Contour integration with a branch cut

In summary: Have you had a proper course, or part-course, on contour integration? Without that, this will be very difficult.
  • #1
mercenarycor
6
0

Homework Statement


-11 dx/(√(1-x2)(a+bx)) a>b>0

Homework Equations


f(z0)=(1/2πi)∫f(z)dz/(z-z0)

The Attempt at a Solution


I have absolutely no idea what I'm doing. I'm taking Mathematical Methods, and this chapter is making absolutely no sense to me. I understand enough to tell I'm supposed to do contour integration on this with a branch cut on the singularity, but actually doing it is another thing. Also, I have no idea what to do with the second term in the denominator. If you can explain this to me, I would be grateful; and please, try to dumb it down. I can't even figure out how to find residues.

The farthest I got was K=∫-11dx/(√(1-x2)(a+bx)) + lim r->∞0π reidθ/(√(1-r2ei2θ)(a+bre))
I stopped there, however, because I'm fairly certain I'm embarking on several hours of barking up the wrong tree.
 
Physics news on Phys.org
  • #2
Have you had a proper course, or part-course, on contour integration? Without that, this will be very difficult.

I could tell you that you're supposed to place a branch cut between ##x = \pm 1##, and use a "dog bone contour" (aka "dumbbell contour"), but that won't be much help if you don't know how to do easier contour integrals and compute basic residues. [Google for "dog bone contour" to see what this looks like.]
 
  • #3
strangerep said:
Have you had a proper course, or part-course, on contour integration? Without that, this will be very difficult.

The OP is taking a math methods course right now and learning about contour integration and complex analysis right now.
 
  • #4
mercenarycor said:
I have absolutely no idea what I'm doing. I'm taking Mathematical Methods, and this chapter is making absolutely no sense to me. I understand enough to tell I'm supposed to do contour integration on this with a branch cut on the singularity, but actually doing it is another thing. Also, I have no idea what to do with the second term in the denominator. If you can explain this to me, I would be grateful; and please, try to dumb it down. I can't even figure out how to find residues.

As strangerep suggested, you should probably go back to trying to do easier problems first and getting a handle on those before trying to tackle this one.

You could try finding a similar but simpler example in your textbook (one singularity, single branch cut) and asking questions about that first.
 

What is contour integration with a branch cut?

Contour integration with a branch cut is a technique used in complex analysis to evaluate integrals along a path in the complex plane that avoids a branch cut, a type of discontinuity in a complex-valued function. This technique involves deforming the contour of integration in a way that avoids the branch cut and allows for the use of Cauchy's integral theorem.

What is a branch cut?

A branch cut is a line or curve in the complex plane where a multi-valued function is discontinuous. This occurs when the function has different values depending on the path taken to approach a point on the branch cut. Branch cuts are often found in functions with logarithms, roots, and trigonometric functions.

Why is it important to use contour integration with a branch cut?

Contour integration with a branch cut allows us to evaluate integrals of complex functions that would otherwise be difficult or impossible to calculate using traditional methods. It also helps us to avoid discontinuities in the function, which can cause problems in the evaluation of the integral.

How do you choose the contour for contour integration with a branch cut?

The choice of contour for contour integration with a branch cut depends on the specific function being integrated and the location and nature of the branch cut. In general, the contour should be chosen in a way that avoids the branch cut and maintains the analyticity of the function.

What are some applications of contour integration with a branch cut?

Contour integration with a branch cut is used extensively in complex analysis and has numerous applications in physics, engineering, and other fields. It is particularly useful in the evaluation of integrals involving special functions, such as the error function, and in solving differential equations with complex solutions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
869
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
Replies
1
Views
611
  • Calculus and Beyond Homework Help
Replies
2
Views
986
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
Back
Top