$$f(s) = \zeta(s)\zeta(s+1)\Gamma(s) $$
This has a double pole at $s=0$ , from $\zeta(s+1)$ and $\Gamma(s)$ respectively, and $\lim_{s\to0}s^2f(s) = -1/2$
Then, the next step, I have difficulty with understanding the following (from my book without any explanation)
$$ \lim_{s\to0} s\left(f(s) - \frac{-1/2}{s^2}\right) = -\log\sqrt{2\pi} \;(= \zeta'(0)) $$
Can anyone explain why?
Firstly, observe that $$ \frac{{{\rm d}(t^2 f(t))}}{{{\rm d}t}} = t^2 f(t)\frac{{{\rm d}\log (t^2 f(t))}}{{{\rm d}t}} = t^2 f(t)\left( {\frac{2}{t} + \frac{{\zeta '(t)}}{{\zeta (t)}} + \frac{{\zeta '(t + 1)}}{{\zeta (t + 1)}} + \frac{{\Gamma '(t)}}{{\Gamma (t)}}} \right). $$ Utilising the Laurent series expansions of the zeta and gamma functions, we obtain $$ \frac{{\zeta '(t + 1)}}{{\zeta (t + 1)}} = - \frac{1}{t} + \gamma + o(1),\quad \frac{{\Gamma '(t)}}{{\Gamma (t)}} = - \frac{1}{t} - \gamma + o(1) $$ as $t\to0$. Taking into account that $$ \lim _{t \to 0} t^2 f(t) = - \frac{1}{2},\quad \frac{{\zeta '(0)}}{{\zeta (0)}} = - 2\zeta '(0) = 2\log \sqrt {2\pi } , $$ we arrive at the desired result.