Complex Integration-Residue Theory

  • Context: Graduate 
  • Thread starter Thread starter jameson2
  • Start date Start date
  • Tags Tags
    Complex Theory
Click For Summary
SUMMARY

The discussion centers on the application of Complex Integration-Residue Theory to compute the integral \(\int_{-\infty}^{+\infty} \frac{x^2+3}{(x^2+1)(x^2+4)}\). The user correctly identifies the singularities at \(\pm i\) and \(\pm 2i\) and computes the residues at the enclosed singularities in the upper half-plane, yielding a final answer of \(\frac{5\pi}{6}\). The user questions the choice of contour and its implications, confirming that the semicircular contour is standard due to its ability to exclude certain singularities while ensuring the integral over additional paths tends to zero.

PREREQUISITES
  • Understanding of Complex Analysis concepts, specifically the Residue Theorem.
  • Familiarity with contour integration techniques in the complex plane.
  • Knowledge of meromorphic functions and their singularities.
  • Ability to compute residues at poles in complex functions.
NEXT STEPS
  • Study the Residue Theorem in detail, focusing on its applications in complex integration.
  • Learn about different contour choices in complex analysis and their effects on integral evaluation.
  • Explore examples of contour integration that involve singularities on the real axis.
  • Investigate the behavior of integrals over semicircular contours as the radius approaches infinity.
USEFUL FOR

Students and professionals in mathematics, particularly those specializing in complex analysis, as well as anyone looking to deepen their understanding of integration techniques involving residues.

jameson2
Messages
42
Reaction score
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 \int_{-\infty}^{+\infty} \frac{x^2+3}{(x^2+1)(x^2+4)}<br />

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

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 \frac{1}{3i} at i and \frac{1}{12i} 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 2\pi i times the sum of the residues, and got the answer to the intgral as \frac{5\pi}{6} 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
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K