Integral calculation I have a power spectral density function of the form
$$S(\omega) = \frac{a}{(\omega^2-K^2)^2 + K^2 h^2 \omega^2},$$
in which $a, K, h$ are some positive constants.
I want to calculate the corresponding spatial correlation 
function which is defined as follows:
$$R(x) = \int_{0}^{+\infty} \frac{\sin(\omega x) \omega \, d\omega}{(\omega^2-K^2)^2 + K^2 h^2 \omega^2}.$$
Does anyone have an idea about how to calculate this integral?
Thanks in advance.
 A: I would use the residue theorem on a related integral, the results of which I would differentiate to get the desired integral.  To wit, consider
$$\oint_C dz \frac{e^{i x z}}{(z^2-K^2)^2+K^2 h^2 z^2} $$
where $C$ is a semicircle of radius $R$ in the upper half plane when $x \gt 0$.  The denominator may be simplified a bit by expanding and collecting powers of $z$:
$$z^4 - (2-h^2)K^2z^2 + K^4$$
It will be in our interest to know the roots of this quartic:
$$z^2 =\left ( 1-\frac12 h^2\right ) K^2 \pm i h K^2\sqrt{1-\frac14 h^2}$$
Not knowing any constrains in $h$, I will assume for now that the term inside the square root is positive.  In this case, the roots of the quartic lie on a circle of radius $K$ centered at the origin.  Let's focus on just the two zeroes in the upper half plane:
$$z_{\pm} = \pm K \, e^{\pm i \frac12 \arctan{\left (\frac{h \sqrt{1-h^2/4}}{1-h^2/2} \right )}} = \pm K e^{\pm i \arcsin{h/2}}$$
Because the integral vanishes about the circular arc as $R\to\infty$, we have
$$\int_{-\infty}^{\infty} d\omega \frac{e^{i \omega x}}{(\omega^2-K^2)^2+K^2 h^2 \omega^2} = i 2 \pi \sum_{\pm} \operatorname*{Res}_{z=z_{\pm}} \frac{e^{i x z}}{(z^2-K^2)^2+K^2 h^2 z^2}$$
Since the poles are simple, the residue calculation is relatively straightforward.  For example,
$$ \operatorname*{Res}_{z=z_{+}} \frac{e^{i x z}}{(z^2-K^2)^2+K^2 h^2 z^2} = \frac{e^{i k x \sqrt{1-h^2/4}} e^{-k h x/2}}{4 K^3 e^{i \arcsin{h/2}} \left [e^{i 2 \arcsin{h/2}}-(1-h^2/2) \right ]} $$
$$ \operatorname*{Res}_{z=z_{-}} \frac{e^{i x z}}{(z^2-K^2)^2+K^2 h^2 z^2} = -\frac{e^{-i k x \sqrt{1-h^2/4}} e^{-k h x/2}}{4 K^3 e^{-i \arcsin{h/2}} \left [e^{-i 2 \arcsin{h/2}}-(1-h^2/2) \right ]} $$
At this point, the algebra gets a little tough, but it is doable.  I'll leave it at that, though.
Once done, you need to keep in mind that you have computed
$$2 \int_0^{\infty} d\omega \frac{\cos{\omega x}}{(z^2-K^2)^2+K^2 h^2 z^2}$$
To get the integral you want, you need to take the negative derivative with respect to $x$.
You then would consider the case $x \lt 0$ and use a contour in the lower half plane, following all of the above steps.
A: (Too long for comment. Not a solution, but probably a useful first step towards developing a solution.)
As dfeuer noted in his comment, this integral looks to be solvable in principle, but WolframAlpha's attempt at a presentation is one ugly SOB. It's always a good idea with integrals like these to clean 'em up as much as possible beforehand. The first thing I would do is transform our three parameter integral $R(x;K,h) = \int_{0}^{+\infty} \frac{\sin(\omega x) \omega \, d\omega}{(\omega^2-K^2)^2 + K^2 h^2 \omega^2}$ using the substitution $\omega=K\omega'$, with the aim of reducing the problem to an equivalent two parameter integral
\begin{align*}R\left(x;K,h\right) &= \int_{0}^{+\infty} \frac{\sin(\omega x) \omega \, d\omega}{(\omega^2-K^2)^2 + K^2 h^2 \omega^2}\\
&= \frac{1}{K^2}\int_{0}^{+\infty} \frac{\sin(\omega' Kx) \omega' \, d\omega'}{(\omega'^2-1)^2 + h^2 \omega'^2}\\
&= \frac{1}{K^2} R\left(p;1,h\right),
\end{align*}
where I've introduced the parameter $p:=Kx$.

Now we can turn attention the simpler integral
$$R(x;h) = \int_{0}^{+\infty} \frac{\sin(\omega x) \omega \,d\omega}{(\omega^2-1)^2 + h^2 \omega^2}.$$
A special case: $R(x;\sqrt{2}) = \frac{\pi}2 e^{-\frac{x}{\sqrt{2}}}\sin{\left(\frac{x}{\sqrt{2}}\right)}$.
A: Thanks to Ron Gordon, I did the algebraic manipulations and the final result is as follows:
$R(x) = \int_0^{\infty} \frac{\omega sin(\omega x) d \omega}{ (\omega^2 - K^2)^2 + K^2 h^2 \omega^2} = \frac{\pi}{2K^2h\sqrt{1-h^2/4}} e^{-Kxh/2} sin(kx\sqrt{1-h^2/4})$
For the special case in which $h = \sqrt{2}$, the solution is seen to be exactly the one proposed by David H.
Thanks You again,
