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The Riemann Zeta function $\zeta(s)$ satisfies the functional equation $$ \zeta(s)=2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s) $$Because of this, it is obvious ("trivial") that the zeta function has zeros at $s=-2n, n\in\mathbb{Z}^+$. At least that's what it says on Wikipedia, anyway. To me it seems like there would also be trivial zeros at the positive even integers, since in the case $s=2$, a factor would be $\sin(2\pi/2)$, which is just $\sin(\pi)$, or $0$. Clearly this can't be right because that would be a counterexample to the Riemann hypothesis and besides it's well known that $\zeta(2)=\frac{\pi^2}{6}$. And sure enough that's the result when you put $n=1$ into the formula $$ \zeta(2n)=\frac{(-1)^{n+1}B_{2n}(2\pi)^{2n}}{2(2n)!} $$ Can somebody explain to me why the zeta function does not have zeroes at $s=2n,n\in\mathbb{Z}$?

My initial thought was that this is only valid for $\Re(s)<1$ but Wikipedia says [of the function equation]:

This is an equality of meromorphic functions valid on the whole complex plane

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    $\begingroup$ In short: while $\sin(\pi s/2)$ has zeros, $\Gamma(1-s)$ has poles, and those two cancel out. $\endgroup$ – Wojowu May 18 at 21:54
  • $\begingroup$ well, this is embarrassing. I spent a good amount of time thinking about it before asking so it's kind of irritating that the answer is something simple like that. thanks for the help. $\endgroup$ – sam-pyt May 18 at 21:57
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You might have read onto the end of the paragraph in your Wikipedia article:

When s is an even positive integer, the product $\sin(\frac{\pi s}{2})\Gamma(1 − s)$ on the right is non-zero because $\Gamma(1 − s)$ has a simple pole, which cancels the simple zero of the sine factor.

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  • $\begingroup$ oh, wow. the answer was right in front of me the whole time. thanks for your answer $\endgroup$ – sam-pyt May 18 at 21:59

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