A calculation of the dark matter density profile in a dissipative dark matter model leads to the integral
$$f(x,\theta)=\int\limits\_0^\infty\frac{y\,e^{-y}\,dy}{\sqrt{x^4+y^4+2x^2y^2\cos{2\theta}}}.$$
Is it possible to calculate this integral in a closed form? What is its limit in the $x\ll 1$ case?



Closed form of an dark matter related Integral
Denote $y=tx$, we get a Laplace transform $$f(x,\theta)=\int\_0^{\infty}\frac{t}{\sqrt{t^4+2t^2\cos 2\theta+1}}e^{-xt}dt.$$
Integral over $[0,1]$ is bounded uniformly in $x$, and for $[1,\infty)$ we have
$$
\int\_1^{\infty}\frac{t}{\sqrt{t^4+2t^2\cos 2\theta+1}}e^{-xt}dt=
\int\_{1}^\infty \frac1t e^{-xt}dt+\int\_1^{\infty}\left(\frac{t}{\sqrt{t^4+2t^2\cos 2\theta+1}}-\frac1t\right)e^{-xt}dt.
$$
The second term is again uniformly bounded, since the expression in brackets decays as $1/t^2$. Next,
$$\int\_{1}^\infty \frac1t e^{-xt}dt=\int\_x^\infty \frac{e^{-s}}sds=\int\_x^1\frac1sds+\int\_x^1\frac{e^{-s}-1}sds+\int\_1^{\infty}\frac{e^{-s}}sds,$$
the second and the third terms are uniformly bounded and the first term equals $-\log x$. Thus, $f(x,\theta)=-\log x+O(1)$ when $\theta$ is fixed (and $\cos 2\theta\ne -1$).


