$\log(1-\cos (x-t)) \in \mathcal{L}^2([0,2\pi]^2)$ 
I'm currently studying on symmetric kernels of integral equations and here is the problem from Raisinghania textbook "Integral Equation and Boundary Value Problems". It is obvious that the kernel is symmetric since $$K(x,t)=\log(1-\cos (x-t))=\log(1-\cos (t-x))=K(t,x).$$
My question here is how can we know that $\log(1-\cos (x-t)) \in \mathcal{L}^2([0,2\pi]^2)$. It's not in the exercise questions, it's just my thoughs. We see that $\forall x,t:$ $x=t$ we have $K(x,x)=\log(0)= - \infty$. This part confuses me the most to prove that $K(x,t) \in \mathcal{L}^2([0,2\pi]^2)$. Any explanation would be helpful.

From the decomposition
$$ K(x,t)=-\log(2)-2\sum_{n\geq 1}\frac{\cos(nx)\cos(nt)}{n}-2\sum_{n\geq 1}\frac{\sin(nx)\sin(nt)}{n} $$
it is straightforward that $K\in \mathcal{L}^2((0,2\pi)^2)$, since $\left\{\frac{1}{n}\right\}_{n\geq 1}\in\ell_2$.
