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First I would just like to clarify I understand that the Riemann Zeta Function is only defined for $s>1$ and that we have to use other formulas that use analytic continuation to "define" the other values of $s$. What I don't understand however is why does the analytically continued function give us zero when $s$ is a negative even integer. I couldn't seem to find any proofs online and I would love to know the answer.

Badr B
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  • I think you enjoy watching this video about Riemann Zeta function https://www.youtube.com/watch?v=d6c6uIyieoo – Leandro Nov 25 '17 at 02:14
  • @Leandro I actually watched it twice already haha. Watched it the first time about a week ago and again about an hour ago. I feel like I have a basic understanding of the function, but I don't really know how the trivial zeros are proven to actually be zeros and how the output of the function is calculated in the first place. – Badr B Nov 25 '17 at 02:17
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    You can show it using the functional equation, or from $$\Gamma(s)\zeta(s) = \sum_{n=1}^\infty \int_0^\infty x^{s-1} e^{-nx}dx = \int_0^\infty x^{s-2}\frac{x}{e^x-1}dx \= \int_1^\infty x^{s-2}\frac{x}{e^x-1}dx+\sum_{k=0}^\infty \frac{B_k}{k!} \frac{1}{s+k-1}$$ where $B_k$ are the Bernouilli numbers defined by $\sum_{k=0}^\infty \frac{B_k}{k!}x^k=\frac{x}{e^x-1}$ for $|x| < 2\pi$, and $B_{2k} = 0$ because $\frac{1}{e^x-1}-\frac{1}{x}$ is even @Leandro – reuns Nov 25 '17 at 03:12
  • @reuns Thank you very much. I'll spend the majority of tomorrow analyzing the equation. – Badr B Nov 25 '17 at 03:20

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