Complex Integration / Residue Problems

In summary: Goodnight everyone.In summary, G01 attempted to solve three homework problems, but failed on the first two. He is lost on how to compute the residues for functions with denominators that are easy to factor. He plans to work on this problem and get back to the community.
  • #1
G01
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Homework Statement



Hi everyone. I'm currently taking a graduate math physics course and complex integrals are beating the crap out of me. Some of my questions may be relatively basic. Forgive me, I'm trying to teach myself and am regretting not taking a course on complex analysis as an undergrad. I have several problems. I'll list them all at once and mark the corresponding work for each separate problem below.

Thank you in advance. Any help is greatly appreciated!

1. Solve the following integral twice. Once by closing the contour in the Upper Half Plane, and once by closing in the Lower Half Plane:

[tex]I=\int_{-\infty}^{\infty} \frac{1}{x^3 + i} dx[/tex]

2. Solve: [tex]I=\int_0^{\infty} \frac{1}{1+x^3} dx[/tex]

HINT for #2: Use the following contour in the complex plane: 1. The real axis from 0 to R. 2. The arc, [itex]\theta=0..\frac{2\pi}{3}[/itex]. 3. The line from the end of that arc back to the origin. Take the limit as R approaches infinity.

3. Show [tex]I=\int_{-\infty}^{\infty}\frac{\sin(x)}{x}dx=\pi[/tex] NOT by computing a principle value, but instead by letting [itex]x-0\,\rightarrow\, x-(0+i\epsilon)\,}[/itex] and then compting the integral and taking the limit as epsilon approaches 0.

Homework Equations



Work to come, still editing post.

The Attempt at a Solution



1.

Work with the complex integral:

[tex]I_c=\oint\frac{dz}{z^3+i}[/tex]

Close the contour in the UHP:

[tex]I_c=\lim_{R\rightarrow\infty}\int_{-R}^{R}\frac{dx}{x^3 + i}+\lim_{R\rightarrow\infty}\int_{C_R}\frac{dz}{z^3+i}[/tex]

where [itex]C_R[/itex] is the half circle of radius R in the UHP.

Since [itex]z^3=R^3\exp{3i\theta}[/itex], the second integral goes to zero in the limit. Then, using the residue theorem:

[tex]I_c=2\pi i \Sigma (Residues)=\lim_{R\rightarrow\infty}\int_{-R}^{R}\frac{dx}{x^3 + i}[/tex]

If my work is correct up until here, then I think my only problem is with computing the residues. Up until now I have only needed to compute residues for functions with denominators that easily factored, making the residue computation easy. Here, I'm lost.

I know there are zeros when z=-i. The zeroes are all along the unit circle at [itex] \theta=\frac{\pi}{6},\frac{\pi}{2}, \frac{5\pi}{6},...[/itex] (Is this correct?)

How do I compute find the residues at these points in the easiest way? Find the Laurent series? I'm sorry if I'm missing something stupid. I am just pressed for time and could be missing a multitude of things.

2.

Dick, Way too tired tonight and left this problem in my office! I'll post this work tomorrow morning! Sorry, I know you need something to do.

3.

So, let [itex]x-0\,\rightarrow\, x-(0+i\epsilon)\,}[/itex] :

Therefore:

[tex]I=\lim_{\epsilon\rightarrow0}\int_{-\infty}^{\infty}\frac{\sin(x)}{x-i\epsilon}dx[/tex]

[tex]I=\frac{1}{2i}\lim_{\epsilon\rightarrow0}\int_{-\infty}^{\infty}\frac{e^{ix}dx}{x-i\epsilon}-\frac{1}{2i}\lim_{\epsilon\rightarrow0}\int_{-\infty}^{\infty}\frac{e^{-ix}dx}{x-i\epsilon}[/tex]

Now if I let epsilon go to zero now, I have an integral that we discussed in class. I know the first integral is equal to [itex]i\pi[/itex] and the second [itex]-i\pi[/itex].

This simplifies to the correct result, but technically I used a principle value to compute the exponential integrals, which was against the rules of the problem. Is there another easy way to find those integrals?
 
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  • #2
Hi G01! I can see the texing is going to take a while. But what went wrong with the first one? That should be easy. There are two poles in the lower half plane and one in the upper. The sum of all the residues is zero. But one contour is clockwise and the other is counterclockwise. You have to get the same answer either way. Sorry, I couldn't wait. I'm not doing anything else right now.
 
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  • #3
z=(-i) isn't a root of z^3+i. z=i is. z^3+i=(z-exp(-i*pi/6))*(z-exp(i*pi/2))*(z-exp(i*pi*(7/6))). You haven't got the angles right. But you can factor it. Just like any other polynomial. And, yes, the roots are on the unit circle at an angular distance of 2pi/3 from each other.
 
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  • #4
Dick said:
z=(-i) isn't a root of z^3+i. z=i is. z^3+i=(z-exp(-i*pi/6))*(z-exp(i*pi/2))*(z-exp(i*pi*(7/6))). You haven't got the angles right. But you can factor it. Just like any other polynomial. And, yes, the roots are on the unit circle at an angular distance of 2pi/3 from each other.

Ahh yeah. I see now. I think I have some mental block on complex algebra! I'll work on this and get back to you. Sorry, but it is late here and I'm about to crash. Sorry to keep you waiting on that third problem!:smile:
 
  • #5
For number 3, use the residue theorem to compute the two integrals. There's single pole at i*epsilon (take epsilon>0). The only tricky bit is that one contour has to be closed in the upper half plane and one in the lower half plane because of the sign of the iz in the exponentials. (Think what happens for z's with large imaginary parts).
 
  • #6
Hey, Dick! Sorry for the wait. More bad news for you if you wanted another problem:

I figured out the 2nd problem so I won't post it. You hints helped with the other two, as well! Thanks alot!
 
  • #7
G01 said:
Hey, Dick! Sorry for the wait. More bad news for you if you wanted another problem:

I figured out the 2nd problem so I won't post it. You hints helped with the other two, as well! Thanks alot!

Dang. I was looking forward to 2. You are so easy to help.
 

1. What is complex integration?

Complex integration is a mathematical technique used to calculate the integral of complex functions. It involves integrating complex-valued functions over a closed curve in the complex plane.

2. What are the applications of complex integration?

Complex integration has many applications in physics, engineering, and mathematics. It is used in the study of electric and magnetic fields, fluid dynamics, and signal processing. It also has applications in the evaluation of certain real integrals.

3. What is a residue in complex integration?

A residue in complex integration refers to the value of a function at a singular point inside a contour. It is used to evaluate complex integrals by using Cauchy's residue theorem.

4. How is the residue calculated?

The residue of a complex function at a singular point can be found by using the formula Res(f,z0) = limz→z0(z-z0)f(z), where z0 is the singular point. This formula can be derived from the Laurent series expansion of the function around the singular point.

5. What is the significance of residues in complex integration problems?

The calculation of residues is essential in solving complex integration problems. It allows us to evaluate complex integrals using the residue theorem, which states that the integral of a function over a closed contour is equal to 2πi times the sum of the residues of the function inside the contour.

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