<s>[INST] Exact expression of a trigonometric integral Let $a>2$ be a real number and consider the following integral
$$
I(a)=\int_0^\pi\int_0^\pi \frac{\sin^2(x)\sin^2(y)}{a+\cos(x)+\cos(y)} \mathrm{d}x\,\mathrm{d}y
$$

My question. Does there exist a closed-form expression of $I(a)$?

Some comments. Since $a-2<a+\cos(x)+\cos(y)<a+2$ and $\int_0^\pi \int_0^\pi \sin^2(x)\sin^2(y)\ \mathrm{d}x\, \mathrm{d}y=\frac{\pi^2}{4}$, we have the following bounds
$$
\frac{\pi^2}{4(a+2)} < I(a) < \frac{\pi^2}{4(a-2)},
$$ 
however I didn't manage to find an exact expression for $I(a)$. Any help is welcome!
 [/INST] Partial answer:
As the integrand is an even function, write $$I(a)=\frac14\int_{-\pi}^\pi\int_{-\pi}^\pi\frac{\sin^2x\sin^2y}{a+\cos x+\cos y}\,dx\,dy$$ and let $w=e^{ix}$. Then for $C_1=\{w:|w|=1\}$, we have \begin{align}\int_{-\pi}^\pi\frac{\sin^2x}{a+\cos x+\cos y}\,dx&=\oint_{C_1}\frac{\frac{(w-1/w)^2}{-4}}{b+\frac{w+1/w}2}\frac{dw}{iw}=\frac i2\oint_{C_1}\frac{(w^2-1)^2}{w^2(w^2+2bw+1)}\,dw\end{align} where $b:=a+\cos y$. Inside $C_1$, there is a double pole at $0$, so letting $f(w)$ be the integrand, \begin{align}\operatorname{Res}(f,0)=\lim_{w\to0}\frac d{dw}(w^2f(w))=-2b.\end{align} Solving the quadratic in the denominator of $f$ gives $w=-b\pm\sqrt{b^2-1}$ but only the positive root lies in $C_1$. As this is a simple pole, \begin{align}\operatorname{Res}(f,-b+\sqrt{b^2-1})=\lim_{w\to-b+\sqrt{b^2-1}}(w+b-\sqrt{b^2-1})f(w)=2\sqrt{b^2-1}.\end{align} Hence \begin{align}\int_{-\pi}^\pi\frac{\sin^2x}{a+\cos x+\cos y}\,dx=\frac i2\cdot2\pi i(2\sqrt{b^2-1}-2b)=-2\pi(\sqrt{b^2-1}-b)\end{align} so that \begin{align}I(a)&=-\frac\pi2\int_{-\pi}^\pi\sin^2y\left(\sqrt{(a+\cos y)^2-1}-a-\cos y\right)\,dy\\&=-\frac\pi2\int_{-\pi}^\pi\sin^2y\sqrt{(a+\cos y)^2-1}\,dy+\frac{a\pi^2}2.\end{align} Now let $z=e^{iy}$. Then for $C_2=\{z:|z|=1\}$, we have \begin{align}\int_{-\pi}^\pi\sin^2y\sqrt{(a+\cos y)^2-1}\,dy&=\oint_{C_2}\left(\frac{z-1/z}{2i}\right)^2\sqrt{\left(a+\frac{z+1/z}2\right)^2-1}\cdot\frac{dz}{iz}\\&=\frac i8\oint_{C_2}\frac{(z^2-1)^2}{z^4}\sqrt{(z^2+2az+1)^2-4z^2}\,dz.\end{align} Evidently, there is a pole of order $4$ at $0$ which lies in $C_2$. However, due to the presence of $\sqrt\cdot$, branch points also need to be considered, and solving the equation $z^2+2az+1=\pm2z$ reveals that only $z=\pm1-a+\sqrt{a^2\mp2a}$ are the two branch points in $C_2$.
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