Contour Integral

Quote:

Originally Posted by nmatthies1

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 ?????