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Why is $$\zeta(1 - s) = -\frac{1}{s} + \cdots$$ for small negative values of $s$?

A detailed explanation would be appreciated.

glebovg
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  • $\zeta$ is meromorphic with a simple pole at $s=1$. – anon Sep 30 '12 at 04:14
  • @DonAntonio $\zeta(1-s)$ gets turned into a series expansion between the last two lines in the linked article. – anon Sep 30 '12 at 04:15
  • Thanks @anon , the link works now. – DonAntonio Sep 30 '12 at 04:19
  • @anon How? Can you show? – glebovg Sep 30 '12 at 04:21
  • Off the top of my head no, but googling gives me http://www.math.utah.edu/~milicic/zeta.pdf for instance. – anon Sep 30 '12 at 04:24
  • @anon I don't get it. How does $\zeta (1 - s)$ become $(- 1/s + \cdots)$? In the wiki article it seems like a trivial step because there is no detailed explanation. – glebovg Sep 30 '12 at 04:37
  • @GerryMyerson Yes, but so far, no one has fully answers any of my questions. – glebovg Sep 30 '12 at 04:38
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    A meromorphic function $f(z)$ with a simple pole at $z=a$ with residue $r$ admits (locally) a power series expansion $$f(z)=\frac{r}{z-a}+c_0+c_1(z-a)+\cdots.$$ For reference, consult, say, any text or comprehensive notes on complex analysis / variables / functions. The fact that zeta is meromorphic with simple pole of residue $1$ at $s=1$ is often considered common knowledge (it should be listed on just about any serious reference on the Riemann zeta function on the web), and the proof contained in any good introductory notes on $\zeta$ - or, for instance, the link I gave. – anon Sep 30 '12 at 04:44
  • Can you actually show how $\zeta (1 - s)$ become $(- 1/s + \cdots)$? – glebovg Sep 30 '12 at 04:49
  • Me personally? No, I don't recall that stuff, but I know I can easily find references for this sort of thing, which I've then passed on to you. – anon Sep 30 '12 at 04:55
  • I just accepted Old John's answer. Frankly, this reputation concept is quite sad. – glebovg Sep 30 '12 at 05:18

1 Answers1

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For $s < 0$ $$ \begin{eqnarray} \zeta(1-s) &=& \sum_{n=1}^\infty n^{s-1} = \int_1^\infty x^{s-1} \mathrm{d}x + \sum_{n=1}^\infty \left( n^{s-1} - \int_{n}^{n+1} x^{s-1} \mathrm{d}x \right) \\ &=& -\frac{1}{s} + \sum_{n=1}^\infty \left( n^{s-1} - \frac{(n+1)^s - n^s}{s} \right) \end{eqnarray} $$ Now, observe that the following limit is finite $$ \lim_{s \to 0^-} \zeta(1-s) + \frac{1}{s} = \sum_{n=1}^\infty \left( \frac{1}{n} - \log\frac{n+1}{n} \right) = \gamma \tag{1} $$ where $\gamma$ is the Euler-Mascheroni constant. The sum on the right-hand-side of $(1)$ converges , since for large $n$ $$ \frac{1}{n} - \log\left( 1+\frac{1}{n}\right) = \frac{1}{2 n^2} + \mathcal{o}\left(\frac{1}{n^2}\right) $$ Alternatively: $$ \sum_{n=1}^\infty \left( \frac{1}{n} - \log\frac{n+1}{n} \right) = \lim_{m \to \infty} \sum_{n=1}^m\left( \frac{1}{n} - \log\frac{n+1}{n} \right) = \lim_{m \to \infty} \left( \sum_{n=1}^m \frac{1}{n} - \log(m+1) \right) = \gamma $$ Hence, this establishes that, for small negative $s$: $$ \zeta(1-s) = -\frac{1}{s} + \gamma + \mathcal{o}(1) $$

Sasha
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