Still having trouble solving integrals via residues

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In summary, the conversation discusses how to solve closed integrals around a given circle, specifically \oint\frac{dz}{sinz} around |z-6|=4 and \oint \frac{sinz}{sinh^{2}z}dz around |z|=3. It is determined that the integral in the first case is equal to zero because the singularities (n*pi) are not within the circle, while in the second case, the integral can be solved by finding the residues of the singularities inside the circle. The conversation also touches on the confusion regarding the poles of 1/(sinz) and 1/(sinh^2(z)) according to Wolfram, with the conclusion being that the poles for
  • #1
saybrook1
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Homework Statement


Hi guys, I don't quite understand how to solve closed integrals "around a given circle." I was given [tex] \oint\frac{dz}{sinz} [/tex] around [tex]|z-6|=4 [/tex] and said that the integral is equal to zero because the singularity [tex] n\pi i [/tex] is not within the circle. Is this correct? Also I need to solve [tex]\oint \frac{sinz}{sinh^{2}z}dz [/tex] around [tex]|z|=3[/tex]. In this case do I just need to solve for the residues of the singularities which are inside of this circle? Or am I going about this completely wrong? Any help would be greatly appreciated.

Homework Equations


All are given above.

The Attempt at a Solution


I have attempted to solve the second integral by expanding sinhz and looking for the residues of this function within the given circle. I'm not sure if I've done that right or not. Thanks in advance for any help or a point in the right direction; if anyone knows where I could find some similar examples that would also be awesome.
 
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  • #2
It seems like the second integral will also be zero since the singularity is n*pi*i which is also out of the circle for any value of n other than zero? What am I missing here?
 
  • #3
saybrook1 said:

Homework Statement


Hi guys, I don't quite understand how to solve closed integrals "around a given circle." I was given [tex] \oint\frac{dz}{\sin z} [/tex] around [tex]|z-6|=4 [/tex] and said that the integral is equal to zero because the singularity [tex] n\pi i [/tex] is not within the circle. Is this correct?
Why do you think the singularities are on the imaginary axis?

Also I need to solve [tex]\oint \frac{\sin z}{\sinh^{2}z}dz [/tex] around [tex]|z|=3[/tex]. In this case do I just need to solve for the residues of the singularities which are inside of this circle? Or am I going about this completely wrong? Any help would be greatly appreciated.
Yes, you only care about the poles inside of the contour.
 
  • #4
vela said:
Why do you think the singularities are on the imaginary axis?Yes, you only care about the poles inside of the contour.
Okay right, so the singularities in the first case are actually n*pi and there are three of them within that circle. Thanks a bunch, I've got that problem down now. For the second problem the singularities are also n*pi but they are second order. So I can either do kind of a nasty derivative or try and expand sinz in powers of (1-n*pi)? Any idea about how I would go about this expansion in order to find the necessary residues? Thanks a ton!
 
  • #5
saybrook1 said:
Okay right, so the singularities in the first case are actually n*pi and there are three of them within that circle. Thanks a bunch, I've got that problem down now. For the second problem the singularities are also n*pi but they are second order. So I can either do kind of a nasty derivative or try and expand sinz in powers of (1-n*pi)? Any idea about how I would go about this expansion in order to find the necessary residues? Thanks a ton!
Also for some reason, wolfram tells me that the poles of 1/(sinz) are 2*n*pi and 2*n*pi +pi. And for 1/(sinh^2(z)) it gives poles at 2*n*pi*i and 2*n*pi*i +i*pi. Are these correct? Sorry for all of the questions and thanks again.
 
  • #6
You need to be able to figure out where the poles are. What's the condition for the integrand to have a pole at some point? When is it met for this particular integrand?
 
  • #7
The denominator must equal zero at that point.
 
  • #8
I'm just confused because wolfram tells me that the zero's of sinhz are n*pi*i. I'm not sure as to why when I ask for the poles it adds a factor of 2 and a phase. Shouldn't that be irrelevant? The factor of 2 still makes it go to zero. I think I will just stick with saying the poles for 1/sinhz are at n*pi*i and the poles for 1/sinz are at n*pi.
 

Related to Still having trouble solving integrals via residues

1. What are residues in integration?

Residues in integration refer to the complex numbers that are obtained by evaluating a function at its singular points (poles) within a closed contour. They are used to evaluate integrals that cannot be solved using traditional methods.

2. How do you identify the poles of a function?

To identify the poles of a function, you must first factor the function into its numerator and denominator. The poles will be the values of the variable that make the denominator equal to zero.

3. Can all integrals be solved using residues?

No, not all integrals can be solved using residues. The function must have a finite number of singular points within the contour and the integral must be over a closed contour. Additionally, the function must be analytic (differentiable) on and within the contour.

4. How do you use residues to solve an integral?

To use residues to solve an integral, you must first identify the poles of the function within the closed contour. Then, you must calculate the residue at each pole. Finally, you can use the Residue Theorem to evaluate the integral by summing the residues of the poles within the contour.

5. Are there any limitations to using residues in integration?

Yes, there are limitations to using residues in integration. As mentioned before, the function must have a finite number of singular points within the contour, and the integral must be over a closed contour. Also, the function must be analytic (differentiable) on and within the contour, and the contour must be simple (no self-intersections).

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