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The Riemann zeta function $ζ(s)$ is defined for all complex numbers $s ≠ 1$ with a simple pole at $s = 1$. It has zeros at the negative even integers, i.e., at $s = −2, −4, −6, ...$.

My question: How one can obtain these roots.

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2 Answers 2

up vote 3 down vote accepted

These roots arise from that the Riemann zeta function satisfies the following functional equation

$$ \zeta(s)=2^s \pi^{s-1} sin(\frac{\pi s}{2}) \Gamma(1-s)\zeta(1-s) $$

Thus, when s=-2n, we get that

$$ \zeta(-2n)=2^{-2n} \pi^{-(2n+1)} sin(-n\pi ) \Gamma(1+2n)\zeta(1+2n) $$

which gives us

$$ \zeta(-2n) = 0 $$ as $sin(-n \pi)=0 $ and $\Gamma(1+2n) $ is finite for all n $\in \mathbb{Z}^+ $.

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$$\zeta(-2)=\sum_{k=1}^\infty \frac{1}{k^{-2}}$$ Consider the example: Zeta at negative integer $-2$. To me it is never zero, it is ever growing.

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