Why does the following not show $\zeta(0) = -\frac{1}{2}$? 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}
 A: The functional equation under consideration yields:
$$\zeta\left(\text{s}\right)=\frac{\pi^\text{s}}{\sqrt{\pi}}\frac{\Gamma\left(\frac{1-\text{s}}{2}\right)}{\Gamma\left(\frac{\text{s}}{2}\right)}\zeta(1-s)\tag1$$
So the fact that $\zeta(0)=\lim_{\text{s}\to0}\zeta(\text{s})=-\frac{1}{2}$ follows from the following three ingredients:
$$\zeta(\text{s})\sim_1\frac{1}{\text{s}-1}\qquad \Gamma(\text{s})\sim_0\frac{1}{\text{s}}\qquad \sqrt{\pi}=\Gamma\left(\frac{1}{2}\right)=\lim_{\text{s}\to\frac{1}{2}} \,\Gamma(s)\tag2$$
I assume you know the first two ones. The last one can be computed directly from the integral definition of $\Gamma$ by variable change $u=\sqrt{t}$, using $\int_0^{+\infty}e^{-\text{u}^2}\space\text{du}=\frac{\sqrt{\pi}}{2}$.

The so-called functional equation of Riemann zeta function is:
$$\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)\tag3
$$
Using:
$$\zeta(\text{s})\sim_1\frac{1}{s-1}\qquad\sin\left(\frac{\pi\text{s}}{2}\right)\sim_0\frac{\pi\text{s}}{2}\qquad1=\Gamma(1)=\lim_{\text{s}\to1}\,\Gamma(\text{s})\tag4$$
So, it follows that:
$$\zeta(\text{s})\sim_02^0\pi^{-1}\frac{\pi\text{s}}{2}\Gamma(1)\left(\frac{-1}{\text{s}}\right)=-\frac{1}{2}\tag5$$
A: 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.
