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Complex integration over a square contour

  1. Nov 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Let [itex]\Gamma[/itex] be the square whose sides have length 5, are parallel to the real and imaginary axis, and the center of the square is i. Compute the integral of the following function over [itex]\Gamma[/itex] in the counter-clockwise direction. You must use two different methods to solve the problem in order to receive full credit. Show all work.

    [itex]\frac{e^{z}}{(z^{2} + 2*z + 1)}[/itex]

    2. Relevant equations

    Residue for pole of order 2: [itex]Res(f(z),z_{0}) = \stackrel{lim}{_{z \rightarrow z_{0}}}\frac{d}{dz}(z-z_{0})^{2}*f(z)[/itex]

    Residue Theorem: [itex]\oint_{C} f(z)dz = i2\pi\sum(Residues) [/itex]

    3. The attempt at a solution

    When we rewrite the equation as: [itex]\frac{e^{z}}{(z+1)^{2}}[/itex] we can see that there is a pole of order 2 at z = -1. Since the singularities only have Real components, the function is meromorphic and the poles are isolated. The [itex]\Gamma[/itex] contour is analytic everywhere else on and inside [itex]\Gamma[/itex] so the poles are removable and we can use the Residue theorem to perform the integration.

    First calculate the Residues: [itex]Res(f(z),z_{0}) = \stackrel{lim}{_{z \rightarrow -1}}\frac{d}{dz}((z+1)^{2}*\frac{e^{z}}{(z+1)^{2}})[/itex] = [itex]\frac{1}{e}[/itex]

    Now we can use the Residue theorem to perform the integration:

    [itex]\oint_{C} \frac{e^{z}}{(z+1)^{2}} = i2\pi\frac{1}{e} = i2.3115 [/itex]

    ----

    I am absolutely stuck on finding a second method to solve this problem! I thought maybe I could parametrize it but no symbolic integration exists for this function. I tried integration by parts, the int() function in MATLAB and my TI-89 all without success. The quad() function in MATLAB succeeded but unfortunately my prof is not allowing numeric integration methods..

    Please help!
     

    Attached Files:

  2. jcsd
  3. Nov 23, 2011 #2

    Dick

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    Science Advisor
    Homework Helper

    Doing contour integrals isn't only "other way" around. And I'll agree that looks hard. How about if you shift the variable of integration to z'=z+1. Why can you do this? Now you've got exp(z'-1)/(z')^2. That's (1/e)*exp(z')/(z')^2. Now expand exp(z') in a power series and look for the 1/z' term. Is that enough of an "other way"?
     
  4. Nov 24, 2011 #3


    You know, there's another singular point right? Why not integrate "around" it and compute it's residue and equate the integral over the box to the integral around this singular point.
     
  5. Nov 24, 2011 #4
    OK I just heard about the following hint that came from the prof:

    set [itex]z = ε e^{it}[/itex]

    Maybe this is an alternate way to compute the residue method? I cannot find this in my textbooks
     
  6. Nov 24, 2011 #5
    . . . bet a dollar he wants you to use the residue at infinity for the second approach. I don't know about you, but that's what I'm turning in for the second one.
     
  7. Nov 24, 2011 #6
    I'm with you, but I am unfamiliar with the technique. How is this accomplished?
     
  8. Nov 24, 2011 #7
    If a function is analytic in the finite plane except for a finite number of singular points inside a closed contour (like yours is) then you can use the residue at infinity to evaluate the integral around a closed contour:

    [tex]\oint_C f(z)dz=2\pi i\; \text{Res}\;\left[\frac{1}{z^2}f\left(\frac{1}{z}\right),0\right][/tex]

    You can do that right? Just plug it all in, make that 1/z substitution, then make like you're taking the residue at z=0. That's the function at the value of 1/z ok? Gonna' get a little involved with a double product of infinite series to figure out what the residue of e^(1/z) is but just do the first few, extract the 1/z terms, notice the sequence, and you already know what the answer is so you'll know what that infinite sum is.
     
  9. Nov 25, 2011 #8
    Thanks, I will give this a shot
     
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