[proofplan] We decompose $f$ near the simple pole $a$ as $f(z) = c_{-1}/(z - a) + g(z)$, where $g$ is holomorphic (hence bounded) near $a$ and $c_{-1} = \operatorname{Res}(f, a)$. The singular part integrates exactly to $i(\beta - \alpha) c_{-1}$ by direct computation, and the holomorphic remainder is controlled by the ML inequality, which forces its contribution to zero as $\varepsilon \to 0$. [/proofplan] [step:Decompose $f$ into its singular part and a holomorphic remainder] Since $a$ is a simple pole, the [Laurent series expansion](/theorems/350) gives \begin{align*} f(z) = \frac{c_{-1}}{z - a} + g(z), \end{align*} where $c_{-1} = \operatorname{Res}(f, a)$ and $g: B(a, r) \to \mathbb{C}$ is holomorphic (the sum of the non-negative-power terms of the Laurent series). Since $g$ is continuous on the compact set $\overline{B(a, r/2)}$, there exists $K > 0$ such that $|g(z)| \leq K$ for all $|z - a| \leq r/2$. [/step] [step:Compute the integral of the singular part exactly] On the arc $\gamma_\varepsilon(t) = a + \varepsilon e^{it}$, $t \in [\alpha, \beta]$: \begin{align*} \int_{\gamma_\varepsilon} \frac{c_{-1}}{z - a} \, dz = c_{-1} \int_\alpha^\beta \frac{i\varepsilon e^{it}}{\varepsilon e^{it}} \, dt = c_{-1} \cdot i(\beta - \alpha). \end{align*} [/step] [step:Bound the holomorphic remainder by the ML inequality and send $\varepsilon \to 0$] The arc $\gamma_\varepsilon$ has length $\ell(\gamma_\varepsilon) = \varepsilon(\beta - \alpha)$. For $\varepsilon < r/2$, the bound $|g(z)| \leq K$ holds on $\gamma_\varepsilon^*$. The ML inequality gives \begin{align*} \left|\int_{\gamma_\varepsilon} g(z) \, dz\right| \leq K \cdot \varepsilon(\beta - \alpha) \to 0 \quad \text{as } \varepsilon \to 0. \end{align*} Combining: \begin{align*} \lim_{\varepsilon \to 0} \int_{\gamma_\varepsilon} f(z) \, dz = c_{-1} \cdot i(\beta - \alpha) + 0 = i(\beta - \alpha) \cdot \operatorname{Res}(f, a). \end{align*} [/step]