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Integrals of Complex Functions

  1. Mar 4, 2014 #1
    1. The problem statement, all variables and given/known data
    What is the integral from negative infinity to positive infinity of the following functions?

    a) f(z) = [itex]\frac{e^{-i5z}}{z^{2}+1}[/itex]

    b) f(z) = [itex]\frac{e^{-i5z}}{z^{2}-1}[/itex]

    c) f(z) = [itex]\frac{1}{π}[/itex][itex]\frac{a}{z^{2}+a^{2}}[/itex]

    d) f(z) = [itex]e^{\frac{-(z-ia)^{2}}{2}}[/itex]

    e) f(z) = [itex]\frac{sinz}{z}[/itex]

    2. Relevant equations

    All my professor showed was an example, which would be (c) above

    [itex]\frac{1}{π}[/itex][itex]\int\frac{a}{z^{2}+a^{2}}[/itex]dz
    [itex]\frac{1}{π}[/itex][itex]\int\frac{1}{2ai}[/itex]([itex]\frac{1}{z-ia}[/itex]-[itex]\frac{1}{z+ia}[/itex])dz
    [itex]\frac{1}{2iπ}[/itex][itex]\int\frac{1}{z-ia}[/itex]-[itex]\frac{1}{z+ia}[/itex] dz=1
    dz[itex]\Rightarrow[/itex][itex]Re^{iθ}idθ[/itex]

    limits of integration are now from 0 to π
    [itex]\int\frac{1}{R^{2}e^{2iθ}+a^{2}}[/itex][itex]Re^{iθ}iθdθ[/itex]
    [itex]R^{2}[/itex][itex]\rightarrow[/itex]∞
    [itex]\oint[/itex]f(z)dz=0
    f(z) is analytic
    f(z) = [itex]\frac{1}{2πi}[/itex]

    and this is where the class time expired and he said "you can figure out the rest"

    3. The attempt at a solution
    I have no clue where to start with these. The class is Mathematical Methods of Physics and this year there weren't enough students in the class so they offered it as an independent study. The professor that "observes" our work didn't tell us how to do these or where to start. I'm just looking for guidance on what to do, that an idiot can understand. Often I read threads on here and the way some of you convey your knowledge is out of my league. I would appreciate any help! Thanks
     
  2. jcsd
  3. Mar 4, 2014 #2
    So the general approach is to use the Cauchy Integral Formula and Cauchy Integral Theorem. The theorem has a wiki page, as does the formula: http://en.wikipedia.org/wiki/Cauchy's_integral_formula.

    You are basically integrating about the boundary of a disk, where the values contained in the disk are equal to the values on the boundary. This should seem familiar, as it is the basis for Stokes' Theorem, and consequently Green's Theorem.

    If you look at the Cauchy Integral Theorem and its proof, you can see ∫f(z)dz = 0. This is to say that for any closed path, no work is done.

    My suggestion would be to compute ∫(1/z) dz on the unit circle, draw a graph of this (it will be a circle of radius 1 with a hole at R = 0), and examine its physical significance. From here, compare with the Cauchy Integral Formula and Theorem, and understand why this works.

    Your reasoning behind factoring these denominators into a form of 1/(z-a) is necessary because it is just a translation of the solution to ∫(1/z) dz.
     
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