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Cauchys integral formula.

  1. Sep 24, 2010 #1
    If i have a closed pathintegral of the form:
    (sin(z)+3cos(z) + 3e^z)/((z-(pi/2)^2)
    How is cauchys integral formula applicable? If I split the integral into partial fractions wont i still get A/(z-pi/2) + B/(z-pi/2)^2, the B part of which wont be applicable to cauchys formula since the denominator is still squared.

    In short, how do i apply cauchys formula to integrals with a denominator with a power higher than 1.
     
  2. jcsd
  3. Sep 24, 2010 #2

    HallsofIvy

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    Apparently, by "Cauchy's formula" you are referring to
    [tex]f(a)= \frac{1}{2\pi i}\cint_\gamma \frac{f(z)}{z- a} dz[/tex]

    However, it can be easily generalized to
    [tex]f^{(n)}(a)= \frac{n!}{2\pi i}\cint_\gamma \frac{f(z)}{(z- a)^{n+1}}dz[/tex]
    where the left sided is the nth derivative of f evaluated at z= a.

    In your case, that integral is equal to the derivative of [itex]f(z)= sin(z)+ 3cos(z)+ 3e^z[/itex] evaluated at [itex]z= \pi/2[/itex], multiplied by [itex]2\pi i[/itex].
     
    Last edited by a moderator: Sep 25, 2010
  4. Sep 25, 2010 #3
    ...multiplied by [itex]2\pi i[/itex].
     
  5. Sep 25, 2010 #4

    HallsofIvy

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    Right, thanks. I will now edit my post so I can pretend I didn't make that foolish mistake!
     
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