Theorem. Suppose $\lim_{|z|\to\infty} |f(z)| = 0$ and $a > 0$. Then $\lim_{R\to \infty} \int_{C_R} e^{iaz} f(z) \, dz = 0$ where $C_R$ is the semicircle in the upper half-plane with radius $R$ centered on $0$.
Proof. Our assumption implies that $M_R \equiv \max_{|z| = R} |f(z)|$ obeys $\lim_{R\to\infty} M_R = 0$.
Fix R and parametrize $z$ by $Re^{it} = R\cos(t) + iR\sin(t)$ for $t \in [0, \pi]$. Then we have
$$
\begin{eqnarray}
I_R &=& \int_{C_R} e^{iaz} f(z) \, dz \\
&=& \int_0^\pi e^{ia(R\cos(t) + iR\sin(t))} f(z) iRe^{it} \, dt \\
&=& iR \int_0^\pi f(z) e^{-aR\sin(t)} e^{i(t + aR\cos(t))} \, dt
\end{eqnarray}
$$
so that $|I_R| \leq RM_R \int_0^\pi e^{-aR\sin(t)} \, dt = 2RM_R \int_0^\frac{\pi}{2} e^{-aR\sin(t)} \, dt$. But note that on $[0, \frac{\pi}{2}]$ we have $\sin(t) \geq \frac{2}{\pi}t$, so the latter integral is bounded by
$$
\int_0^\frac{\pi}{2} e^{-\frac{2aR}{\pi}t} \, dt = \frac{\pi}{2aR}(1 – e^{-aR}) \leq \frac{\pi}{2aR}.
$$
Hence $|I_R| \leq 2RM_R \cdot \frac{\pi}{2aR} \to 0$. $\blacksquare$
You can also do this lemma with integrals with a negative exponent by taking the clockwise semicircle in the lower half-plane.