Show that if [imath]0<b<a[/imath]

[math]\int_{-\infty}^\infty \frac{\cos{x}}{(x^2+a^2)(x^2+b^2)}\,dx=\frac{\pi}{a^2-b^2}(\frac{e^{-b}}{b}-\frac{e^{-a}}{a})[/math]

I started out by expressing [imath]\cos{x}[/imath] in terms of complex exponentials. We also know that the integrated function has simple poles at [imath]\pm ai[/imath] and [imath]\pm bi[/imath]. I created a contour from -R to R along the real axis, and going about a circle of radius R in the positive sense from R to -R. Thus, there are only two poles in the interior of the contour ([imath]ai[/imath] and [imath]bi[/imath]), and finding the residues at poles yields:

[math]Res(f,ai)=\frac{e^{-a}+e^a}{-4i(a^2-b^2)a}[/math]

[math]Res(f,bi)=\frac{e^{-b}+e^b}{4i(a^2-b^2)b}[/math]

Using the Cauchy Integral Formula, we know that the original integral is equal to the sum of the residues multiplied by [imath]2\pi i[/imath] minus the line integral along the semi-circle contour (not including the real axis).

The first part with the residues I found was equal to

[math]\frac{\pi}{a^2-b^2}(\frac{e^{-b}+e^b}{2b}-\frac{e^{-a}+e^a}{2a})[/math]

And I showed that the contour integral went to zero as R went to infinity (which is what the value of the bounds need to be) by saying f is bounded by the cosine (between 1 and -1), so the max value that f gets to with R is approximately 1/R

^{4}. I then said that the max value that the integral goes to is the length of the contour (which is simply [imath]\pi R[/imath] multiplied by the max value of f, which is [imath]\frac{\pi}{R^3}[/imath], which goes to zero as R goes to infinity.

Obviously, the result I arrived at is not what is required, and this is why I come to you for a bit of help. Did I go wrong in approximating/bounding the contour integral, or did I screw up my residues?