I haven't absorbed this post fully yet; but I agree with everything. I will point out one thing, that perhaps you haven't studied yet.
Any integral of the form:
\[
\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} g(s)F(s,z)\,ds\\
\]
Is kind of like a "modified" Mellin transform--and they appear frequently in the study of Hyper Geometric series.
I will explain something pretty cool; you may not have seen before. Let's take the standard sum:
\[
e^{-z} = \sum_{n=0}^\infty \frac{(-z)^n}{n!}\\
\]
Then, the sum of residues:
\[
\sum_{n=0}^\infty \text{Res}\left(\Gamma(s)z^{-s}, s = -n\right) = e^{-z}\\
\]
But this sum of residues is just the integral:
\[
e^{-z} = \frac{1}{2\pi i} \int_{1-i\infty}^{1 + i\infty} \Gamma(s)z^{-s}\,ds\\
\]
I'm sure you are well aware of this. But what happens if we add in a pole? Let's say that: \(p(s)\) is a rational function, with poles at \(s = s_j\) for \(1 \le j \le K\); and they are located for \(\Re(s) < 1\). Then:
\[
H(z) = \sum_{n=0}^\infty p(n) \frac{(-z)^n}{n!} + \sum_{j=1}^K \text{Res}\left(\Gamma(s)p(s)z^{-s}, s = s_j\right) = \frac{1}{2\pi i} \int_{1-i\infty}^{1 + i\infty} \Gamma(s)p(s)z^{-s}\,ds\\
\]
Let's for the sake of the argument, assume they are simple poles; so \(p\) is a rational function with only simple poles. (This just saves me from writing a bunch of derivatives and double sums, lol). Then we can reduce our function to:
\[
H(z) = \sum_{n=0}^\infty p(n) \frac{(-z)^n}{n!} + \sum_{j=1}^K a_j \Gamma(s_j) z^{-s_j}\\
\]
Where:
\[
p(s) = \frac{a_j}{s-s_j} + h(s)\\
\]
Where \(h\) is holomorphic in a neighborhood of \(s_j\). It is the residue of \(p\) at \(s = s_j\). The beauty doesn't end here though. No, no it doesn't. We also get that:
\[
\Gamma(s) p(s) = \int_0^\infty H(x)x^{s-1}\,dx\\
\]
BUT; we get this in a very restricted value \(A < \Re(s) < 1\). Where \(A = \text{max}_{1\le j \le K}(\Re(s_j))\). What happens if we remove that singularity--so pick \(\Re(s_J) = A\); then:
\[
\Gamma(s) p(s) = \int_0^\infty \left(H(x) - a_J \Gamma(s_J)x^{-s_J}\right)x^{s-1}\,dx\\
\]
But this is only true for \(B < \Re(s) < A\), where \(B = \text{max}_{j \neq J}(\Re(s_j))\).
This result is especially famous for the exponential function. Where, a little less famous result of Euler...
\[
\Gamma(s) = \int_0^\infty \left(e^{-x}-1\right) x^{s-1}\,dx\\
\]
But this is only true for \(-1 < \Re(s) < 0\). This just equates to deleting the first residue of \(\Gamma\); in which:
\[
e^{-z}-1 = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Gamma(s)z^{-s}\,ds
\]
For \(-1 < c < 0\); which can be reduced to:
\[
\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Gamma(s)z^{-s}\,ds = \frac{1}{2\pi i}\int_{1-i\infty}^{1+i\infty} \left(\Gamma(s) - \frac{1}{s}\right)z^{-s}\,ds\\
\]
In short, I do believe you are rediscovering some things. But I also think this was a great post--and a great line of research
I wish you nothing but luck. But I do think you should look at the Mellin transform. I still believe this is the missing key. You have used like 4 Mellin type transforms in this post, and if that isn't enough to push you in that direction, I don't know what is! Plus the idea of "adding residues" to make a divergent sum converge is nothing new. But I do believe you are being novel about it
Think of it, as all the steps you are doing with your "rearrangement"--which is technically wrong--the reason it works is because if you cast it in a Mellin Transform light; everything works; and the Residues pop out !!!!
