Evaluating Integral with CIF: Residue Calculation

[Mechdude]

Homework Statement

evaluate
$$\int_{-\infty}^{\infty} \frac{dx}{x^2 - \sigma^2}$$

Homework Equations

CIF: $$\int f(z) = 2 \pi i \times \sum Res f(z) |_a$$

The Attempt at a Solution

let
$$\int_c \frac {dz}{z^2 - \sigma^2} = \int_c \frac{1}{(z - \sigma)(z+ \sigma)}$$
with poles at $z = \pm \sigma$ but i take only those in the upper half plane, and thus the residue:
$$\lim_{ z \to \sigma} (z-\sigma) \frac{1}{( z- \sigma)( z + \sigma)} = \frac{1}{\sigma + \sigma} = \frac{1}{2 \sigma}$$
with the integral becoming :
$$2 \pi i \sum Res f(z) = 2 \pi i \frac{1}{2 \sigma} = \frac{\pi i }{\sigma}$$
please correct me!
im trying out some stuff in residues and their applications to evaluation of integrals, the question had these three parts:
1. let $\sigma \to \sigma +i \gamma$
2. let $\sigma \to \sigma - i \gamma$
3. Take the cauchy principle values.
i think i just did the third part, if i did it properly to begin with... but that aside, how do i begin on the other two parts?

 

[mighty2000]

Hi there,

Firstly, your solution for the third part is correct. Good job!

For the first part, you can use the same method as you did for the third part, but instead of taking the residue at z = \sigma, you take it at z = \sigma + i\gamma. This will give you a different value for the integral, which will also depend on \gamma. Then, you can take the limit as \gamma \to 0 to get the final result.

For the second part, you can use a similar approach, but this time, take the residue at z = \sigma - i\gamma. This will give you another value for the integral, which will also depend on \gamma. Again, you can take the limit as \gamma \to 0 to get the final result.

I hope this helps! Let me know if you have any other questions. Keep up the good work with evaluating integrals using residues. It's a very useful technique in many areas of science and mathematics.