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Contour Integral

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data

    dvj2mo.jpg

    3. The attempt at a solution

    We can parametrise the contour [itex]\gamma[/itex] (the positively oriented unit circle) by [itex]\gamma(t) = e^{it}[/itex] for [itex]t \in [0, 2\pi ][/itex]

    So by the definition of a contour integral

    10y07kp.jpg

    [itex]\displaystyle I = \frac{1}{2\pi i} \int^{2\pi}_0 \frac{2e^{it}}{e^{2it} + w^2} ie^{it} \; dt[/itex]

    [itex]\displaystyle \;\;\;= \frac{1}{\pi} \int^{2\pi}_0 \frac{e^{2it}}{e^{2it} + w^2} \; dt[/itex]

    How do I evaluate this?
     
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  3. Jan 21, 2012 #2

    HallsofIvy

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    Let [itex]y= e^{2it}+ w^2[/itex]
     
  4. Jan 21, 2012 #3
    In fact, noticing that the numerator of the integrand of [tex]\displaystyle I = \frac{1}{2\pi i} \int^{2\pi}_0 \frac{2ie^{2it}}{e^{2it} + w^2} \; dt[/tex] is the derivative of the denominator: [tex]\displaystyle I = \frac{1}{2\pi i} \left[ \log(e^{2it} + w^2) \right]^{2\pi}_0 = \frac{1}{2\pi i} [\log(1+w^2) - \log(1+w^2)] = 0[/tex]

    But what is this thing about the complex parameter [itex]w[/itex] that the question asks about? Will the result ever not be 0?
     
  5. Jan 21, 2012 #4

    Dick

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    Use the residue theorem! Don't mess with antiderivatives if there are singularities around.
     
  6. Jan 21, 2012 #5
    OK, so turning to the residue theorem:

    1fa3qx.jpg

    The singularities of [itex]\displaystyle f(z) = \frac{2z}{z^2+w^2}[/itex] are [itex]\pm iw[/itex].

    [itex]\text{res}(f,\pm iw) = 1[/itex].

    The contour [itex]\gamma[/itex] is a positively oriented unit circle.

    So if [itex]w \in [-1,1][/itex] then the winding number [itex]n ( \gamma , \pm iw ) =1[/itex]. If [itex]w \notin [-1,1][/itex] then the winding number [itex]n ( \gamma , \pm iw ) =0[/itex].

    So [tex]2\pi i \times I = 2\pi i \left( \pm iw \times 1 \right)[/tex] for [itex]w\in [-1,1][/itex] and [itex]I=0[/itex] otherwise?
     
  7. Jan 21, 2012 #6

    Dick

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    The residues aren't [itex]\pm iw[/itex]. They are 1, as you said. Fix your final answer. And they want you to consider w as a complex parameter, describing it using interval notation is for real numbers. And you might want to think about the case |w|=1 separately. But that's a good start. Just needs a little fixing.
     
    Last edited: Jan 21, 2012
  8. Jan 21, 2012 #7
    The singularities are [itex]\pm iw[/itex] ; the residues are 1 - correct?

    [itex]I = (1 \times 1 + 1 \times 1) = 2[/itex] if [itex]-1 < \text{Re}(w) < 1[/itex] and [itex]-i <\text{Im}(w) < i[/itex].

    and [itex]I=0[/itex] if [itex]\text{Re}(w) < -1[/itex] or [itex]\text{Re}(w) > 1[/itex] or [itex]\text{Im}(w) < -i[/itex] or [itex]\text{Im}(w) > i[/itex]

    If [itex]|w|=1[/itex] is the winding number still 1?
     
  9. Jan 21, 2012 #8

    Dick

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    So the integral is 4*pi*i if w inside of the contour right? The region you are describing sounds like a square to me. And the contour is round. If |w|=1 then your contour integral is singular, isn't it? The poles will be on the contour. You can still assign the integral a value if you interpret it as a Cauchy principal value, if you covered that.
     
  10. Jan 21, 2012 #9
    I think I=2 as that factor of [itex]2\pi i[/itex] gets cancelled by that of the original integral. Why is the integral 'singular' if [itex]|w|=1[/itex]? And yes, I know the contour is a circle but I'm having trouble describing it - I now see I was describing a square before!
     
  11. Jan 21, 2012 #10

    Dick

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    Right. The original definition cancels the 2*pi*i. The integral is singular if |w|=1 because then iw is on the unit circle. You'll get a zero in the denominator as you integrate. What's wrong with |w|<1 as a description of the interior of the unit circle?
     
  12. Jan 21, 2012 #11
    I don't know why I didn't think to describe [itex]w[/itex] like that. Does this description of [itex]I[/itex] look OK:

    [tex]I = \left\{ \begin{array}{lr}
    2 & : \;|w|< 1\\
    0 & : \;|w|>1\\
    \text{undefined} & : \; |w|=1
    \end{array}
    \right.[/tex]
     
  13. Jan 21, 2012 #12

    Dick

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    Looks fine. Like I said before, if you take the principal value sense of the integral you could show you get 1 for |w|=1, but if you haven't covered that, don't worry about it.
     
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