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Evaluate real integral using residue theorem, where did I go wrong?

  1. Mar 16, 2012 #1
    Edit: Never mind I found my error, moderator can lock this.


    1. The problem statement, all variables and given/known data

    Evaluate the integral [itex]\int_0^{\pi} \frac{dt}{(a+cost)^2}[/itex] for a > 1.



    2. Relevant equations

    [tex]\int_0^{\pi}\frac{dt}{(a+cost)^2} = \pi i\sum_{a\epsilon \mathbb{E}}Res(f;\alpha)[/tex]

    Where [itex]\mathbb{E}[/itex] is the open unit disk, and [itex]f(z) = \frac{1}{iz(a+\frac{1}{2}(z+\frac{1}{z}))^2}[/itex].



    3. The attempt at a solution

    [tex]f(z) = \frac{1}{iz(a+\frac{1}{2}(z+\frac{1}{z}))^2} = \frac{1}{iz\frac{(z^2 + 2az + 1)^2}{4z^2}} = \frac{-4iz}{(z^2 + 2az + 1)^2} = \frac{-4iz}{[(z + a + \sqrt{a^2 - 1})(z + a - \sqrt{a^2 - 1})]^2}[/tex]

    Thus f has two roots of multiplicity two. It's fairly easy to see that only the root [itex]z = -a + \sqrt{a^2 - 1}[/itex] lies inside the open unit disk for a > 1, thus we set [itex] b = -a + \sqrt{a^2 - 1}[/itex] and obtain:

    [tex]\pi i\sum_{\alpha\epsilon\mathbb{E}}Res(f;\alpha) = Res(f;b) = lim_{z\rightarrow b}\frac{d}{dz}\frac{-4iz(z+a-\sqrt{a^2 - 1})^2}{(z + a + \sqrt{a^2 - 1})^2(z + a - \sqrt{a^2 - 1})^2} = lim_{z\rightarrow b}4\pi\frac{d}{dz}\frac{z}{(z+a+\sqrt{a^2 - 1})^2} = lim_{z\rightarrow b}4\pi \frac{a + \sqrt{a^2 - 1} -z}{(z + a + \sqrt{a^2 - 1})^3}[/tex]

    From here I take the limit and after simplification obtain [itex]\frac{\pi\sqrt{a^2 - 1}}{a^3}[/itex]. Yet the book says my answer should be [itex]\frac{\pi a}{(a^2 - 1)\sqrt{a^2 - 1}}[/itex]. Where have I gone wrong? Thanks.
     
    Last edited: Mar 16, 2012
  2. jcsd
  3. Mar 16, 2012 #2
    Explain your mistake so that everyone can learn from it!
     
  4. Mar 16, 2012 #3
    It wasn't really a mistake in my understanding of the theory it was just a sign error. I was accidentally plugging in -b instead of b.
     
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