I am currently looking into Dimer coverings and my next step is to find how the following limit is calculated: $$\begin{align*} L &= \lim_{n \to \infty}\frac{1}{n^2}\ln\left(2^{n^2/2}\prod_{j=1}^{n/2} \prod_{k=1}^{n/2}\left(\cos^2(\frac{j \pi}{n+1}) + \cos^2(\frac{k \pi}{n+1})\right)\right) \\[1ex] &= \frac{1}{16 \pi^2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\ln[4+2\cos(x) + 2\cos(y)] \, dx \, dy \end{align*}$$
I think I got quite far, but there is some error in my calculation, which gives the required result, but the lower bound of my integrals is 0 instead of $- \pi$. I will write my calculations down here. If anyone finds what I did wrong, please tell me. It would help a lot.
$$\begin{align*} L &= \lim_{n \to \infty} \frac{1}{n^2} \ln\left(2^{n^2/2}\prod_{j=1}^{n/2} \prod_{k=1}^{n/2} \left(\cos^2\left(\frac{j \pi}{n+1}\right) + \cos^2\left(\frac{k \pi}{n+1}\right)\right)\right) \\[1ex] &= \lim_{n \to \infty} \frac{1}{n^2} \sum_{j=1}^{n/2} \sum_{k=1}^{n/2}\ln\left(4\left(\cos^2\left(\frac{j \pi}{n+1}\right) + \cos^2\left(\frac{k \pi}{n+1}\right)\right)\right) \\[1ex] &= \frac{1}{4 \pi^2} \int_0^{\pi/2} \int_0^{\pi/2} \ln(4(\cos^2(x)+\cos^2(y)) \, dx \, dy \\[1ex] &= \frac{1}{4 \pi^2} \int_0^{\pi/2} \int_0^{\pi/2} \ln\left(4\left(\frac12\cos(2x)+\frac12+\frac12\cos(2y)+\frac12\right)\right) \, dx \, dy\\ &= \frac{1}{4 \pi^2} \int_0^{\pi/2} \int_0^{\pi/2} \ln(2\cos(2x)+2\cos(2y)+4) \, dx \, dy\\ &= \frac{1}{16 \pi^2} \int_0^{\pi} \int_0^{\pi} \ln(2\cos(x)+2\cos(y)+4) \, dx \, dy \end{align*}$$