Using the functional equation of the Zeta function to compute positive integer values I was reading this article by Ivic. In the introduction, he mentions the functional equation of the Riemann Zeta function, which he says is valid for all complex $s$:
$$
\zeta(s)=\chi(s)\zeta(1-s),
$$
where
$$
\chi(s)=2^s\pi^{s-1}\sin(\frac{\pi s}{2})\Gamma(1-s).
$$
From this we get that its zeroes are at the negative even integers. But then we have incorrect values at positive odd integers $s=2n-1$ for $n\in\mathbb{Z}, n>1$, i.e.,
$$
\zeta(2n-1)=\chi(2n-1)\zeta(-2(n-1)),
$$
which obviously gives $\zeta(2n-1)=0$ since the RHS is $\chi(2n-1)\times 0$. This is of course false, since all values of $\zeta$ converge to a nonzero value for odd positive integers $>1$.
Something is amiss here, what is it? Am I to understand that the functional equation of the Riemann Zeta function is not valid for all complex $s$?

Your conclusion: "... which obviously gives $\,\zeta(2n-1)\color{red}{=0}\,$ since the RHS is $\,\chi(2n-1)\times0\,$ ..." is incorrect. 
$$ \zeta(2n-1)=\chi(2n-1)\,\zeta(-2(n-1))=\infty\times0\,\color{red}{\ne0} $$ 
And the functional equation of Riemann Zeta function is valid $\,\forall\,s\in\mathbb{C}\,$ including $\,s=1\,$.
$$ \begin{align} 
& \lim_{n\to2}\left[\,\chi(2n-1)\,\zeta(-2(n-1))\,\right]=\zeta(3) \\ 
& \lim_{n\to3}\left[\,\chi(2n-1)\,\zeta(-2(n-1))\,\right]=\zeta(5) \\ 
& \dots\,\dots \\[4mm] 
& \zeta(s)=\chi(s)\zeta(1-s)\implies\zeta(1-s)=\zeta(s)/\chi(s) \\
& \lim_{s\to1}\left[\,\zeta(s)\,/\,\chi(s)\,\right]=\zeta(0)=-1/2 \\[2mm] 
& \lim_{s\to1}\left[\,(1-s)\,\zeta(s)\,\right] =\lim_{s\to1}\left[\,(1-s)\,2^s{\pi}^{s-1}\sin\left({\frac{\pi s}{2}}\right)\Gamma(1-s)\zeta(1-s)\,\right] = \\ 
& \lim_{s\to1}\left[\,2^s{\pi}^{s-1}\sin\left({\frac{\pi s}{2}}\right)\Gamma(2-s)\zeta(1-s)\,\right] =2^1{\pi}^0\sin\left({\frac{\pi}{2}}\right)\Gamma(1)\zeta(0) =-1 \\ 
& \dots\,\dots 
\end{align} $$ 
