Riemann zeta function at zero

Question

Can the value of Riemann zeta function at 0, $\zeta(0)=-1/2$, be deduced from the identity $E(z)=E(1-z)$, where $$E(z)=\pi^{-z/2}\Gamma(z/2)\zeta(z)?$$

Answer 1

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 $\int_0^{+\infty}e^{-u^2}du=\frac{\sqrt{\pi}}{2}$.

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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 $$ \zeta(s)\sim_02^0\pi^{-1}\frac{\pi s}{2}\Gamma(1)\left(\frac{-1}{s}\right)=-\frac{1}{2}. $$