I am trying to prove the result stated in this Wikipedia page that $$\lim_{s\to 0} s^{-r} \zeta_K(s) = -\frac{h_k\cdot R}{w_K}$$
The formulae I have are: $$\begin{aligned}\xi_K(s) &=\zeta_K(s)(D^{s/(2n)}\Gamma_\Bbb R(s))^{r_1}(D^{s/n}\Gamma_\Bbb C(s))^{r_2} \\&= \zeta_K(s) |D_K|^{s/2}\pi^{-sr_1/2}(2\pi)^{-sr_2}\Gamma(s/2)^{r_1}\Gamma(s)^{r_2}\tag{1}\\ &=\Psi(s)+\Psi(1-s)+\mu\left(K^* \backslash \Bbb A^{(1)}_K\right)\left(\frac{1}{s-1}-\frac 1 s\right)\end{aligned}$$ where $\Psi$ is analytic in $s$, and
$$\mu \left(K^* \backslash \Bbb A^{(1)}_K\right) = \frac {2^{r_1}(2\pi)^{r_2}}{w_k\sqrt{|D_K|}}h_K\cdot R,$$
which I am actually still trying to prove, c.f. this other question. My attempt consisted in equating the residue of $\xi(s)$ that I get from the third line of (1), with the one I get from the second line using $\Gamma(s) \sim \frac 1 s$ for $s\to 0$. We thus obtain:
$$-\mu \left(K^* \backslash \Bbb A^{(1)}_K\right) =- \frac {2^{r_1}(2\pi)^{r_2}}{w_k\sqrt{|D_K|}}h_K\cdot R=\text{leading coefficient of } \zeta_K(s)|_{s=0}\cdot \text{constant coefficient of }|D_K|^{s/2}\pi^{-sr_1/2}(2\pi)^{-sr_2}|_{s=0}\cdot 2^{r_1},$$
where the last $2^{r_1}$ comes from $\Gamma(s/2)^{r_1}$. But this gives us
$$-\frac {(2\pi)^{r_2}h_K\cdot R}{w_k\sqrt{|D_K|}}=\text{leading coefficient of } \zeta_K(s)|_{s=0}$$
Where was my mistake?
