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Verify that my contour integral is correct

  1. May 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate [itex]\oint[/itex][itex]\frac{1}{a-cos\theta}[/itex] d[itex]\theta[/itex] from [0:2[itex]\pi[/itex]]

    2. Relevant equations
    Method of contour integration.


    3. The attempt at a solution
    I make the substitution z=exp(i[itex]\theta[/itex]) [itex]\Rightarrow[/itex]
    giving me the integral:

    [itex]\oint[/itex]1/(a-½(z+z-1)) dz/iz from [0:2[itex]\pi[/itex]]
    And rearranging:
    2/i [itex]\oint[/itex]1/(-z2+2az-1) dz from [0:2[itex]\pi[/itex]]
    The quadratatic equation in the denominator has the roots z0 = a ± √(a2-1). Only the one with negative sign is inside the unit circle for a>1 (actually, how can you know that it will be that for all a>1?) which gives:

    Pole at z0 = a - √(a2-1)

    So we can write f(z) as:

    f(z) = 2i/((z-a + √(a2-1))(z-a - √(a2-1)))

    And since it's a simple pole we're dealing with the residue can be found as:

    limz->z0{(z-z0)f(z)} = limz->a - √(a2-1){(z-a+√(a2-1) 2i/((z-a + √(a2-1)(z-a - √(a2-1))} = limz->a - √(a2-1){2i/(z-a - √(a2-1)} = 2i/-2√(a2-1) = -i/√(a2-1)

    And the residue theorem gives the final value for the integral, I:

    I = 2[itex]\pi[/itex]/√(a2-1)

    Is this correct?
     
    Last edited: May 15, 2012
  2. jcsd
  3. May 15, 2012 #2

    vela

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    Yes, that's correct.

    Try writing ##a=1+\delta## and plug that expression into ##a-\sqrt{a^2-1}##. You should be able to see then that the pole will lie inside the unit circle.
     
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