Free Boson on Torus - Partition function

Question

I am currently reading Section 10.2 of the Yellow book by Francesco et al. They have shown in Equation 10.16 that \begin{equation} Z_{bos}(\tau) = \sqrt{A} \ \prod_{n} \left (\frac {2\pi}{\lambda_n}\right )^{1/2} \end{equation} where $\lambda_n \neq 0$. From now we will ignore the $2\pi$ term in the numerator. To calculate this, the book says we use Zeta function regularization technique. They define \begin{equation} G(s) = \sum_{n} ' \frac{1}{\lambda^s_n} \end{equation} where prime sum means that $\forall n, \lambda_n \neq 0 $. As it is clear, at $s = 0$, $G(s)$ is not properly defined so we have to analytically continue to lower value of $s$. However, the book says in Equation 10.18 that \begin{equation} Z_{bos}(\tau) = \sqrt{A} \exp\left(\frac{1}{2}G'(0) \right) \end{equation} I am not sure how they arrived at this conclusion.

Answer 1

Note that: \begin{equation} G'(s)=-\sum_n ' \ln( \lambda_n)\times e^{-s\,\ln(\lambda_n)} \end{equation} Then you have trivially: \begin{align} Z_\text{bos.}&=\sqrt{A}\, \left.\exp \left( \frac{1}{2}G'(s)\right)\right|_{s=0} \\ &=\sqrt{A} \exp \left( \frac{1}{2}G'(0)\right) \end{align}