TITLE: Zeta function integral
QUESTION [6 upvotes]: How can  I show$$\frac{1}{2\pi i}\oint_{c}\frac{1}{\zeta(s)s(s-1)^2}ds=-1$$ Where C is a closed curve encircling all of the zeros of $\zeta(s)$,
Perhaps can someone just help me show it exists (the integral)
Doesn't the fact the real parts of the zeros of the zeta function are less then 1 imply its existence?

REPLY [3 votes]: Here is the sum of residues at the trivial zeroes of the Zeta function
$$ -\sum _{k=1}^{\infty }\,\frac{1}{{2k\zeta}'(-2k)(2k+1)^2} \sim 0.9998418292,$$
where the residue at $s=-2k$ is given by
$$ \lim_{s \to -2k}\frac{(s+2k)}{\zeta(s)s(s-1)^2}=-\frac{1}{{2k\zeta}'(-2k)(2k+1)^2}. $$
Note: You need to find a suitable sequence of contours $C_n$.