Suppose I'm trying to evaluate the following integral along a large semicircle in the upper half plane: $$ \lim_{R\to \infty} \int_{C_R} e^{iaz} \, f(z) \, , \qquad \text{where } f(z) = \frac{1}{z} \, e^{-ib\sqrt{1+z^2}} \, ,$$ with both $a, b > 0$. The branch cut of $\sqrt{1+z^2}$ can be chosen such that $f(z)$ is analytic on and exterior to $C_R$. But since $$ \lim_{R\to\infty} |f(z \in C_R)| \leq \frac{1}{R} \, e^{b \sqrt{1+R^2}} \quad \text{is unbounded} \, ,$$ the integral seems to be unbounded as well.
However, I'm wondering if I can do the following trick. First, by writing $\sqrt{1+z^2} \xrightarrow[|z|\to\infty]{} z$, I could perhaps replace the integral by $$ \lim_{R\to \infty} \int_{C_R} e^{i(a-b)z} \, \frac{1}{z} \, .$$ With $a > b$, we can now apply Jordan's lemma and conclude that the integral actually vanishes.
Is this a valid argument?