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$$\lim_{\epsilon\to 0} \frac{\zeta(1+\epsilon) + \zeta(1-\epsilon)}{2} =\gamma$$

I am somewhat familiar with the zeta function, but have not taken complex analysis, yet. I saw this on Wikipedia and was wondering if someone can explain what the equation means. Does it say that $\zeta(1)$ would equal $\gamma$ if the function didn't go to infinity and negative infinity or something completely different?

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up vote 10 down vote accepted

Consider the function $f(x) = \dfrac{1}{x-1} + c,$ which has a singularity (simple pole with residue $1$, if you like) at $x = 1$ and is well-defined everywhere else.

A trivial computation gives $\lim_{\epsilon \to 0} \dfrac{ f(1+ \epsilon) + f(1 - \epsilon) }{2} = c.$

So the equation you ask about says that in a n.h. of $s = 1$, the function $\zeta(s)$ looks like $$\dfrac{1}{s-1} + \gamma + \text{higher order terms in } (s-1).$$

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Can you give an example of what higher order terms in s-1? –  zerosofthezeta Dec 3 '13 at 4:48
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@zeta: Dear zeta, There will be a power series in $(s-1)$, of which $\gamma$ is the constant term. See here for the formula for the coefficient of $(s-1)^n$. Regards, –  Matt E Dec 3 '13 at 4:55
    
(+1) nice answer. –  Mhenni Benghorbal Dec 3 '13 at 8:20

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