How do you integrate$\int_{-\infty}^{\infty} dw \frac{1}{(\alpha^2-w^2)^2+ w^2\beta^2} $ I am having trouble integrating the following expression appearing in a mechanical problem:
$$\int_{-\infty}^{\infty} dw \frac{1}{(\alpha^2-w^2)^2+ w^2\beta^2} $$
I tried using the residue theorem, but having a polynom of degree 4 in the denominator causes me trouble. Mathematica doesn't seem to take it as I type it.
 A: After some manual simplification of square roots of complex numbers, Mathematica solves the original integral to
$$
\int_{-\infty }^{\infty } \frac{1}{\left(\alpha ^2-\omega ^2\right)^2+\beta ^2} \, d\omega=\frac{\pi  \sqrt{\frac{\alpha ^2+\sqrt{\alpha ^4+\beta ^2}}{\alpha ^4+\beta ^2}}}{\sqrt{2} \beta }.
$$
The revised integral can also be done by Mathematica, with the result
$$
\int_{-\infty }^{\infty } \frac{1}{\left(\alpha ^2-\omega ^2\right)^2+\omega ^2\beta ^2} \, d\omega = \begin{cases}\frac{\sqrt{2} \pi }{\alpha ^2 \left(\sqrt{-2 \alpha ^2+\beta ^2-\beta  \sqrt{-4 \alpha ^2+\beta ^2}}+\sqrt{-2 \alpha ^2+\beta  \left(\beta +\sqrt{-4 \alpha ^2+\beta ^2}\right)}\right)}, & 2\alpha\leq\beta, \\
\frac{\pi }{\alpha ^2 \beta }, & 2\alpha >\beta.
\end{cases}
$$
Using Harald Hanche-Olsen's suggestion, this simplifies to
$$
\int_{-\infty }^{\infty } \frac{1}{\left(\alpha ^2-\omega ^2\right)^2+\omega ^2\beta ^2} \, d\omega = \frac{\pi }{\alpha ^2 \beta }
$$
for all real numbers $\alpha$, $\beta$.
A: Here is how we might go about it while minimizing the use of computer algebra tools. The idea is to compute the integral on a contour consisting of the line segment from $-R$ to $R$ on the real axis and a semicircle of radius $R$ in the upper half plane, letting $R$ go to infinity. Since our function $$f(w) = \frac{1}{(\alpha^2-w^2)^2 + \beta^2 w^2}$$ is $O(1/R^4)$ on the semicircle its contribution is $O(1/R^3)$ and vanishes in the limit. So to calculate the integral we only need to find the sum of the residues at the poles in the upper half plane.
First, calculate the location of the poles:
$$ (\alpha^2 - w^2)^2 = - \beta^2 w^2 \\
\alpha^2 - w^2 = \pm i \beta w $$ so that
$$ w^2 \pm i \beta w - \alpha^2 = 0$$ or
$$ \rho_{0, 1, 2, 3} = \pm \frac{1}{2} i\beta \pm \frac{1}{2} \sqrt{4\alpha^2-\beta^2}.$$ We assume that $\alpha$ and $\beta$ are positive and that $2\alpha>\beta$ and take the poles in the upper half plane which correspond to 
$$\alpha^2 - w^2 = - i \beta w $$
and have the values
$$ \rho_{0, 1} = \frac{1}{2} i\beta \pm \frac{1}{2} \sqrt{4\alpha^2-\beta^2}.$$
We now need to find the residues at these poles. Take $\rho$ as a placeholder that will be instantiated to $\rho_{0,1}$ later on. The poles are simple and the residues are
$$\lim_{w\to \rho} \frac{w-\rho}{(\alpha^2-w^2)^2 + \beta^2 w^2} =
\lim_{w\to \rho} \frac{1}{2(\alpha^2-w^2)(-2w) + 2\beta^2 w} =
\lim_{w\to \rho} \frac{1}{2(-i\beta w)(-2w) + 2\beta^2 w}$$
or
$$\frac{1}{\beta}\lim_{w\to \rho} \frac{1}{4iw^2 + 2\beta w} =
\frac{1}{\beta}\lim_{w\to \rho} \frac{1}{4i(\alpha^2+i\beta w) + 2\beta w}
= \frac{1}{\beta}\lim_{w\to \rho} \frac{1}{4i\alpha^2 -2\beta w}$$
or $$q = \operatorname{Res}_{w=\rho} f(w) =
\frac{1}{\beta} \frac{1}{4i\alpha^2 -2\beta \rho}.$$
Observe that
$$\frac{1}{x+iy} + \frac{1}{-x+iy} = - \frac{2iy}{x^2+y^2}$$ and apply this to the second factor in $q_0$ and $q_1$ to get
$$ x^2 + y^2 = 4\beta^2 \frac{1}{4} (4\alpha^2-\beta^2) +  (4\alpha^2-\beta^2)^2
= (4\alpha^2-\beta^2) (4\alpha^2-\beta^2 + \beta^2) = (4\alpha^2-\beta^2) 4\alpha^2$$
and $$- \frac{2iy}{x^2+y^2} = - \frac{2i(4\alpha^2-\beta^2)}{(4\alpha^2-\beta^2)4\alpha^2}=
-\frac{i}{2\alpha^2}.$$
It follows that the end result is
$$2\pi i (q_0+q_1) = - 2\pi i \frac{1}{\beta} \frac{i}{2\alpha^2} =
\frac{\pi}{\alpha^2\beta}.$$
