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Can anyone please explain to me how to prove that Riemann zeta function is 0 for all negative even numbers. In many references , they have just given the statement without any proof. Any explanation or hint will be a great help. Thanks in advance.

justpraveen
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1 Answers1

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This follows from the functional equation:

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

Because $\sin(n\pi) = 0$ for integers $n$, the RHS is equal to zero.

Note that this does not work to show that $\zeta(2n) = 0$ for positive $n$, as $-k$ is a pole of $\Gamma$ (so the RHS becomes '$0 \cdot \infty$'). It was essential above that $1+2n$ was not a pole of either $\Gamma$ or $\zeta$.

(That's good, because $\zeta(2n)$ is decidedly not zero for positive $n$.)