$$\lim_{\epsilon\to 0} \frac{\zeta(1+\epsilon) + \zeta(1-\epsilon)}{2} =\gamma$$
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?
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 $$\dfrac{1}{s-1} + \gamma + \text{higher order terms in } (s-1).$$