Complex Integration-Residue Theory

In summary, the integral \int_{-\infty}^{+\infty} \frac{x^2+3}{(x^2+1)(x^2+4)} can be solved using the residue theorem, by choosing a contour that encloses all the singularities (in this case, \pm i and \pm 2i). The usual choice is a semicircle in the upper or lower half of the complex plane, but any contour that encloses the singularities can be used. The final answer is obtained by computing the residues at each singularity and taking the sum of them.
  • #1
jameson2
53
0
I'm doing this question, and I have the right answer, but I'm not sure is the method is right. Also, I have a couple of general questions about this method of integration.

Compute [tex] \int_{-\infty}^{+\infty} \frac{x^2+3}{(x^2+1)(x^2+4)}
[/tex]

I factorized the bottom line to find the singularities, which are at [tex] \pm i[/tex] and [tex]\pm 2i [/tex].

First question: in my book, there are examples where the contour is taken as a semicircle in the upper or lower halves of the complex plane. Is it ok to take a contour in the right or left hand side of the complex plane instead? I ask this because all of the singualrities in this question are along the imaginary axis, but say they were all along the real axis? My understanding is that the coutour must enclose the singularities, but a semicircle in the upper or lower halves of the complex plane would run through these (hypothetical) singularities. If it is right to take the the semicircle in the left or right planes, how does this change the way to answer the question, if at all? Or if this is not the right way to take the contour, how should it be done?

Back to the question:
Taking the contour as a semicircle in the upper half of the complex plane excludes the singularities at -i and -2i. So computing the residues at each of the enclosed singularities gives a residue of [tex]\frac{1}{3i}[/tex] at i and [tex] \frac{1}{12i}[/tex] at 2i.

Second question: why is a semicircle the usual choice? How is it any different to a contour that encloses maybe only 1, or 3 of the singularities in this question instead of 2? Each of these give diffenent final answers but I'm wondering what's so special about this particular contour, and it's answer?

Anyway, I filled in the two residues into the formula stating that the integral is equal to [tex]2\pi i[/tex] times the sum of the residues, and got the answer to the intgral as [tex]\frac{5\pi}{6}[/tex] which is what the answer in the book is.(Also, I don't trust this book as I've found a lot of mistakes, so this could easily be wrong.)

So basically I think I'm right with the answer, I'm just a little shaky on the method. Any tips are much appreciated.
Thanks
 
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  • #2
You can choose the contour in any way you like. You can apply the residue theorem only if the function is meromorphic inside the contour (i.e. analytic everywhere, except at some poles or essential singularities). Then you get a result of the form:

Integral over desired interval + integral over additional paths = 2 pi i sum of residues

Then, the integral over the additional paths can tend to zero, as is usually the case for the semi-circle.
 

Related to Complex Integration-Residue Theory

1. What is complex integration-residue theory?

Complex integration-residue theory is a mathematical concept that deals with calculating integrals of complex functions by using the residues, or singularities, of the function. It is a powerful tool used in complex analysis to evaluate integrals that would be difficult or impossible to solve using traditional methods.

2. How is complex integration-residue theory different from regular integration?

Regular integration deals with real-valued functions, while complex integration-residue theory deals with complex-valued functions. In complex integration, the path of integration can be taken in any direction, whereas in regular integration, the path is limited to the real axis. Additionally, complex integration takes into account the singularities of the function, which can greatly affect the value of the integral.

3. What are residues in complex integration-residue theory?

In complex integration-residue theory, residues are the values of the function at its singularities. They play a crucial role in calculating the value of complex integrals, as they can determine whether the integral is finite or infinite. Residues are calculated using the Cauchy Residue Theorem, which states that the integral of a function around a closed path is equal to the sum of the residues within the path.

4. How is complex integration-residue theory used in practical applications?

Complex integration-residue theory is used in a variety of fields, including physics, engineering, and finance. It is commonly used to solve problems involving electric circuits, fluid mechanics, and signal processing. In finance, it is used to calculate complex integrals in option pricing models. Additionally, it is used in the study of conformal mappings and the theory of complex variables.

5. What are some common techniques used in complex integration-residue theory?

Some common techniques used in complex integration-residue theory include the Cauchy Residue Theorem, the Laurent series expansion, and the method of partial fractions. These techniques allow for the calculation of residues and the evaluation of complex integrals. Other techniques such as the method of residues and the argument principle are also commonly used in more advanced applications of complex integration-residue theory.

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