Suppose that $\int_{a}^{\infty} f (x) \, d x$ converges absolutely.
This looks like the Riemann-Lebesgue Lemma. Can I use a similar proof? However, the assumption in the Lemma is that $f (x)$ is Riemann integrable, whereas in this question we are given absolute convergence.
Edit: Proof of the Riemann-Lebesgue Lemma:
Suppose $f$ is Riemann integrable on $[a,b]$. then given $\epsilon > 0$, there is a partition $\{ a = x_{0}, x_{1}, \dots, x_{n} = b \}$, so that $\displaystyle \epsilon / 2 > \int_{a}^{b} f - \sum_{1}^{n} m_{i} \Delta x_{i} \geq 0$, where $m_{i}$ is the minimum of $f$ on $[x_{i-1}, x_{i}]$. But the sum can be written as $\displaystyle \sum_{1}^{n} m_{i} \Delta x_{i} = \int_{a}^{b} g$, where $\displaystyle g = \sum m_{i} \chi_{x_{i - 1, x_{i}}}$, and the inequality takes the form $$ \epsilon/2 > \int_{a}^{b} (f - g) \geq 0 .$$ Now we use the fact that $f-g \geq 0$ to get \begin{align*} \left\rvert \int_{a}^{b} f(x) e^{i \lambda x} \, dx \right\rvert &\leq \left\rvert \int_{a}^{b} \left[ f(x) - g(x) \right] e^{i\lambda x} \, dx \right\rvert + \left\rvert \int_{a}^{b} g(x) e^{i\lambda x} \, dx \right\rvert \\ &\leq \int_{a}^{b} (f - g) \, dx + (\lambda/1) \left\rvert \sum m_{i} (e^{i\lambda x_{i}} - e^{\lambda x_{i - 1}}) \right\rvert \end{align*}
The function $g$ has been fixed. Take $\lambda$ large enough that $$ (\lambda/1) \left\rvert \sum m_{i} (e^{i\lambda x_{i}} - e^{\lambda x_{i - 1}}) \right\rvert < \epsilon / 2 $$
Therefore, $$ \left\rvert \int_{a}^{b} f(x) e^{i \lambda x} \, dx \right\rvert < \epsilon $$