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