Consider an extension of the Riemann zeta function $\zeta(s)$ where $s$ now runs over a non-standard enlargement of the complex plane. 


Observe that if $s=\sigma + it$ with $\sigma>1$ real and finite (or at least infinitesimally close thereto), but $t$ infinite, then the summands $n^s= n^\sigma(\cos \ln(n)t + i\sin \ln(n)t)$ still have values infinitesimally close to finite complex numbers. Indeed, by fixing an infinite real $T$, we can obtain from $\zeta(s+iT)$, by passing to standard parts, a convergent standard Dirichlet series.


At least with a sufficiently saturated non-standard enlargement, I believe these same Dirichlet series arise from starting with the standard Euler product and shifting, arbitrarily and independently, all the various factors vertically by various amounts.


My questions: What can one say about the possibility of finding analytic continuations for any or all of these Dirichlet series to larger domains? Does the functional equation speak to this matter? If any of these functions have natural boundary at $s=1$ on the standard view, but analytic continuation one the non-standard sense, I would welcome
any insight into how such could happen.


(Corrections welcome if my question betrays any basic misunderstanding!) 



Non-standard enlargements, $\zeta(s)$ and analytic continuation
I just saw this on arXiv: [*Nonstandard Mathematics and New Zeta and L-Functions*](http://arxiv.org/abs/0808.1965), the PhD thesis of one Benjamin Clare of U. Nottingham.



> 
> This Ph.D. thesis, prepared under the supervision of Professor Ivan Fesenko, defines new zeta functions in a nonstandard setting and their analytical properties are developed. Further, p-adic interpolation is presented within a nonstandard setting which enables the concept of interpolating with respect to two, or more, distinct primes to be defined.
>  The final part of the dissertation examines the work of M. J. Shai Haran and makes initial attempts of viewing it from a nonstandard perspective.
> 
> 
> 


(Corrections welcome if my answer betrays any basic misunderstanding!)


