Proper Way to Calculate Value of Riemann Zeta function?

Question

I understand that an Analytic Continuation of a function will extend its domain into areas that it previously wasn't defined in.

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

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

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. If I wanted to calculate the value of the $\Gamma$ function in this case, can I pick any definition of the $\Gamma$ function that's using the same definition planes as the Riemann Zeta function I'm using? Or is there a specific variant that I should use?

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 $\operatorname{Re}(s) \neq 1$. For instance, if I supply an $s$ with $\operatorname{Re}(s) = 0.7$, I would need to evaluate $\zeta(1 - 0.7)$, which does not converge... Does this mean I have to use one of the other definitions of the Riemann Zeta functions, such as its Integral Representations? 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.

Answer 1

I think you may be mixing up two related things.

1. Sometimes one way of defining a function works only on part of its domain, and you can address this by giving different definitions for different regions. For instance, the Dirichlet series for the zeta function only works where $\Re s>1$.

2. Sometimes a "function" is actually multi-valued and you need to pick a particular branch if you want it to have a definite value everywhere. For instance, for any $z\not=0$, $\log z$ has an infinite number of values differing by integer multiples of $2\pi i$.

The zeta function and gamma function have property 1, but they don't have property 2. So for any given value of $s$ there is a definite value of $\zeta(s)$, and a definite value of $\Gamma(s)$.

There is only a single zeta function. (Well, actually there are a bunch of other things called zeta functions, but there is only a single Riemann zeta function.) There is only a single gamma function. You might use different ways to calculate them depending on what value you're interested in, but it's the same function regardless.

The truth of the formula you quote only depends on the values of the various functions in it, not on how they're computed. So it "doesn't care" how you evaluate $\zeta(s)$ or $\zeta(1-s)$ or $\Gamma(1-s)$ or whatever.

If you want to use that formula to calculate values of $\zeta(s)$ and all you know to begin with is the Dirichlet series, then indeed it won't help you in the region $0<\Re s<1$. But it's still true there.

You can, by the way, use the formula as a tool for doing such calculations, because there's a trick to turn the Dirichlet series into something that works in $\Re s>0$. It goes like this. First of all, although the original series $\sum n^{-s}$ only converges when $\Re s>1$, the slightly-modified series $\sum (-1)^{n-1}n^{-s}$ converges when $\Re s>0$, because any alternating series $\sum (-1)^{n-1}b_n$ where $b_n \rightarrow 0$ monotonically converges. Call that sum $f(s)$, and now note that $f(s)=\zeta(s)-2\sum(2n)^{-s}$, and that correction term can be rewritten as $-2\sum2^{-s}n^{-s}=-2^{1-s}n^{-s}$. So $f(s)=(1-2^{1-s})\zeta(s)$ and so $\zeta(s)=(1-s^{1-s})^{-1}f(s)$. And $f$ is analytic where $\Re s>0$, for the same general reason as $\zeta$ (defined by the Dirichlet series) is analytic where $\Re s>1$, so this thing is an analytic continuation of $\zeta$ to $\Re s>0$. And now you can use the reflection formula to get the values everywhere else.

Answer 2

The functional equation is just a property of the Riemann zeta function. It does not provide the definition of the Riemann zeta function for all complex numbers $s\neq 1$.

There are several proofs that the original zeta function, defined by the series $\zeta(s) = \sum_{n=1}^{\infty}n^{-s}$ for $\Re s>1$, has the analytic continuation to complex numbers $s\neq 1$. For instance, for $\Re s>1$ one has the formula \begin{align}\zeta(s)\Gamma(s/2)\pi^{-s/2} = -\frac{1}{s(1-s)} + \int_1^\infty (u^{s/2}+u^{(1-s)/2})\psi(u)\frac{du}{u}\end{align} where $\psi(u) = \sum_{n=1}^{\infty}e^{-\pi n^2u} = O(e^{-\pi u})$ as $u\to\infty$. What is important is that, the last improper integral converges locally uniformly on $\mathbb{C}$, hence defines an entire function. Therefore, we may give new definition of the Riemann zeta function $\zeta(s)$ for all $s\neq 1$, by means of the above formula.

The Gamma function is treated in the same way. For $\Re s>0$, one has the formula $\Gamma(s+1) = s\Gamma(s)$, hence one can give new definition of the Gamma function on $\Re s>-1$ (except for $s\neq 0$) by $\Gamma(s) = \Gamma(s+1)/s$. This is analytic for $\Re s>-1$ and $s\neq 0$. One can continue in this fashion, to obtain analytic continuation of the Gamma function for all $s$, apart from non-positive integers.

To conclude, you may rely on the above integral formula (although it may not be easy to compute) or other formula for $0\le\Re s\le 1$ to compute its value. There are many such known formula of this kind.