Riemann zeta function at zero Can the value of Riemann zeta function at 0, $\zeta(0)=-1/2$, be deduced from the identity $E(z)=E(1-z)$, where
$$E(z)=\pi^{-z/2}\Gamma(z/2)\zeta(z)?$$
 A: The functional equation under consideration yields:
$$
\zeta(s)=\frac{\pi^s}{\sqrt{\pi}}\frac{\Gamma\left(\frac{1-s}{2}\right)}{\Gamma\left(\frac{s}{2}\right)}\zeta(1-s).
$$
So the fact that $\zeta(0)=\lim_0\zeta(s)=-\frac{1}{2}$ follows from the following three ingredients:
$$
\zeta(s)\sim_1\frac{1}{s-1}\qquad \Gamma(s)\sim_0\frac{1}{s}\qquad \sqrt{\pi}=\Gamma\left(\frac{1}{2}\right)=\lim_{1/2} \,\Gamma(s).
$$
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 $\int_0^{+\infty}e^{-u^2}du=\frac{\sqrt{\pi}}{2}$.

The so-called functional equation of Riemann zeta function is:
$$
\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s).
$$
Using 
$$
\zeta(s)\sim_1\frac{1}{s-1}\qquad\sin\left(\frac{\pi s}{2}\right)\sim_0\frac{\pi s}{2}\qquad1=\Gamma(1)=\lim_1\,\Gamma(s)
$$
it follows that 
$$
\zeta(s)\sim_02^0\pi^{-1}\frac{\pi s}{2}\Gamma(1)\left(\frac{-1}{s}\right)=-\frac{1}{2}.
$$
