I am trying to evaluate the $\zeta(s)$ at $s=0$, but I am not sure what is incorrect about the following?
\begin{equation*} \label{eq:RiemannzetaFinal}
\zeta(s) = 2^s\pi^{s-1}\sin \Bigl(\frac{s\pi}{2}\Bigr)\Gamma(1-s)\zeta(1-s), \text{ for } \Re(s) \leq 1 \text{ and } s \neq 0,1. \tag{1}
\end{equation*}
Taking the limit of $(1)$, as $s \to 0$,
\begin{align} \label{eq:Limit1}
\zeta(0) & = \lim_{s \to 0} \Bigl[ 2^s\pi^{s-1}\sin \Bigl(\frac{s\pi}{2}\Bigr)\Gamma(1-s)\zeta(1-s) \Bigr] \nonumber \\
& = \frac{1}{\pi} \cdot \lim_{s \to 0}\Bigl[ \sin \Bigl(\frac{s\pi}{2}\Bigr) \zeta(1-s) \Bigr].
\end{align}
Writing $\zeta(1-s)$ in terms of its Laurent expansion, about the simple pole $s_0=1$ with residue $1$ and writing the series expansion of $\sin \bigl(\frac{s\pi}{2} \bigr)$:
\begin{align}
\zeta(0) & = \frac{1}{\pi} \cdot \lim_{s \to 0} \Biggl[\Bigl( \frac{s\pi}{2} – \frac{(s\pi)^3}{48} + \dotsb \Bigr) \Bigl(\frac{1}{s-1}+ a_0 + a_1(s-1) + a_2(s-1)^2 + \dotsb \Bigr)\Biggr] \nonumber \\
& = \frac{1}{\pi} \cdot \frac{\pi}{2} \cdot \lim_{s \to 0} \Biggl[\Bigl( s – \frac{s^3\pi^2}{24} + \dotsb \Bigr) \Bigl(\frac{1}{s-1}+ a_0 + a_1(s-1) + a_2(s-1)^2 + \dotsb \Bigr)\Biggr] \nonumber \\
& = \frac{1}{2} \cdot \lim_{s \to 0} \Biggl[ \frac{s}{s-1} + \mathcal{O}(s) \Biggr] \nonumber \\
& = 0?\nonumber
\end{align}
Best Answer
The actual mistake is when you render
$\zeta(1-s)=\Bigl(\frac{1}{s-1}+ a_0 + a_1(s-1) + a_2(s-1)^2 + \dotsb \Bigr).$
But this is an expansion about $s=1$ when you should be expanding about $s=0$, which corresponds to $1-s=1$. Properly,
$\zeta(1-s)=\Bigl(\frac{-1}{s}+ a_0 + a_1(s) + a_2(s)^2 + \dotsb \Bigr).$
Put that in for the zeta function and all else will follow as you expect.