Recently I came across the Riemmann representation of the Zeta function as follows: $$\zeta (s) = (2^s)(\pi^s-1) \sin(\frac {\pi s} 2) \Gamma(1-s) \zeta(1-s) .$$
Now, I went ahead to calculate the term $\Gamma(1-s)$ for $s=2$, knowing beforehand the values of the otehr terms, namely, $\zeta(2) = \dfrac {\pi^2} 6$ (Basel problem) and $\zeta(1-2) = \zeta(-1) = \dfrac 1 {12}$. So plugging in the values I got:
$$ \frac {\pi^2} 6 = (2^2) (\pi^2-1) \sin(\frac {2 \pi} 2) \Gamma(1-2) \frac {-1} {12} \\ \frac {\pi^2} 6 = 4 \pi (-1) \Gamma(-1) \frac 1 {12} .$$
Simplifying we get that $\Gamma(-1) = \frac \pi 2$!
However, $\Gamma (-1)$ is infinite (pole). So can someone explain me where is the problem with the above formula? Thanks.