Is it possible to evaluate analytically the following double integral? In a mathematical physical problem related to the search of Green's functions in an anisotropic medium, I came across a non-trivial double integral. 
Is it possible to evaluate analytically the following double integral?
$$
\varphi(h, A, B) = -\frac{1}{(2\pi)^2} \int_0^{\infty} \int_{0}^\pi \frac{\sin^3\theta \, e^{ihk\cos\theta}}{\cos^4\theta-A\cos^2\theta-B}\, \mathrm{d}\theta \, \mathrm{d}k \, , 
$$
where $h$, $A$ and $B$ are all positive real numbers. 
Making the change of variable $q=\cos\theta$, the above integral can conveniently be written as
$$
\varphi(h, A, B) = -\frac{1}{(2\pi)^2} \int_0^{\infty} \int_{-1}^1 \frac{(1-q^2) \, e^{ihkq}}{q^4-Aq^2-B}\, \mathrm{d}q \, \mathrm{d}k \, .
$$
Now, by changing the integration range for $q$ we obtain
$$
\varphi(h, A, B) = -\frac{1}{2\pi^2} \int_0^{\infty} \int_{0}^1 \frac{(1-q^2) \, \cos(hkq)}{q^4-Aq^2-B}\, \mathrm{d}q \, \mathrm{d}k \, .
$$
By making use of Maple, I have noticed that the function $\varphi$ depends solely on $h$ and $B$ and does not depend on $A$. More precisely 
$$
\varphi (h,A,B) = \varphi(h,B) = \frac{1}{4\pi B h} \, .
$$
Is there a way to show that mathematically in a rigorous way? 
Is that true?
Any help / hints / indication would be highly appreciated. 
Thank you,
Fede
 A: How to motivate the solution
If to convert the integral to the form
$$
\varphi(h, A, B) = -\frac{1}{8\pi^2} \int\limits_{-\infty}^{\infty} \int\limits_{-1}^1 \frac{(1-q^2) \, e^{-ihkq}}{q^4-Aq^2-B}\, \mathrm{d}q \, \mathrm{d}k
$$
and then change the order of integration in the integral
$$
\varphi(h, A, B) = -\frac{1}{8\pi^2} \int\limits_{-1}^1 \frac{1-q^2}{q^4-Aq^2-B} \left(\int\limits_{-\infty}^{\infty}e^{-2\pi i k\, \dfrac{qh}{2\pi}} \, \mathrm{d}k \right) \, \mathrm{d}q,$$
then there is an opportunity to use the representation of the Dirac delta function as a Fourier transform in the form of
$$\delta(x) = \int\limits_{-\infty}^{\infty}e^{-2\pi i k x} \, \mathrm{d}k $$
(see also Wolfram MathWorld),
where 
$$x={qh\over2\pi},$$
so
$$\int\limits_{-\infty}^{\infty}e^{-i hkq} \, \mathrm{d}k = \delta\left({qh\over 2\pi}\right) = {2\pi\over|h|}\delta(q)$$
and
$$\boxed{\varphi(h, A, B) = {1\over4Bh}}.$$
How to prove the change in the order of integration
Really, the issue integral is improper and exists as a limit
$$\varphi(h, A, B) = -\lim_{M\to\infty}\frac{1}{8\pi^2} \int\limits_{-M}^{M} \int\limits_{-1}^1 \frac{(1-q^2) \, e^{-ihkq}}{q^4-Aq^2-B}\, \mathrm{d}q \, \mathrm{d}k.$$
Note that the inner integral converges for all positive values of parameters $A$ and $B.$ The external integral in finite limits has no singularities and therefore converges for all possible values of $M,$ so 
$$\frac{1}{8\pi^2} \int\limits_{-M}^{M} \int\limits_{-1}^1 \frac{(1-q^2) \, e^{-ihkq}}{q^4-Aq^2-B}\, \mathrm{d}q \, \mathrm{d}k$$
$$ = -\frac{1}{8\pi^2} \int\limits_{-1}^1 \frac{1-q^2}{q^4-Aq^2-B} \left(\int\limits_{-M}^{M}e^{-2\pi i k\, \dfrac{qh}{2\pi}} \, \mathrm{d}k \right) \, \mathrm{d}q.$$
Now the passage to the limit is possible, since as the value of $M$ increases, there are no singularities in the vicinity of the maximum of the function. The properties of the function corresponding to the value of the inner double integral approach, to within a constant, the properties of the delta function (an intense burst with the constant area) and eventually lead to a formally obtained answer. In this case, the possibility of passage to the limit shows that getting integral converges, and this proves the correctness of the change in the order of integration.
