Source: Standard complex-analysis result; treatment follows J. E. Santos' Cambridge IB lectures.
Statement. Let $f(z) = e^{iaz}g(z)$ with $a > 0$. Show that if
$g(z)\to 0$ uniformly on the upper half-plane as $|z|\to\infty$, i.e.
\[\max_{\theta\in[0,\pi]}|g(Re^{i\theta})| \to 0 \quad \text{as } R\to\infty,
\tag{1.1}\]
then the integral over the upper semicircular arc $\mathcal{C}_R =
\{z = Re^{i\theta}, \theta\in[0,\pi]\}$ vanishes:
\[\lim_{R\to\infty}\int_{\mathcal{C}_R}f(z)\,dz = 0.\tag{1.2}\]
Find an upper bound for $\left|\int_{\mathcal{C}_R}f(z)\,dz\right|$ and
give an example of a class of functions $g$ satisfying (1.1).
A typical example satisfying
(1.1) is any rational function $g(z) = P(z)/Q(z)$ with
$\deg Q \geq \deg P + 1$, for which
$\max|g(Re^{i\theta})| = \mathcal{O}(R^{-1})\to 0$.
Parametrise $z = Re^{i\theta}$, so $dz = iRe^{i\theta}d\theta$. The
integral becomes
\[\int_{\mathcal{C}_R}f(z)\,dz
= iR\int^\pi_0 g(Re^{i\theta})\,e^{iaRe^{i\theta}}\,e^{i\theta}\,d\theta.\tag{1.3}\]
Taking the absolute value and using $|e^{i\theta}|=1$ and
$|e^{iaRe^{i\theta}}| = e^{-aR\sin\theta}$:
\[\begin{aligned}\left|\int_{\mathcal{C}_R}f(z)\,dz\right|
&\leq R\int^\pi_0 |g(Re^{i\theta})|\,e^{-aR\sin\theta}\,d\theta \notag\\
&\leq M(R)\cdot R\int^\pi_0 e^{-aR\sin\theta}\,d\theta,\end{aligned}\tag{1.4}\]
where $M(R)\equiv\max_{\theta\in[0,\pi]}|g(Re^{i\theta})|$. By symmetry of
$\sin\theta$ about $\theta=\pi/2$:
\[R\int^\pi_0 e^{-aR\sin\theta}\,d\theta
= 2R\int^{\pi/2}_0 e^{-aR\sin\theta}\,d\theta.\tag{1.5}\]
Since $\sin\theta$ is concave on $[0,\pi/2]$, it lies above the chord
connecting $(0,0)$ and $(\pi/2,1)$:
\[\sin\theta \geq \frac{2\theta}{\pi}, \qquad \theta\in\left[0,\tfrac{\pi}{2}\right].\tag{1.6}\]
Therefore:
\[2R\int^{\pi/2}_0 e^{-aR\sin\theta}\,d\theta
\leq 2R\int^{\pi/2}_0 e^{-2aR\theta/\pi}\,d\theta
= \frac{\pi}{a}\!\left(1-e^{-aR}\right)
\leq \frac{\pi}{a}.\tag{1.7}\]
Combining, we obtain the bound
\[\left|\int_{\mathcal{C}_R}f(z)\,dz\right| \leq \frac{\pi}{a}\,M(R).
\tag{1.8}\]
Taking $R\to\infty$: by assumption (1.1), $M(R)\to 0$, so
the right-hand side of (1.8) vanishes by the squeeze
theorem, and hence
\[\lim_{R\to\infty}\int_{\mathcal{C}_R}f(z)\,d z = 0.\tag{1.9}\]