Either way, beautiful post Caleb. If you say you aren't that mathematically inclined--I'd hate to meet the Caleb who is!
Any integral of the form:
\[
\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} g(s)F(s,z)\,ds\\
\]
Is kind of like a "modified" Mellin transform--and they appear frequently in the study of Hyper Geometric series.
I will explain something pretty cool; you may not have seen before. Let's take the standard sum:
\[
e^{-z} = \sum_{n=0}^\infty \frac{(-z)^n}{n!}\\
\]
Then, the sum of residues:
\[
\sum_{n=0}^\infty \text{Res}\left(\Gamma(s)z^{-s}, s = -n\right) = e^{-z}\\
\]
But this sum of residues is just the integral:
\[
e^{-z} = \frac{1}{2\pi i} \int_{1-i\infty}^{1 + i\infty} \Gamma(s)z^{-s}\,ds\\
\]
I'm sure you are well aware of this. But what happens if we add in a pole? Let's say that: \(p(s)\) is a rational function, with poles at \(s = s_j\) for \(1 \le j \le K\); and they are located for \(\Re(s) < 1\). Then:
\[
H(z) = \sum_{n=0}^\infty p(n) \frac{(-z)^n}{n!} + \sum_{j=1}^K \text{Res}\left(\Gamma(s)p(s)z^{-s}, s = s_j\right) = \frac{1}{2\pi i} \int_{1-i\infty}^{1 + i\infty} \Gamma(s)p(s)z^{-s}\,ds\\
\]
Let's for the sake of the argument, assume they are simple poles; so \(p\) is a rational function with only simple poles. (This just saves me from writing a bunch of derivatives and double sums, lol). Then we can reduce our function to:
\[
H(z) = \sum_{n=0}^\infty p(n) \frac{(-z)^n}{n!} + \sum_{j=1}^K a_j \Gamma(s_j) z^{-s_j}\\
\]
Where:
\[
p(s) = \frac{a_j}{s-s_j} + h(s)\\
\]
Where \(h\) is holomorphic in a neighborhood of \(s_j\). It is the residue of \(p\) at \(s = s_j\). The beauty doesn't end here though. No, no it doesn't. We also get that:
\[
\Gamma(s) p(s) = \int_0^\infty H(x)x^{s-1}\,dx\\
\]
BUT; we get this in a very restricted value \(A < \Re(s) < 1\). Where \(A = \text{max}_{1\le j \le K}(\Re(s_j))\). What happens if we remove that singularity--so pick \(\Re(s_J) = A\); then:
\[
\Gamma(s) p(s) = \int_0^\infty \left(H(x) - a_J \Gamma(s_J)x^{-s_J}\right)x^{s-1}\,dx\\
\]
But this is only true for \(B < \Re(s) < A\), where \(B = \text{max}_{j \neq J}(\Re(s_j))\).
This result is especially famous for the exponential function. Where, a little less famous result of Euler...
\[
\Gamma(s) = \int_0^\infty \left(e^{-x}-1\right) x^{s-1}\,dx\\
\]
But this is only true for \(-1 < \Re(s) < 0\). This just equates to deleting the first residue of \(\Gamma\); in which:
\[
e^{-z}-1 = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Gamma(s)z^{-s}\,ds
\]
For \(-1 < c < 0\); which can be reduced to:
\[
\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Gamma(s)z^{-s}\,ds = \frac{1}{2\pi i}\int_{1-i\infty}^{1+i\infty} \left(\Gamma(s) - \frac{1}{s}\right)z^{-s}\,ds\\
\]
In short, I do believe you are rediscovering some things. But I also think this was a great post--and a great line of research
I wish you nothing but luck. But I do think you should look at the Mellin transform. I still believe this is the missing key. You have used like 4 Mellin type transforms in this post, and if that isn't enough to push you in that direction, I don't know what is! Plus the idea of "adding residues" to make a divergent sum converge is nothing new. But I do believe you are being novel about it Think of it, as all the steps you are doing with your "rearrangement"--which is technically wrong--the reason it works is because if you cast it in a Mellin Transform light; everything works; and the Residues pop out
!!!!Either way, beautiful post Caleb. If you say you aren't that mathematically inclined--I'd hate to meet the Caleb who is!
