# Trick for calculating the thermal entropy of the SYK model in the large $N$ limit

Reading about the SYK model I encounter a trick that should help to calculate the thermal entropy of the system. I am not able to understand what they are doing though. Particularly the part from eq.135 to eq.137 of Gabor Sarosi is giving me problems. My attempt to understand this part:

$$e^{-\beta F}=Z\sim e^{-NI[G_*,\Sigma_*]}$$

Where

$$I[G,\Sigma]=-\frac{1}{2}\log\det[\partial_{\tau}-\Sigma]+\frac{1}{2}\int\int d\tau d\tau'\left(\Sigma(\tau,\tau') G(\tau,\tau')-\frac{1}{q}J^2G(\tau,\tau')^q\right)$$

Since we don't want to deal with the determinant we write

$$\begin{equation} \begin{split} J\partial_J\left(-\frac{\beta F}{N}\right)&=J\partial_J(-I)\\ &=\frac{J^2}{q}\int\int_0^\beta d\tau'd\tau G_*(\tau,\tau')^q\\ &=\frac{\beta J^2}{q}\int_0^\beta d\tau G_*(\tau)^q \end{split} \end{equation}$$

Where in the last step I used translational invariance $$G(\tau,\tau')=G(\tau-\tau')$$. I use the Swinger-Dyson equations

$$\begin{split} &G(i\omega_n)=[-i\omega_n-\Sigma(i\omega_n)]^{-1}\\ &\Sigma(\tau)=J^2G(\tau)^{q-1}. \end{split}$$

from this one gets that

$$G(\tau)^q=\frac{1}{J^2}\Sigma(\tau)G(\tau)$$

Plugging this in to our integral gets us:

$$\begin{split} J\partial_J\left(-\frac{\beta F}{N}\right)&=\frac{\beta}{q}\int_0^\beta d\tau \Sigma(\tau)G(\tau)\\ &=\frac{1}{q}\sum_{\omega_n,\omega_m}\Sigma(i\omega_n)G(i\omega_m)\frac{1}{\beta}\int_0^\beta d\tau e^{-i\tau(\omega_n+\omega_m)}\\ &=\frac{1}{q}\sum_{\omega_n}\Sigma(i\omega_n)G(-i\omega_n)\\ &=-\frac{1}{q}\sum_{\omega_n}\Sigma(i\omega_n)G(i\omega_n)\\ &=\frac{1}{q}\sum_{\omega_n}(i\omega_n+G(i\omega_n)^{-1})G(i\omega_n)\\ &=\frac{1}{q}\sum_{\omega_n}i\omega_n G(i\omega_n)+\frac{1}{q}\sum_{\omega_n}1\\ &=-\frac{\beta}{q}\lim_{\tau\rightarrow0^+}\partial_\tau G(\tau)+\infty \end{split}$$

This is the point where I run in to trouble. Gabor Sarosi states that

$$\frac{J^2\beta}{q}\int_0^\beta d\tau G(\tau)^q=-\frac{\beta}{q}\lim_{\tau\rightarrow0^+}\partial_\tau G(\tau)$$

Why doesn't he have this infinite constant? Note that one can proof that $$\frac{\beta}{q}\lim_{\tau\rightarrow0^+}\partial_\tau G(\tau)=\frac{\beta}{N}\langle H \rangle$$ Is he absorbing the infinite constant into the energy somehow? Maybe using some sort of regularization scheme I am unaware of? Or did I just make a mistake? Hopefully one of you guy's can tell me what's going on here.

$$-\frac{\partial}{\partial \tau}G(\tau)-\int d\tau' \Sigma(\tau-\tau')G(\tau')= \delta(\tau) \xrightarrow[\delta(0^+)=0]{\tau \to 0^+} -\lim_{\tau \to 0^+}^{}\frac{\partial}{\partial \tau}G(\tau)-\int d\tau' \Sigma(0^+ -\tau')G(\tau') = 0.$$