Consider a square centered at $(0,0)$ of side length $2x_1$ and the contour $\gamma_{x_1}$ consisting of the upper half of this square oriented counterclockwise. This is a contour that often appears in complex integration.
Inspired by these great notes, I wanted to find a sufficient condition for a function so that its integral on the path $\gamma_{x_1}$ tends to zero as $x_1\to +\infty$. This is my attempt:
Consider a continuous function $f$ such that for every $z\in \gamma_{x_1}$ we have $$|f(z)|\leq \frac{M}{|z|^a},$$ where $M>0$ and $a>0$. Then, $$\int_{\gamma_{x_1}}f(z)e^{ibz}dz\xrightarrow{x_1\to+\infty}0,$$ for every $b>0$.
Proof: Notice that \begin{align*}\big|f(z)e^{ibz}\big|\leq M\frac{\big|e^{ibz}\big|}{|z|^a}=M\frac{e^{-b\text{Im} z}}{|z|^a}\leq M\frac{e^{-bx_1}}{x_1^a},\end{align*} for $z\in \gamma_{x_1}$. Therefore, we can bound the integral as follows: \begin{align*}\left|\int_{\gamma_{x_1}}f(z)e^{ibz}dz\right|&\leq \text{length}(\gamma_{x_1})\max_{z\in \gamma_{x_1}}\big|f(z)e^{ibz}\big|\\ &\leq 4x_1M\frac{e^{-bx_1}}{x_1^a}.\end{align*} Now, the last expression tends to $0$ as $x_1\to +\infty$ since the exponential goes faster to zero. Done.
Is my proposition correct? Thanks!
EDIT: As @FShrike commented, the bound $e^{-b\text{Im}z}\leq e^{-bx_1}$ is not correct. So we must assume $a>1$ rather than $a>0$ so that the correct integral bound $$\left|\int_{\gamma_{x_1}}f(z)e^{ibz}dz\right|\leq \frac{4M}{x_1^{a-1}},$$ tends to zero when $x_1\to +\infty$.