$\frac{e^{ixt}-1}{ix} \to \frac{i}{x+i0}$ as $t\to\infty$

Question

In a physics paper, I saw an expression $$\frac{e^{ixt}-1}{ix} \to \frac{i}{x+i0}, \quad t\to\infty,$$ without any further explanation. I think this expression should be regarded as a distribution sense, but I cannot figure out why it is true. Introducing a test function $f\in C^\infty_0(\mathbb R)$, we should evaluate $$\lim_{t\to\infty}\int_{\mathbb R} \frac{e^{ixt}-1}{ix} f(x)dx.$$ How should one proceeded? Writing $f(x) = f(0) + xg(x)$ for some function $g$ does not help.

Answer 1

One way to think of it is as a limit of a Fourier transform: $$ \frac{e^{ixt}-1}{ix} = \int_{0}^{t} e^{ixs} \, ds = \int_{-t}^{0} e^{-ixs} \, ds = \mathcal{F}\{ \mathbf{1}_{[-t,0]}(s) \}(x) \\ \to \mathcal{F}\{ 1-u(s) \}(x) = 2\pi\delta(x) - \pi \left( \operatorname{pv}\frac{1}{i\pi x} + \delta(x) \right) \\ = \pi\delta(x) - \operatorname{pv}\frac{1}{ix} , $$ where $\mathbf{1}_{[-t,0]}(s)$ is the indicator function on the interval $[-t,0]$ and $u$ is the Heaviside step function.