The Riemann Zeta Function satisfies the functional equation $\zeta(s)=2^s\pi^{s-1}\sin\left(\dfrac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)$. But when $s$ is a positive even number, $\sin\left(\dfrac{\pi s}{2}\right)$ and $\Gamma(1-s)$ are both $0$, which would imply that $\zeta(2)=0$, but of course, $\zeta(2)=\dfrac{\pi^2}{6}$. What am I missing?
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I asked a similar question here/ the third comment helped: http://math.stackexchange.com/questions/606321/does-the-funcitonal-equation-of-the-zeta-function-apply-for-all-reals – zerosofthezeta Jun 24 '14 at 20:01
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$\Gamma$ has a pole, not a zero, at each negative integer.
anomaly
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Oh, whoops. Does the product of the sine and the gamma functions "cancel" to some other function at the even integers? – Nishant Jun 24 '14 at 02:26
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