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The Riemann zeta function is the function of the complex variable $s=α+iβ$, defined in the half-plane $α>1$ by the absolutely convergent series

$$ζ(s)=\sum_{n=1}^\infty \frac{1}{n^s}$$

In many books, the authors speak about the derivative $ζ′(s)$ in the critical strip. My question is: How we can calculate this derivative in despite that $ζ(s)$ is defined in the half-plane $α>1$, thus no convergence in the critical strip.

DER
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2 Answers2

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The function is defined everywhere else in the complex plane by analytic continuation, which is a technique to extend in a natural way any function that can be defined as a power series that converges in some open subset of the complex plane.

The technique is based on a theorem of complex analysis that says that any two analytic functions (i.e. they can be expressed as convergent power series) that are equal at infinitely many points in a bounded area of the complex plane are the same everywhere they are defined. Which means that if we have a function $f$ that we know is analytic on for instance some open subset $U\subset\Bbb C$, then there is in some sense at most one function $g$ which is analytic on (almost all of) $\Bbb C$, and at the same time agrees with $f$ when restricted to $U$.

Thus the series $\zeta(s)$ you quote, which is an analytic function on the part of the complex plane where the real part exceeds $1$, can be continued uniquely to a function defined almost everywhere on $\Bbb C$. And that continuation is what we call the Riemann zeta function.

Arthur
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  • @ Arthur: Does there exists an explicit formula for that continuation. – DER Sep 24 '13 at 10:58
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    @AR1 The wikipedia article on the Riemann hypothesis lists this identity: $$ \left(1 - \frac{2}{2^s}\right)\zeta(s) = \sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n^s} $$ where the series on the right converges for all complex $s$ with real part greater than $0$, so you can extend the domain of $\zeta$ accordingly. Furthermore, the functional equation $$ \zeta(s) = 2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right) \Gamma(1-s)\zeta(1-s) $$ relates the value of the function at $s$ with the value of the function at $1-s$, so you can use that to define it on the left half of the complex plane. – Arthur Sep 24 '13 at 11:09
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    @Ar1 from the functional equation, and setting $s$ to negative even integers, you can see the trivial zeros of the function. It's also clear to see from it why the line $\frac{1}{2} + \beta i$ is distinguished from any other line, since that's the axis of "reflection", that relates two different values of the function with eachother. – Arthur Sep 24 '13 at 11:13
  • @ Arthur: So, for the first identity we can make derivation as in the case of the product of two functions without any problem. – DER Sep 24 '13 at 11:13
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    @Ar1 Yes, for numbers with real part between $0$ and $1$, you just define $$ \zeta(s) = \frac{\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n^s}}{\left(1 - \frac{2}{2^s}\right)} $$ – Arthur Sep 24 '13 at 11:14
  • @ Arthur: Thank you very much. – DER Sep 24 '13 at 11:15
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In addition to Arthur's excellent answer, if you look at Edwards's great book , "Riemann's Zeta Function," p. 10-11, you can see $\zeta(s)$ defined as an integral which "remains valid for all $s$"

$$\zeta (s) = \frac{\Pi(- s)}{2 \pi i} \int_{+ \infty}^{+ \infty} \frac{(- x)^s}{e^x - 1}\frac{dx}{x}$$

and for Re($s$) > $1$ is equal to the Dirichlet series you present above.

Here is a link: http://books.google.com/books?id=ruVmGFPwNhQC&pg=PA10