Learn How to Evaluate Improper Integrals with Residues | Residues Homework Help

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The discussion focuses on evaluating the improper integral of cos(x) divided by the product of two quadratic terms using residue theory. The integral is defined from negative to positive infinity, and the solution involves identifying the poles at x = ±ai and x = ±bi, where the integrand is not analytic. A contour integral approach is suggested, where the integral over a semicircular contour vanishes, allowing the evaluation of the integral along the real axis. The final result is derived from the sum of the residues at the identified poles, leading to the specified formula. Understanding the application of residues is emphasized as crucial for solving the problem.
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Homework Statement



Use residues to evaluate the improper integral:

\int^{\infty}_{- \infty} \frac{cos(x)dx}{(x^{2} + a^{2})(x^{2} + b^{2})} = \frac{\pi}{a^{2} - b^{2}} ( \frac{e^{-b}}{b} - \frac{e^{-a}}{a} )

Homework Equations



a>b>0

The Attempt at a Solution



If someone could just explain the concept of residues and how to apply them, that would be great. They give the answer, but you have to use residues to show how to get there. I've looked over my notes and through the book and I don't understand them at all.
 
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The integrand, cos(x)/(x^2+ a^2)(x^2+ b^2) is analytic everywhere except at x=\pm ai and x= \pm bi. If you take a contour integral where the contour is the x-axis from (-R, 0) to (R,0) together with the half circle from (R,0) to (R, 0), you should be able to show that the integral over the half circle is 0 so that the integral over that contour is just the integral from -\infty to \infty as you want. Of course, that is equal to 1/(2\pi i) times the sum of the residues at x= ai and x= bi which you should be able to caculate.
 
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