1. Contour Integral

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

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

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

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

2. Originally Posted by nmatthies1
For $\displaystyle \varepsilon >0$ let $\displaystyle \gamma_{\varepsilon}$ be the piece of circular arc of radius $\displaystyle \varepsilon$ parameterised by $\displaystyle \gamma_{\varepsilon}(t)=\varepsilon e^{it}$ for $\displaystyle t \in [\alpha , \beta] \subseteq [0,2\pi]$. Show, for continuous $\displaystyle f: \mathbb{C} \to \mathbb{C}$, that

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

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

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

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