{|z| = R} \frac{F(z)}{z^{k+1}}\,dz = \int_{|z| = 1/R} \left(F(z)-1\right)z^{k-1}\,dz\\
\]
If \(k \ge 1\)... And therefore this is a natural analytic continuation!
This shit is highly non-trivial BRO!
