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?

1$\begingroup$ Asymptotics for fixed $\theta\ne \pi/2+\pi k$ and small positive $x$ is $\log x+O(1)$. $\endgroup$– Fedor PetrovOct 27 '16 at 7:01
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 $\cos 2\theta\ne 1$).

$\begingroup$ A nice trick to obtain the estimate on the integral $\int_{x}^{\infty}\frac{e^{s}}{s}ds$ is via L'Hôpital's rule: $$\lim_{x\to 0}\frac{\int_{x}^{\infty}\frac{e^{s}}{s}ds}{\log x} = \lim_{x\to 0}\frac{e^{x}/x}{1/x} = \lim_{x\to 0}e^{x}=1$$ $\endgroup$– T. LeOct 27 '16 at 17:49

$\begingroup$ @T.Le Of course, but it gives slightly worse result than $\log x+O(1)$ $\endgroup$ Oct 27 '16 at 17:53