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Complex Integration / Residue Problems

  1. Sep 30, 2009 #1

    G01

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    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    Work to come, still editing post.

    3. 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?
     
    Last edited: Sep 30, 2009
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  3. Sep 30, 2009 #2

    Dick

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    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.
     
    Last edited: Sep 30, 2009
  4. Sep 30, 2009 #3

    Dick

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    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.
     
    Last edited: Sep 30, 2009
  5. Sep 30, 2009 #4

    G01

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    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:
     
  6. Oct 1, 2009 #5

    Dick

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    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).
     
  7. Oct 1, 2009 #6

    G01

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    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!
     
  8. Oct 1, 2009 #7

    Dick

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