[proofplan] We parametrise the semicircular arc $\gamma_R(\theta) = Re^{i\theta}$, $\theta \in [0, \pi]$, and bound the integral using the decay hypothesis $|zf(z)| \leq M$ and the exponential damping $e^{-\alpha R \sin\theta}$ in the upper half-plane. Jordan's inequality $\sin\theta \geq 2\theta/\pi$ for $\theta \in [0, \pi/2]$ replaces $\sin\theta$ by a linear lower bound, making the resulting integral explicitly computable. The bound $M\pi/(\alpha R) \to 0$ as $R \to \infty$. [/proofplan] [step:Parametrise the semicircular arc and extract the exponential decay factor] On the semicircular arc $\gamma_R(\theta) = Re^{i\theta}$, $\theta \in [0, \pi]$, parametrise with $z = Re^{i\theta}$, $dz = iRe^{i\theta} \, d\theta$: \begin{align*} \left|\int_{\gamma_R} f(z) e^{i\alpha z} \, dz\right| = \left|\int_0^\pi f(Re^{i\theta}) e^{i\alpha R e^{i\theta}} \, iRe^{i\theta} \, d\theta\right| \leq R \int_0^\pi |f(Re^{i\theta})| \cdot e^{-\alpha R \sin\theta} \, d\theta. \end{align*} The factor $e^{-\alpha R \sin\theta}$ arises because $\operatorname{Re}(i\alpha R e^{i\theta}) = -\alpha R \sin\theta$, and $\sin\theta \geq 0$ for $\theta \in [0, \pi]$, so $e^{i\alpha z}$ is exponentially damped in the upper half-plane when $\alpha > 0$. [/step] [step:Apply the decay hypothesis $|zf(z)| \leq M$ and use symmetry] Since $|zf(z)| \leq M$ for $|z| > r_0$ and $R > r_0$, we have $|f(Re^{i\theta})| \leq M/R$. Therefore \begin{align*} \left|\int_{\gamma_R} f(z) e^{i\alpha z} \, dz\right| \leq M \int_0^\pi e^{-\alpha R \sin\theta} \, d\theta. \end{align*} By the symmetry $\sin(\pi - \theta) = \sin\theta$, the integral over $[0, \pi]$ equals twice the integral over $[0, \pi/2]$: \begin{align*} M \int_0^\pi e^{-\alpha R \sin\theta} \, d\theta = 2M \int_0^{\pi/2} e^{-\alpha R \sin\theta} \, d\theta. \end{align*} [/step] [step:Apply Jordan's inequality $\sin\theta \geq 2\theta/\pi$ to obtain the explicit bound] **Jordan's inequality:** For $\theta \in [0, \pi/2]$, $\sin\theta \geq 2\theta/\pi$. This holds because $\sin$ is concave on $[0, \pi/2]$ (since $\sin'' = -\sin \leq 0$ there), so the graph of $\sin$ lies above the chord from $(0, 0)$ to $(\pi/2, 1)$, which has equation $y = 2\theta/\pi$. Substituting this lower bound: \begin{align*} 2M \int_0^{\pi/2} e^{-\alpha R \sin\theta} \, d\theta \leq 2M \int_0^{\pi/2} e^{-2\alpha R \theta / \pi} \, d\theta. \end{align*} The integral on the right is computed explicitly: \begin{align*} \int_0^{\pi/2} e^{-2\alpha R \theta / \pi} \, d\theta = \left[-\frac{\pi}{2\alpha R} e^{-2\alpha R \theta / \pi}\right]_0^{\pi/2} = \frac{\pi}{2\alpha R}\left(1 - e^{-\alpha R}\right) \leq \frac{\pi}{2\alpha R}. \end{align*} Therefore \begin{align*} \left|\int_{\gamma_R} f(z) e^{i\alpha z} \, dz\right| \leq 2M \cdot \frac{\pi}{2\alpha R} = \frac{M\pi}{\alpha R} \to 0 \quad \text{as } R \to \infty. \end{align*} [guided] Jordan's inequality is the essential ingredient. Without it, we cannot evaluate $\int_0^{\pi/2} e^{-\alpha R \sin\theta} \, d\theta$ in closed form because $\sin\theta$ is transcendental. The inequality $\sin\theta \geq 2\theta/\pi$ replaces $\sin$ by a linear function, turning the integral into a simple exponential integral. Why does $\sin\theta \geq 2\theta/\pi$ hold on $[0, \pi/2]$? The function $\sin$ is concave on this interval (since $\sin'' = -\sin \leq 0$ there). A concave function lies above any chord, and the chord from $(0, 0)$ to $(\pi/2, 1)$ has equation $y = 2\theta/\pi$. Why is the bound $M\pi/(\alpha R)$ sharp enough? The factor $1/R$ ensures the integral vanishes as $R \to \infty$. The hypothesis $|zf(z)| \leq M$ is optimal: if $f$ decays slower than $1/|z|$ (say $|f(z)| \sim 1/|z|^\beta$ with $\beta < 1$), the bound becomes $M\pi/(\alpha R^\beta)$, which still vanishes, but the proof requires $\beta > 0$. If $f$ does not decay at all, the lemma fails. [/guided] [/step]