integration - Calculating a Complicated Integral of Two Variables

I have encountered the following integral
$$I=\int_{0}^{2\pi} \int_{0}^{2\pi} \frac{c_1 + c_2 \cos\theta_2}{-a + \cos\theta_1 + \cos\theta_2} \cos(m\theta_1) \cos(n\theta_2) d\theta_1 d\theta_2$$

where $a>2$ and I am having a hard time moving forward. I would be happy with an analytic solution, converting this to some non-elementary but known special function, or finding the asymptotic behavior of the integral as $a \rightarrow 2$ from the positive reals.

I was able to perform the $\theta_1$ integration by looking in Gradshteyn and Ryzhik and this yields something proportional to
$$\int_{0}^{2\pi} \frac{c_1 + c_2\cos\theta_{2}}{\sqrt{a^2 - 2 a\cos(\theta_{2})-\sin(\theta_{2})^2}} \cos\left[n\theta_{q'}\right]\\ \times \left(\sqrt{a^2 - 2 a\cos(\theta_{2})-\sin(\theta_{2})^2} -a + \cos(\theta_{2}) \right)^{m}d\theta_{2}.$$
Trying to find this form in a table was hopeless, so I changed variables to $x=\cos\theta_2$. Having to expand $\cos(n\theta_2)$ introduced a sum, and the resulting integral is proportional to
$$\int_{-1}^{1} \frac{dx}{\sqrt{1-x^2}} \frac{c_1 + c_1 x}{\sqrt{\left(-a + x \right)^2 -1 }} \\ \times \left(\sqrt{\left(-a + x\right)^2 -1} -a + x) \right)^{m}\sum\limits_{\substack{r=0 \\ 2r \leq n}} (-1)^r \binom{n}{2r} x^{n-2r} (1-x^2)^r.$$
I am pretty stumped at this point.

I tried doing some further expansions, but I got infinite sums of infinite sums, so this didn't seem like progress (the hypergeometric ${}_{3}F_{2}$ and the regularized ${}_{2}\tilde{F}_{1}$ appeared in this approach).

There are reduction formulas for integrals, but many of them don't seem to apply, as I have two distinct square roots ($\sqrt{1-x^2}$ and $\sqrt{\left(-a + x\right)^2 -1}$). However, my integrand is a rational function of these square roots and $x$.

There is a small parameter in the problem, namely $\epsilon>0$ in $a = 2 +\epsilon$. The integral diverges for $a=2$, so this is a singular perturbation problem (if this approach is even helpful).

I would appreciate any suggestions or advice on this problem!

Answer

Extracting the leading order is not hard. Indeed, let $f : \mathbb{R}^2 \to \mathbb{C}$ be a continuous function such that $f(k_1, k_2) = f(k_1 + 2\pi, k_2) = f(k_1, k_2 + 2\pi)$ for all $k_1, k_2 \in \mathbb{R}$. Then with $\|f\| = \sup |f|$,

$$\int_{0}^{2\pi}\int_{0}^{2\pi} \frac{f(k)}{a-\cos k_1 - \cos k_2} \, dk_1 dk_2 = \int_{B(0,\pi)} \frac{f(k)}{a - 2 + \frac{1}{2}|k|^2} \, dk + \mathcal{O}(\|f\|).$$

Now applying the polar coordinates change followed by the substitution $r=\sqrt{2(a-2)u}$,

$$\begin{align*} \int_{B(0,\pi)} \frac{f(k)}{a - 2 + \frac{1}{2}|k|^2} \, dk & = \int_{0}^{\pi} \frac{r}{a-2+\frac{1}{2}r^2} \left( \int_{S^1} f(r\omega) \, d\omega \right) \, dr \\ &= \int_{0}^{\frac{\pi^2}{2(a-2)}} \left( \int_{S^1} f(\omega\sqrt{2(a-2)u}) \, d\omega \right) \, \frac{du}{u+1} \\ &\sim 2\pi f(0) \log \left( \frac{\pi^2}{2(a-2)} \right) \\ &\sim -2\pi f(0) \log(a-2). \end{align*}$$

As an alternative direction, I guess that a probabilistic interpretation may perhaps help analyze the behavior of $I$. Indeed,

$$G^p(x,y) = \frac{1}{2\pi^2} \int_{0}^{2\pi}\int_{0}^{2\pi} \frac{e^{-ik\cdot y}}{\frac{2}{1-p} - \cos(k_1) - \cos(k_2)} \, dk$$

is the Green function of the 2-D simple random walk on $\mathbb{Z}^2$ killed at rate $p$.