# Calculating the Zeroes of the Riemann-Zeta function

Wikipedia states that

The Riemann zeta function $\zeta(s)$ is defined for all complex numbers $s \neq 1$. It has zeros at the negative even integers (i.e. at $s = −2, −4, −6, ...)$. These are called the trivial zeros. The Riemann hypothesis is concerned with the non-trivial zeros, and states that: The real part of any non-trivial zero of the Riemann zeta function is $\frac{1}{2}$.

What does it mean to say that $\zeta(s)$ has a $\text{trivial}$ zero and a $\text{non-trivial}$ zero. I know that $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$$ what wikipedia claims it that $\zeta(-2) = \sum_{n=1}^{\infty} n^{2} = 0$ which looks absurd.

My question is can somebody show me how to calculate a zero for the $\zeta$ function.

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I think it may help mathworld.wolfram.com/Riemann-SiegelFormula.html – 89085731 Apr 20 '12 at 13:34
The series is not applicable for $\Re(s)\leq 1$; one uses a different formula (an analytic continuation, if you will) of the $\zeta$ function (so yes, it does look absurd until you consider the extension of the function to the rest of the complex plane). – J. M. Apr 20 '12 at 13:37
As Ginger mentions, one uses the Riemann-Siegel formula numerically to compute the nontrivial zeroes (there are no known closed forms for them). – J. M. Apr 20 '12 at 13:39

You are going to need a bit of knowledge about complex analysis before you can really follow the answer, but if you start with a function defined as a series, it is frequently possible to extend that function to a much larger part of the complex plane.

For example, if you define $f(x)=1+x+x^2+x^3+...$ then $f$ can be extended to $\mathbb C\setminus \{1\}$ as $g(x)=\frac{1}{1-x}$. Clearly, it is "absurd" to say that $f(2)=-1$, but $g(2)=-1$ makes sense.

The Riemann zeta function is initially defined as a series, but it can be "analytically extended" to $\mathbb C\setminus \{1\}$. The details of this really require complex analysis.

Calculating the non-trivial zeroes of the Riemann zeta function is a whole entire field of mathematics.

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In particular: $$\zeta(s)=2(2\pi)^{s-1}\sin\frac{\pi s}{2}\Gamma(1-s)\zeta(1-s)$$ Replace $s$ in both sides with a negative even integer and observe... – J. M. Apr 20 '12 at 13:46
Analytic continuation for $\Re s>0$ does not really require so much knowledge other than integrals, and good notion of convergence. Analytic continuation to $s \neq 1$ requires only the Poisson summation formula, not really complex analysis either. – late_learner Apr 20 '12 at 13:50
I don't know about the integrals late_learner is referring to, but the treatment I'm accustomed to for continuing to $\Re\,s > 0$ is to consider the related Dirichlet $\eta$ function... – J. M. Apr 20 '12 at 13:53
Perhaps, but the whole notion of analytic continuations - What is analytic? Is it distinct? Why would we want this type of continuation? - really require beginning complex analysis. – Thomas Andrews Apr 20 '12 at 13:53
Sorry, I was only focusing on the last question. The OP seems really to have difficulties about the notion of analytic continuation. – late_learner Apr 20 '12 at 14:08
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Copied from Wikipedia:

For all $s\in\mathbb{C}\setminus\{1\}$ the integral relation $$\zeta(s) = \frac{2^{s-1}}{s-1}-2^s\!\int_0^{\infty}\!\!\!\frac{\sin(s\arctan t)}{(1+t^2)^\frac{s}{2}(\mathrm{e}^{\pi\,t}+1)}\,\mathrm{d}t,$$ holds true, which may be used for a numerical evaluation of the Zeta-function. http://mo.mathematik.uni-stuttgart.de/kurse/kurs5/seite19.html

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