$$f(s) = \zeta(s)\zeta(s+1)\Gamma(s) $$
This has a double pole at $s=0$ , and I got -1/2 , and then to find the other
$$ \lim_{s\to0} s(f(s) - \frac{-1/2}{s^2}) $$
(Actually, it is $-log\sqrt{2\pi} = \zeta'(0)$ from my book)
In my calcuation, It comes down to
$$ \lim_{s\to0} (s\zeta(s+1)\Gamma(s+1))' = 0 $$
And this should come down again to $ \zeta(1) +\Gamma'(1) = 0 $, where $\Gamma'(1) = -\gamma, $ so it means $$\lim_{s\to0} \zeta(s+1) = \gamma $$
From wikipedia, its Cauchy P.V $$\lim_{\epsilon\to0} \frac{\zeta(1+\epsilon)+\zeta(1-\epsilon)}{2} = \gamma \space\space\space vs. \space\space\space \zeta(1) = \infty $$
I don't know much about Cauchy's P.V, but it's related to 'integral thing'.
Why should I use this when just getting $\lim_{s\to0} \zeta(s+1)$?
OR Is there another way to get $-log\sqrt{2\pi} = \zeta'(0)$ in the first place?