Divergent Series and Analytical Continuation (LONG post)

[tommy1729]

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