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Integral_c tan(z/2) / (z-x_0)^2 dz ~ it's answer has a sec^2 in it how?

  1. Apr 21, 2007 #1
    1. The problem statement, all variables and given/known data

    I have this question in my text and the answer has a sec^2 in it, and according to what I know, there is no where that you need to use derivatives, but maybe I dont understand the formula? Up until now all the answers have just been regular functions, not derivatives.

    ANyways, here is what i've done

    Question is

    C is a positively oriented boundary of the square whose sides lie on lines x & y = +/- 2 so eval

    Int_c tan(z/2) / (z-x_0)^2 dz

    The answer in the text is i*pi*sec^2(x_0)/2


    2. Relevant equations

    The Cauchy Integral formula I'm using is as follows
    f(z_0) = 1/ (2 * pi * i) ⌠_c f(x) / (z-z_0) dz

    3. The attempt at a solution


    modified the intergral so I could use the above formula

    ⌠_c tan(z/2) / (z-x_0)(z+x_0) dz

    so f(x) = tan(z/2) / (z+x_0)

    and z_0 = x_0
    so f(z_0) = tan(x_0/2) / (2x_0)

    Using Cauchy Integral I get
    tan(x_0/2) / (2x_0) * (2 * pi * i)

    so the 2's cancel and I get

    p * i * tan(x_0 / 2) / x_0

    What don't I know about sec^2 (x) that makes me not be able to get

    the answer?
     
  2. jcsd
  3. Apr 21, 2007 #2

    Gib Z

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    Homework Helper

    Perhaps the sec^2 came up naturally and not precisely as the derivative of tan. Or perhaps they did some integration by parts.
     
  4. Apr 21, 2007 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Cauchy's integral formula is more general than that:
    [tex]f^{(n)}= \frac{n!}{2\pi i}\int \frac{f(z)dz}{(z- z_0)^{n+1}}[/tex]
    where the integral is around a closed curve with z0 in its interior.

    Since your integral has a square in the denominator, look at the first derivative of tan(z) evaluated at z0.
     
    Last edited: Apr 21, 2007
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