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

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In summary, the improper integral \int^{\infty}_{- \infty} \frac{cos(x)dx}{(x^{2} + a^{2})(x^{2} + b^{2})} can be evaluated using residues. The concept of residues involves finding the singular points of the integrand and calculating the sum of the residues at those points. By using a contour integral along the x-axis and a half circle, the integral over the half circle can be shown to be 0, simplifying the calculation. The final result is \frac{\pi}{a^{2} - b^{2}} ( \frac{e^{-b}}{b} - \frac{e^{-a}}{a}
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Homework Statement



Use residues to evaluate the improper integral:

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

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, [itex]cos(x)/(x^2+ a^2)(x^2+ b^2)[/itex] is analytic everywhere except at [itex]x=\pm ai[/itex] and [itex]x= \pm bi[/itex]. 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 [itex]-\infty[/itex] to [itex]\infty[/itex] as you want. Of course, that is equal to [itex]1/(2\pi i) times the sum of the residues at x= ai and x= bi which you should be able to caculate.
 

1. What is an improper integral?

An improper integral is an integral where one or both of the integration limits are infinite or the integrand is not defined at certain points within the integration limits.

2. How do residues help in evaluating improper integrals?

Residues are used to evaluate improper integrals by converting them into a complex contour integral, which can then be solved using the residue theorem.

3. What is the residue theorem?

The residue theorem states that the value of a contour integral around a closed loop is equal to the sum of the residues of the function enclosed by the loop.

4. How do I find the residues of a complex function?

To find the residues of a complex function, you can use the formula Res(f, z0) = limz→z0 (z-z0)f(z), where z0 is a singular point of the function f(z).

5. Are there any limitations to using residues to evaluate improper integrals?

Yes, there are certain conditions that must be met in order for the residue theorem to be applicable, such as the function being analytic and the contour being closed and simple. Additionally, the function must have simple poles within the contour. If these conditions are not met, other methods may need to be used to evaluate the improper integral.

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