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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?

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

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  • $\begingroup$ 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. Le
    Oct 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

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