I would like to calculate $\sum\limits_{\omega_{n},\vec{k}}(\ln(-i\omega_{n}+\xi_{\vec{k}})+\ln(-i\omega_{n}-\xi_{\vec{k}}))$, where $\omega_{n}=\frac{2n+1}{\beta}$ and $n=0,\pm1,\pm2,\dots$
Using the fact $\sum\limits_{\omega_{n},\vec{k}}\ln(-i\omega_{n}+\xi_{\vec{k}})=\sum\limits_{\vec{k}}\ln(1+\text{e}^{-\beta\xi_{\vec{k}}})\tag{1}$
Method 1:
$\sum\limits_{\omega_{n},\vec{k}}\ln(-i\omega_{n}-\xi_{\vec{k}})=\sum\limits_{\omega_{n},\vec{k}}\ln(-i\omega_{n}+(-\xi_{\vec{k}}))=\sum\limits_{\vec{k}}\ln(1+\text{e}^{\beta\xi_{\vec{k}}})$
Therefore $$\sum\limits_{\omega_{n},\vec{k}}[\ln(-i\omega_{n}+\xi_{\vec{k}})+\ln(-i\omega_{n}-\xi_{\vec{k}})]=\sum\limits_{\vec{k}}[\ln(1+\text{e}^{-\beta\xi_{\vec{k}}})+\ln(1+\text{e}^{\beta\xi_{\vec{k}}})]=\sum\limits_{\vec{k}}[2\ln(1+\text{e}^{-\beta\xi_{\vec{k}}})+\beta\xi_{\vec{k}}]\tag{2}$$
Method 2:
for $\forall\omega_{n}$, $\exists p$, so that $\omega_{p}=-\omega_{n}$ $\Rightarrow\sum\limits_{\omega_{n},\vec{k}}\ln(-i\omega_{n}-\xi_{\vec{k}})=\sum\limits_{\omega_{p},\vec{k}}\ln(i\omega_{p}-\xi_{\vec{k}})=\sum\limits_{\omega_{p},\vec{k}}\ln[-(-i\omega_{p}+\xi_{\vec{k}})]$
It seems that the branch-cut has been assumed to be positive real axis in obtaining (1) (I am not sure). Thus we have $$\ln[-(-i\omega_{p}+\xi_{\vec{k}})]=\ln[(-i\omega_{p}+\xi_{\vec{k}})]+\ln(-1)\tag{3}$$
Since there's no pole point of $f[z]=\ln{(-1)}$, the part $\ln{(-1)}$ do not contribute to matsubara frequency. Thus following (3), $$\sum\limits_{\omega_{n},\vec{k}}\ln[-(-i\omega_{p}+\xi_{\vec{k}})]=\sum\limits_{\omega_{p},\vec{k}}\ln[-i\omega_{p}+\xi_{\vec{k}}]$$ and $$\sum\limits_{\omega_{n},\vec{k}}(\ln(-i\omega_{n}+\xi_{\vec{k}})+\ln(-i\omega_{n}-\xi_{\vec{k}}))=\sum\limits_{\vec{k}}2\ln(1+\text{e}^{-\beta\xi_{\vec{k}}})\tag{4}$$
Clearly (2) and (4) contradict, could anyone kindly help?