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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?

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Which zeroes? The trivial ones or the non trivial ones. –  Mhenni Benghorbal Jan 10 '13 at 1:45
    
both trivial and non trivial –  Ethan Jan 10 '13 at 1:52
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Dear Ethan, The trivial zeroes extend all the way along the negative real axis. The trivial zeroes extend all the way up and down along the critical strip (presumably even the critical line). Given this, it's hard to encircle them with a single closed curve. So are you sure you have everything straight? Regards, –  Matt E Jan 10 '13 at 2:50
    
Not sure at all lol –  Ethan Jan 10 '13 at 2:54
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What is the source of this statement? –  Antonio Vargas Jan 10 '13 at 3:00
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2 Answers

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 $C_n$.

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I feel like I messed somthing up here,

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

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What is $c$, exactly? Also, could you give a citation for that version of Perron's formula? –  Antonio Vargas Jan 10 '13 at 4:00
    
@Ethan Typically in Perron's formula, you integrate over a line $\text{Real}(s) = c$. Hence, do you mean an integral over the line or is it some closed curve in which case could you clarify, which Perron's formula you are using. –  user17762 Jan 10 '13 at 4:04
    
I thought I could re-write it as a contour integral over some curve involving the roots of the zeta function, but nvm that, you can just assume the countour integral is a definite integral with lower and uper bound replaced with $c-i\infty$, $c+i\infty$, and c>1, and arbitrary otherwise. –  Ethan Jan 10 '13 at 4:11
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