Improper integrals: singularity on REAL axis (complex variab

In summary, the conversation discusses difficulties with solving improper integrals that have singularities on the real axis. The speaker has read books on the topic but has not fully understood the explanations. They then present an exercise involving a complex variable and two singularities, and discuss using a curve to solve it. The conversation concludes with the speaker seeking help on how to proceed with the exercise.
  • #1
juan.
8
0
Hello everyone! I'm having some troubles when I try to solve improper integrals exercises that have singularities on the real axis. I have made a lot of exercises where singularities are inside a semicircle in the upper half side, but I don't know how to solve them when the singularities are on the real axis.
I read some books but I think they are not very good explained (at least, I can't understood them).

This is the exercise:
[tex]\int_{-\infty}^{\infty} \frac{cos(2\pi x)}{x^2-1} dx[/tex]

Using complex variable, I have:
[tex]f(z) =\frac{exp(i2\pi z)}{z^2-1}[/tex]
so there are 2 singularities:
[tex]z_1 = -1[/tex] and [tex]z_2 = 1[/tex]
I use a curve [tex]C[/tex] that is holomorphic inside it, because both singularities are out of it. Of course, I can divide [tex]C[/tex] in 6 curves: [tex]C_R[/tex] that is the "roof" of the curve and, using Jordan's Lemma, I can prove that
[tex]\int_{C_R}^{ } \frac{exp(i2\pi z)}{z^2-1} dx = 0[/tex]
but I don't know what do I have to do now. I saw in some places they said that the Residue Theorem over the semicircle around the singularities was something like [tex]-i\pi\sum{}{}Res[f(z), z_k][/tex] but I didn't understand why.

I hope you can help me, because I don't know what can I do.
Thanks!
 
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  • #2
This situation is covered in detail in Ahlfors' excellent book "Complex Analysis". It goes somewhat like this:

Let [itex]C[/itex] be the curve consisting of a half circle in the upper half plane with center in 0 (radius R) and the real axis - except for two half circles in the lower half plane around -1 and +1 with radius ε. Now let us consider what happens around z = 1 (the case around z = -1 is handled the same way).

For ε<0.1 we can write [itex]f(z)=\frac{e^{2\pi i z}}{z^{2}-1}=\frac{0.5}{z-1}+A(z)[/itex], where [itex]A(z)[/itex] is analytic at z = 1. Now we have [itex]\int_{0}^{R}f(z)dz =\int_{0}^{1-\epsilon}f(z)dz + \int_{-\pi}^{0}0.5\cdot id\phi + \int_{1+\epsilon}^{R}f(z)dz[/itex]. Let ε→0, and we end up with [itex]\int_{0}^{R}f(z)dz +i \frac{\pi}{2}[/itex]. Do the same with z = -1, and you end up with the formula you "didn't understand".
 

1. What is an improper integral with a singularity on the real axis in complex variables?

An improper integral with a singularity on the real axis in complex variables is an integral where the function being integrated has a singularity (a point where the function is undefined) on the real axis. This means that the integral cannot be evaluated using traditional methods and requires special techniques to find the value.

2. How is a singularity on the real axis different from other types of singularities in complex variables?

A singularity on the real axis is different from other types of singularities in complex variables because it is located on the real axis, which is a line in the complex plane. This makes it easier to analyze and calculate the value of the integral compared to singularities located at other points in the complex plane.

3. What are some common examples of functions with singularities on the real axis in complex variables?

Some common examples of functions with singularities on the real axis in complex variables include the logarithmic function, the inverse trigonometric functions, and the square root function. These functions have singularities at specific points on the real axis, making them good examples for studying improper integrals with singularities on the real axis.

4. How do you calculate the value of an improper integral with a singularity on the real axis?

To calculate the value of an improper integral with a singularity on the real axis, you can use techniques such as contour integration or the residue theorem. These methods involve using complex analysis and manipulating the function to find a closed form expression for the integral, which can then be evaluated.

5. What are some real-world applications of improper integrals with singularities on the real axis in complex variables?

Improper integrals with singularities on the real axis in complex variables have various applications in physics, engineering, and other fields. For example, they can be used to model oscillatory systems, calculate the electric field around a point charge, or determine the stability of a system. These integrals also play a role in the study of complex functions and their behavior near singularities.

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