Not getting the right answer in $\zeta(0)$

Question

I was checking some results in analytical number theory and came up with two different functional equations for the Riemann's Zeta function. The first one is

$$\zeta(s) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) = \zeta(1-s) \pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2})$$

and the second is

$$\zeta(s) = 2^s \pi^{s-1} sen(\frac{\pi s}{2}) \zeta(1-s)\Gamma(\frac{1-s}{2})$$

With the second one I get the correct result ($\zeta(0) = - \frac{1}{2}$) by a limiting process, but with first one I keep getting

$$\zeta(0) = \frac{\zeta(1)}{\Gamma(0)} \Gamma(\frac{1}{2}) \pi^{-\frac{1}{2}} = \frac{\zeta(1)}{\Gamma(0)} = 1$$

Where is the error between these equations?

Answer 1

The problem is you've written $\displaystyle\frac{\zeta(1)}{\Gamma(0)}$ and interpreted it as $\displaystyle\lim\limits_{s\to1}\frac{\zeta(s)}{\Gamma(1-s)}$.

But actually, you need to be thinking about $\displaystyle\lim_{s\to1}\frac{\zeta(s)}{\Gamma(\frac{1-s}{2})}$.

Imagine, for example, the functions $f(x)=g(x)=1/x$ near $x=0$ and you wanted to find the limit of $f(x)/g(x/2)$ as $x\to0$, but you wrote it as $f(0)/g(0)$ which you then interpreted as the limit of $f(x)/g(x)$; it's the same thing here.

Answer 2

From the first formula you get $$\zeta(0)=\lim_{s\rightarrow 0}\frac{\zeta(1-s)}{\Gamma(\frac{s}{2})}.$$ The Laurent series expension of $\zeta(s)$ at the simple pole $s=1$ is $\frac{1}{s-1}+...$ and the Laurent series expension of $\Gamma(s)$ at the simple pole $s=0$ is $\frac{1}{s}+...$

So we have $$\zeta(0)=\lim_{s\rightarrow 0}\frac{\frac{1}{(1-s)-1}+...}{\frac{1}{\frac{s}{2}}+...}=\lim_{s\rightarrow 0}\frac{\frac{1}{-s}+...}{\frac{2}{s}+...}=\lim_{s\rightarrow 0}\frac{-\frac{1}{s}}{\frac{2}{s}}=-\frac{1}{2}.$$