Divergent Series and Analytical Continuation (LONG post)

[JmsNxn]

Actually, writing it this way suggest a small mistake in your formula, which is that \(p(n) \) should actually be \( p(-n) \)

Lmao! I always make this typo! sorry,  --I'm glad you caught that at least, lol

Hmm, maybe now I can start to see why the mellin transform idea could be useful-- because the (inverse) mellin transform is invertible. If we rewrite in more suggestive notation again, we have that
\[2\pi i \frac{\csc(\pi s)}{2i} \frac{1}{s!}p(-s) = \int_0^\infty H(x)x^{-s-1}\,dx\]
Okay, I think I maybe have a sense of how this connects now. Lets take some arbitrary invertible integral transform on some series \( H(x). =\sum f(n,x)\) where \( f(n,x) \) is defined only at the non-negative integers \( n\), and \( H(x) \) is the analytical continuation. If we have that
\[ \int_{c - i \infty}^{c + i \infty} f(n,x)F(n) dn =  H(x)\]
Then, we can't compute the LHS, since it involves evaluating \( f(n,x) \) at values of n where it is not defined. However, the RHS is hopefully a well defined complex function. So then, by inverting our integral transformation (denoted \( \mathfrak{I}^{-1}\) we should obtain that
\[ I^{-1} \{\int_{c - i \infty}^{c + i \infty} f(n,x)F(n) dn\} =  I^{-1}\{H(x)\} \implies\]
\[ f(n,x) =  I^{-1}\{H(x)\}\]
But, the RHS makes sense, and provides a definition for f at values it wasn't defined at before. This sounds like a promising approach-- one could study how to continue functions defined on the integers by looking at in what cases it makes sense to talk about the inverse of a given integral transform. 

In no way do I have the answers, Caleb. But so much of your work smells like Mellin transforms. And what you are doing with \(f(n,x)\) and trying to interpolate with an \(f(s,x)\) using sophisticated rules; is precisely what I mean. I'm glad you're getting where I'm coming from. I'm just trying to say, look into it, I think you'll be pleasantly surprised 

To give some background, a lot of my initial attempts in the past have been centered around approaches very similar to the Mellin Transform, but they didn't quite work out. The problem is the Mellin Transform doesn't seem to allow us to get any closer to understanding the inconsistency in picking up residues. Why do some series (like (1) \( \frac{1}{1+k^z} \)) not pick up residues, but others similar series (like (2) \( \frac{1}{1+z^k} \)) do pick up residues. A second trickiness is that if I just take the integral along a certain line, it won't work, sometimes the residues that need to be picked up can only be properly picked up with a more complicated contour. However, this is just from my niave approaches, if you have something more sophisticated in mind in how to get the residues to pop out correctly I'd be interested to know.

Caleb, I wish I had the answers. These are very very deep questions.--but, I'm going to play through one example, to show you what I mean by Mellin transform tricks which "look" like your tricks:

\[
F(z) = \sum_{k=1}^\infty \frac{1}{1+k^z} = \sum_{j=0}^\infty \sum_{k=1}^\infty (-1)^j k^{jz} = \sum_{j=0}^\infty \sum_{n=0}^\infty \sum_{k=1}^\infty (-1)^j \frac{j^n\log(k)^nz^n}{n!} \\
\]

We can write this is as:

\[
F(z) = \frac{1}{2\pi i} \int_{-1/2-i\infty}^{-1/2+i\infty} \pi \csc(\pi s) \frac{e^{\pi i s}}{1+s^{-z}}\,ds
\]

This converges for \(\Re z > 1\); and conditionally converges for \(\Re(z) = 1\)--(you need some finesse from Fourier analysis to prove this). And \(s^{-z}\) has no zeroes/poles in this domain. Already, we have analytically continued \(F(1+it)\).

Now let's frame it in Mellin Transform Format. We must write it in powers of \(z\); by which we get:

\[
F(z) = \sum_{n=0}^\infty \sum_{j=0}^\infty \frac{1}{2\pi i} \int_{-1/2-i\infty}^{-1/2+i\infty} \pi \csc(\pi s)e^{\pi i s}(-1)^j\frac{j^n(-\log(s)z)^n}{n!}\,ds
\]

Let \(u = \log(s)z\); by which \(du = \frac{z}{s} ds\); and \(s = e^{u/z}\). Where now we have a contour \(C\); which all it does is envelope the poles of \(\pi \csc(\pi s)\) to the left of \(\Re(s) = -1/2\).

So now we get:

\[
F(z) = \sum_{j=0}^\infty \frac{1}{2\pi i} \int_C \pi \csc(\pi e^{u/z})e^{\pi i e^{u/z}} (-1)^jze^{-ju} \frac{e^{u/z}}{z}\,du
\]

Where \(C\) is a contour envoloping the poles of \(\csc\) on the natural numbers; only to a first order; despite the variable change.

This again, only converges for \(\Re(z) \ge 1\). But now we have:

\[
F(z) = \frac{1}{2\pi i} \int_C \pi\csc(\pi e^{u/z})e^{\pi i e^{u/z}}\frac{e^{u/z}}{z(1-e^{-u})}\,du\\
\]

IF WE EXTEND \(F\) TO \(0 < \Re(z) < 1\) WE NEED TO PAY ATTENTION TO THE SINGULARITY WHICH APPEARS AT \(z=0\). WHICH IS THE RESIDUE YOU ARE LOOKING FOR!

Let's write:

\[
H(z) =  \frac{1}{2\pi i} \int_C \pi\csc(\pi e^{u/z})e^{\pi i e^{u/z}}\frac{e^{u/z}}{z(1-e^{-u})} -h(u,z)\,du
\]

Where the integrand is now holomorphic at \(z=0\); rather than blowing up..............

Then, what you have shown and are arguing, which is a common argument:

\[
F(z) = H(z) + \sum \text{Res} h\\
\]

You've split the results in to two parts. Where the RHS is holomorphic for \(\Re(z) > 0\); and the LHS was only holomorphic for \(\Re(z) > 1\).

The ultimate state of what you are doing is analytic continuation, but it's the analytic continuation of sums. There's a reason every analytic continuation of sums uses the Mellin transform. All's I'm saying.

I apologize if this post is a little spotty. I can tend to lose clarity in long posts. I'm just trying to show through example the importance of Mellin transforms!

Sincere Regards, James