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Math Help - Contour Integral

  1. #1
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    Contour Integral

    For \varepsilon >0 let \gamma_{\varepsilon} be the piece of circular arc of radius \varepsilon parameterised by \gamma_{\varepsilon}(t)=\varepsilon e^{it} for t \in [\alpha , \beta] \subseteq [0,2\pi]. Show, for continuous f: \mathbb{C} \to \mathbb{C}, that

    \int_{\gamma_{\varepsilon}} \frac{f(z)}{z}dz \to f(0)(\beta - \alpha )i as \varepsilon \to 0. (1)

    When t \in [0, 2\pi] by Cauchy's integral formula \int_{\gamma_{\varepsilon}} \frac{f(z)}{z}dz=2\pi i f(0) and (1) holds.

    Now, letting g(z)=\frac{f(z)}{z} we have \int_{\gamma_{\varepsilon}} \frac{f(z)}{z}dz= \int_{\alpha}^{\beta} g(\gamma_{\varepsilon} (t)) \gamma_{\varepsilon} '(t)dt= \int_{\alpha}^{\beta} \frac{f(\varepsilon e^{it})}{\varepsilon e^{it}}\varepsilon ie^{it}dt= \int_{\alpha}^{\beta} f(\varepsilon e^{it}) idt= i\int_{\alpha}^{\beta} f(\varepsilon e^{it}) dt but I'm not really sure how to get from this to (1).
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  2. #2
    Junior Member
    Joined
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    Quote Originally Posted by nmatthies1 View Post
    For \varepsilon >0 let \gamma_{\varepsilon} be the piece of circular arc of radius \varepsilon parameterised by \gamma_{\varepsilon}(t)=\varepsilon e^{it} for t \in [\alpha , \beta] \subseteq [0,2\pi]. Show, for continuous f: \mathbb{C} \to \mathbb{C}, that

    \int_{\gamma_{\varepsilon}} \frac{f(z)}{z}dz \to f(0)(\beta - \alpha )i as \varepsilon \to 0. (1)

    When t \in [0, 2\pi] by Cauchy's integral formula \int_{\gamma_{\varepsilon}} \frac{f(z)}{z}dz=2\pi i f(0) and (1) holds.

    Now, letting g(z)=\frac{f(z)}{z} we have \int_{\gamma_{\varepsilon}} \frac{f(z)}{z}dz= \int_{\alpha}^{\beta} g(\gamma_{\varepsilon} (t)) \gamma_{\varepsilon} '(t)dt= \int_{\alpha}^{\beta} \frac{f(\varepsilon e^{it})}{\varepsilon e^{it}}\varepsilon ie^{it}dt= \int_{\alpha}^{\beta} f(\varepsilon e^{it}) idt= i\int_{\alpha}^{\beta} f(\varepsilon e^{it}) dt but I'm not really sure how to get from this to (1).
    do you need to know how

    \int_{\alpha}^{\beta} f(\varepsilon e^{it}) dt =   f(0)(\beta - \alpha )i as \varepsilon \to 0.

    is that your Q ?????
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