Zeta function integral 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?
 A: I feel like I messed somthing up here,
$$\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^nf(\frac{k}{n})=\int_{0}^1 f(x) \ dx$$
$$\sum_{k\leq x}\Lambda(k)=\psi(x)$$
$$\sum_{k\leq x}\mu(k)=M(x)$$
$$\frac{1}{n}\sum_{k=1}^n\ln(\frac{k}{n})M(\frac{n}{k})=\frac{\psi(n)}{n}-\frac{\ln(n)}{n},\text{ by Chebyshevs identity}$$
$$\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n\ln(\frac{k}{n})M(\frac{n}{k})=\int_{0}^1\ln(x)M(\frac{1}{x}) \ dx=\lim_{n \to \infty} \frac{\psi(n)}{n}-\frac{\ln(n)}{n}$$
$$\int_{0}^1\ln(x)M(\frac{1}{x}) \ dx=1, \text{ by the prime number theorem}$$
$$\frac{1}{2\pi i}\oint_{c}\frac{1}{x^s\zeta(s)s}ds=M(\frac{1}{x}), \text{by Perron's formula}$$
$$\frac{1}{2\pi i}\oint_{c}\frac{\ln(x)}{x^s\zeta(s)s}ds=\ln(x)M(\frac{1}{x})$$
$$\frac{1}{2\pi i}\oint_{c}\int_{0}^1\frac{\ln(x)}{x^s\zeta(s)s} dx  \ ds=\int_{0}^1\ln(x)M(\frac{1}{x})  \ dx=1$$
$$\frac{1}{2\pi i}\oint_{c}\int_{0}^1\frac{\ln(x)}{x^s\zeta(s)s} dx \ ds=1$$
$$\int_{0}^1\frac{\ln(x)}{x^s} dx = \frac{-1}{(s-1)^2}, \text{for } \text{ } \Re(s)<1$$
$$\frac{1}{2\pi i}\oint_{c}\int_{0}^1\frac{\ln(x)}{x^s\zeta(s)s} dx \ ds=\frac{1}{2\pi i}\oint_{c}\frac{-1}{\zeta(s)s(s-1)^2} ds=1, \text{ because the zeros of the zeta function satisfy } \text{ } \Re(s)<1$$
$$\frac{1}{2\pi i}\oint_{c}\frac{1}{\zeta(s)s(s-1)^2} ds=-1$$
A: 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 $C_n$. 
