# Confusion with the Riemann Zeta function

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

• In short: while $\sin(\pi s/2)$ has zeros, $\Gamma(1-s)$ has poles, and those two cancel out. – Wojowu May 18 at 21:54
• 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. – sam-pyt May 18 at 21:57

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.