# 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.

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What is your problem? Finding the roots of the denominator? (They are given by $w^2=\alpha^2\pm i\beta$.) – Harald Hanche-Olsen Dec 14 '12 at 16:16
I did it: $\omega^2 = \alpha^2+\beta^2\pm\beta \sqrt{\beta^2+\alpha^2}$ but then I can't rule one off because it would be negative (I'm pretty sure it isn't). And I am not totally sure about the formula for integrating the fraction of two polynomials. – Learning is a mess Dec 14 '12 at 16:19
Thank you for your formula. – Learning is a mess Dec 14 '12 at 16:20
Actually I made I mistake and typing the equation, if you could look a the corrected version... – Learning is a mess Dec 14 '12 at 16:24

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 $\alpha$, $\beta$.

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Is that for the original, or the revised version of the problem? – Harald Hanche-Olsen Dec 14 '12 at 16:29
The original, without the $\omega^2$ multiplying $\beta^2$ in the denominator. The revised integral is done by Mathematica as well. – Eckhard Dec 14 '12 at 16:33
I rather suspect that the denominator in the first case can be simplified dramatically through the identity $(\sqrt{u-v}+\sqrt{u+v})^2=2u+2\sqrt{u^2-v^2}$, with $u=-2\alpha^2+\beta^2$ and $v=\beta\sqrt{-4\alpha^2+\beta^2}$. The denominator becomes $\alpha^2\sqrt{2\alpha^2-\beta^2}$, if my calculations are correct. – Harald Hanche-Olsen Dec 14 '12 at 17:05
Typo at the end of my comment (and I just missed the 5 minute editing deadline): Should be $\alpha^2\sqrt{2\alpha^2+\beta^2}$. – Harald Hanche-Olsen Dec 14 '12 at 17:12
I think using your suggestion shows that the integral in the first case simplifies to $\pi/(\alpha^2\beta)$ and it is therefore not necessary to consider different cases at all. – Eckhard Dec 14 '12 at 17:23

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 $$2\pi i (q_0+q_1) = - 2\pi i \frac{1}{\beta} \frac{i}{2\alpha^2} = \frac{\pi}{\alpha^2\beta}.$$

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