Gaussian integral in momentum space My question is related to p. 353 of Altland and Simon (section 6.7) which concerns about the following field integral

where $\beta = 1/T$ and $V_n$ is defined in the following way:

It seems to be a straightforward Gaussian integration. However, shouldn't the factor in the exponential be 

Can somebody please help?  
 A: As you said it is straightforward, but there is a small subtlety because you are coupling $V_n$ to $V_{-n}$. Since $V_n$ and $V_{-n}$ are coupled, we regroup the terms and write the action in a more transparent form
$$
S=\sum_{n>0}-\frac{\beta}{2E_c}V_{n}V_{-n}+\frac{e^{i\omega_n\tau}-1 }{\omega_n}V_{n}+\frac{e^{i\omega_{-n}\tau}-1 }{\omega_{-n}}V_{-n},
$$
we shift the field as $V_{n}=\tilde{V}_{n}+\frac{2E_{c}}{\beta}\frac{e^{i \tau \omega_{-n} }-1 }{\omega_{-n}}$, and get
$$
S=\sum_{n>0}-\frac{\beta}{2E_c}\tilde{V}_{n}\tilde{V}_{-n}+\frac{2E_{c}T}{\omega_n\omega_{-n}}(e^{i\omega_n\tau}-1)(e^{i\omega_{-n}\tau}-1),
$$
which gives for the prefactor
$$
P= \sum_{n>0}\frac{2E_{c}T}{\omega_n\omega_{-n}}(e^{i\omega_n\tau}-1)(e^{i\omega_{-n}.\tau}-1).
$$
Then we use $\omega_{n}=-\omega_{-n}$ and obtain
$$
P = -\sum_{n>0}\frac{2E_{c}T}{\omega_n^2}(1-e^{i\omega_n\tau})-\sum_{n>0}\frac{2E_{c}T}{\omega_n^2}(1-e^{-i\omega_n\tau})=-\sum_{n\neq 0}\frac{2E_{c}T}{\omega_n^2}(1-e^{i\omega_n\tau}),
$$
where i used $\omega_n^2=\omega_{-n}^2$. 
