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I've been looking at one of the Analytic Continuations of the Zeta function, the Riemann Zeta function:

$$\zeta(s) = 2^s \pi^{s-1} \sin \left(\dfrac{\pi s}2\right) \Gamma(1-s) \zeta(1-s)$$

I understand that there are several possible definitions for the Riemann Zeta function, however what I'm confused about is the exact meaning of the $\Gamma(1-s)$ and $\zeta(1-s)$ elements inside the definition above.

As far as I understand, $\Gamma$ is another function that has several possible definitions. Which one of those is used in the Riemann Zeta function above? Is it a specific one, or can multiple ones be applied?

What's also confusing to me is the $\zeta(1-s)$ part. What I initially assumed that would be is an evaluation of the Original Zeta function when $\operatorname{Re}(s) > 1$. However, that wouldn't converge if $0 \leq \operatorname{Re}(s) \leq 1$, and the Riemann Zeta function above is supposed to be defined in $\operatorname{Re}(s) \neq 1$. Is this some sort of recursive evaluation of the Riemann Zeta function, or a completely different function?

Thank you for reading my post, any guidance is appreciated.

Semiclassical
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Runsva
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    It's Riemann, not Reimann. – Pedro Feb 05 '23 at 01:09
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    Search for analytic continuation of the Riemann zeta function, and same for the gamma function. – reuns Feb 05 '23 at 01:10
  • https://math.stackexchange.com/questions/4232892/where-are-the-poles-of-the-riemann-zeta-function-located – Moishe Kohan Feb 05 '23 at 01:11
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    Thing is, all the ‘different’ analytic continuations produce (by uniqueness) the exact same answer. Iirc, you can even just use this functional equation to make an analytic continuation – FShrike Feb 05 '23 at 01:17
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    Note that if $s$ has real part greater than $1$, then $1-s$ has real part less than $0$. So if $s$ is outside the critical strip, then the reflection formula doesn't require considering zeta values inside the critical strip. Even still, there are subtleties to the reflection formula: in particular, it only applies to integer $s>1$ in the sense of a limit. – Semiclassical Feb 05 '23 at 01:18
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    You appear to have a fundemental misunderstanding. There is only one $Γ$ and $ζ$ function. Every definition you have seen of those two functions is equivalent (on some domain), in the same way that $f(x)=(x+1)^2$ and $f(x)=x^2+2x+1$ are equivalent definitions for $f:\mathbb{R}\rightarrow\mathbb{R}$ over the domain $\mathbb{R}$. The only caveat is that some particular definitions of $Γ$ and $ζ$ only apply to some particular domain. – Christian E. Ramirez Feb 05 '23 at 01:20
  • Thanks for all your comments, I really appreciate your replies; I wanted to clarify just a few things; First, I understand that all "versions" of an analytic continuation of a function will result in the same outputs, just on different areas of different planes. I just wasn't sure whether using a certain definition of the RZ function would restrict me to using a certain definition of the Gamma function. Based on your responses, that would mean that I'm able to use any definition of the Gamma function, as long as it's defined planes allign with those of the RZ function, correct? – Runsva Feb 05 '23 at 02:06
  • I just want to make sure I understand how the inputs here work; So does that mean that I'm only able to supply Integers as inputs to the RZ function? Because if I supply an input with $\operatorname{Re}(s) = 0.7$, I will have to evaluate $\zeta(0.3)$, which will not converge... How do I get around that? – Runsva Feb 05 '23 at 02:09
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    The standard infinite series for $\zeta(s)$ does not converge at $s=0.3$, but the zeta function itself is defined at $s=0.3$. The wiki page has a section with various representations, many of which are valid for $0<\text{Re}(s)<1$. In fact, $ζ(s)$ is defined for every complex $s$ except for $s=1$. To evaluate it at any of those points $s$, you just need to use a representation of $ζ(s)$ that is valid for your particular $s$. – Christian E. Ramirez Feb 05 '23 at 02:34
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    The zeta function isn't restricted to integers. For instance, there's no issue with computing $\zeta(3/2)$. From here, the reflection formula implies that $\zeta(-1/2)=2^{-1/2}\pi^{-3/2}\sin(3\pi/4)\Gamma(3/2)\zeta(3/2)=\zeta(3/2)/(4\pi)$ using standard results for the $\Gamma$ function. So non-integers are not a big deal here. (My comment earlier was that integers are actually more annoying: try directly plugging in $s=3$ into the reflection formula to see what I mean!) – Semiclassical Feb 05 '23 at 02:47
  • Thank you for your replies; So does that mean that the $\zeta(1-s)$ element in the RZ expression from my post represents the evaluation of 1-s for any of the RZ functions that can take it as an input? Perhaps I misworded my question(s), however all I really wanted to know is how to evaluate the $\zeta(1-s)$ (and $\Gamma(1-s)$) part(s). So in the specific case that Semiclassical brought up in his comment, how do I evaluate $\zeta(3/2)$? Do I just pick a RZ function that would accept 3/2 as an input, and take what it gives me? – Runsva Feb 05 '23 at 04:24
  • To add onto my previous comment, does that mean I can use one of the integral RZ representations (https://dlmf.nist.gov/25.5) to get a value for $\zeta(3/2)$? – Runsva Feb 05 '23 at 04:27
  • This sounds correct. The following answer has some of these details: https://math.stackexchange.com/a/3274/137524 – Semiclassical Feb 06 '23 at 22:33

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