Evaluate this complex Integral

In summary, the conversation discusses using Cauchy's Integral Theorem to evaluate a given integral and determining the roots of the denominator in polar form. The solution involves using the residue formula for a second order pole and taking the limit, but there is a potential issue with getting 0 in the denominator.
  • #1
bugatti79
794
1

Homework Statement


Use Cauchy's Integral Theorem to evaluate the following integral

##\int_0^{\infty} \frac{x^2+1}{(x^2+9)^2} dx##

Homework Equations



Res ##f(z)_{z=z_0} = Res_{z=z_0} \frac{p(z)}{q(z)}=\frac{p(z_0)}{q'(z_0)}##

The Attempt at a Solution



I determine the roots of the denominator to be ##x=\pm 3i##.
How do I convert these into polar form. I know ##z=re^{i\theta}##

Do I need to convert these into ##z=e^{f(i\theta)}##?
 
Physics news on Phys.org
  • #2
I wouldn't convert this to polar form. You need to figure out how to write [itex]f[/itex] as a rational function like [itex]p/q[/itex] that you have above. Where [itex]p[/itex] is analytic and non-zero at [itex]z_0[/itex].
 
  • #3
bugatti79 said:

Homework Statement


Use Cauchy's Integral Theorem to evaluate the following integral

##\int_0^{\infty} \frac{x^2+1}{(x^2+9)^2} dx##

Homework Equations



Res ##f(z)_{z=z_0} = Res_{z=z_0} \frac{p(z)}{q(z)}=\frac{p(z_0)}{q'(z_0)}##

The Attempt at a Solution



I determine the roots of the denominator to be ##x=\pm 3i##.
How do I convert these into polar form. I know ##z=re^{i\theta}##

Do I need to convert these into ##z=e^{f(i\theta)}##?

Robert1986 said:
I wouldn't convert this to polar form. You need to figure out how to write [itex]f[/itex] as a rational function like [itex]p/q[/itex] that you have above. Where [itex]p[/itex] is analytic and non-zero at [itex]z_0[/itex].

I realize the pole is second order therefore using the residue forumla for second order pole ##z=3i## we get

##=lim_{z \to 3i} \frac{d}{dz} \frac{ z^2+1}{(z^2+9)^2} = \frac{-2z(z^2-7)}{(z^2+9)^3} |_{z=3i}##

but we are going to get 0 in the denominator here..?
 

Related to Evaluate this complex Integral

1. What is a complex integral?

A complex integral is a mathematical concept that involves the integration of complex-valued functions over a certain range or region. It is an extension of the concept of a real-valued integral and is used in various fields of mathematics and physics to solve problems involving complex numbers.

2. Why is it important to evaluate complex integrals?

Evaluating complex integrals is important because it allows us to solve problems in complex analysis, which is a crucial tool in many areas of mathematics and physics. It also helps us understand the behavior of complex functions and their relationship with real functions.

3. How do you evaluate a complex integral?

To evaluate a complex integral, you need to follow the same rules as evaluating a real integral, but with the added consideration of complex numbers. This involves using integration techniques such as substitution, integration by parts, and partial fractions, and applying the properties of complex numbers.

4. What are some common techniques for evaluating complex integrals?

Some common techniques for evaluating complex integrals include contour integration, residue theorem, Cauchy's integral formula, and the method of steepest descent. These techniques take advantage of the properties of complex numbers and allow us to simplify complex integrals into more manageable forms.

5. Can complex integrals have real values as their result?

Yes, complex integrals can have real values as their result. This happens when the function being integrated is analytic, meaning it is differentiable at every point in its domain. In this case, the integral can be evaluated using real integration techniques and will result in a real number.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
279
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
803
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Topology and Analysis
Replies
14
Views
824
  • Topology and Analysis
Replies
2
Views
950
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
32
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Back
Top