Which equation did Riemann use when proving that $\zeta(-2)=0$. I know that the first trivial zero lies on the point $(-2,\,0)$ and would like to prove it using the same equation.
Any help is much appreciated.
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For the "equation" for the trivial zeros see this question. Where did Riemann prove that $\zeta(s)=0$? – Dietrich Burde Dec 10 '19 at 15:29
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I'm sorry, that made no sense. I mean't to say (-2)=0. Thank you for the link. – 2527 Dec 10 '19 at 15:35
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Here’s the MathJax tutorial – gen-ℤ ready to perish Dec 10 '19 at 16:04
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See https://en.wikipedia.org/wiki/Riemann_zeta_function#Riemann's_functional_equation – Barry Cipra Dec 10 '19 at 16:18
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He proved that $$\int_0^\infty x^{s-2}(\frac{x}{e^x-1}-\sum_{k\le K} \frac{B_k}{k!}x^k 1_{x > 1})dx = \Gamma(s)\zeta(s) -\sum_{k\le K} \frac{B_k}{k!}\frac1{s+k-1}$$ is analytic for $\Re(s) > -K$
where $\frac{x}{e^x-1}=\sum_k \frac{B_k}{k!}x^k$ for $|x|< 2\pi$ thus $$\zeta(-k) = (-1)^k k! (s+k)\Gamma(s)\zeta(s)|_{s=-k} = \frac{(-1)^kB_{k+1}}{k+1}$$ As you see this is much easier than the functional equation.
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