Divergent Series and Analytical Continuation (LONG post)

[Caleb]

Apart from the idea of reflection formula's ( which may or may not be a good idea )

I want to take for example the prime zeta function P(s).

It is well known to have formulas that converge for Re(s) > 1 or Re(s) > 0.

There also exists a formula for Re(s) > 1/2 :

Assuming the RH then \[P(s) = s \int_2^\infty \pi(x) x^{-s-1}dx = s \int_2^\infty (\pi(x)-Li(x)) x^{-s-1}dx+  L(s) \\  = s \int_2^\infty (\pi(x)-Li(x)) x^{-s-1}dx+L(2) + \log(s-1)+ \int_2^s \frac{2^{1-u}-1}{u-1}du\] where the latter integrals converge and are analytic for Re(s) > 1/2.

\[ Li(x) = \int_2^x \frac{dt}{\log t}\] \[L(s) = s\int_2^\infty Li(x) x^{-s-1}dx  = \int_2^\infty \frac{x^{-s}}{\log x}dx \] \[ = L(2)+\int_2^s L'(u)du = L(2) + \log(s-1)+ \int_2^s \frac{2^{1-u}-1}{u-1}du\]

since \[L'(s) = -\int_2^\infty x^{-s}dx = \frac{2^{1-s}}{s-1}\]

I guess that is clear to all here.



Now the natural boundary at Re(s) = 0 is made completely of log singularities getting dense.

Maybe we should make distinctions of the type of natural boundaries we are getting.

I mean for instance 

g(x) = (1 - x)(1 - x^3)(1 - x^5)(1 - x^7)... 

or h(x) = 1 + x^2 + x^2^2 + x^2^3 + ...

are having "different" natural boundaries, like accumilations of zeros.

i SAID MAYBE lol



But now,

what is the value of P(s) for Re(s) < 0 ??

OR is this type of natural boundary unsuitable because it has logs instead of poles and zero's ??
AND IF UNSUITABLE , WHAT DOES THAT MEAN ??  no continuation for some but for others we do ?

I will definitely have a talk about that with my friend mick.


I want to point out that the derivative of the prime zeta has an infinite amount of poles instead of logs.

and the inverse of the derivative of prime zeta has an infinite amount of zero's on Re(s) = 0.


Im holding back on making conjectures , im a bit confused.


regards

tommy1729

This is a good question. Let me try to answer it by sharing my motivation in studying the series I decided to study in the post.

Analytical continuation beyond natural boundaries is hard! I don't think anyone knows how to do it in general. I suspect it can't be done in general-- because I don't think it would be meaningful to analytically continue a series that has purely random coefficents for instance. Since the problem is so hard, I'm choosing to study the easiest possible example I could come up with. 

The examples I study have a really nice property. For instance, consider the following series 
\[f(x)=\sum_{n=0}^\infty \frac{x^n}{1+x^n} \frac{1}{2^n}\]
This series has a natural boundary, since 1+x^n provides a dense set of poles. So, it cannot be analytically continued to |x|>1. However, the series is still well-defined outside of |x|>1. In particular, I can compute f(2) by just plugging into the series. This gives me a very natural candidate for a definition of \( f(x)\) outside the boundary.

So, I choose to study these much easier functions, and try to analyze what sort of properties they have. My goal looks like this
\[ \text{Get a bunch of very easy examples of functions with natural boundaries } \to \]
\[\text{ Study those examples in depth, and understand the mechanism underlying how those functions behave } \to \]
\[\text{ Try to generalize that mechanism into harder functions }\]

The prime zeta function is definitely in the category of "harder functions." I don't know if logarithmic singularities will cause the continuation to behave differently than regular poles-- that's something I will only find out once I've studied the easy examples in depth!

So, its not neccesariy that the prime zeta function has properties that make it unsuitable for continuation-- its that I haven't yet figured out the right way to do continuation beyond natural boundaries.