How to calculate $\int\limits_{-\infty}^\infty\frac{e^{ix}}{x}dx$

First you can estimate $\displaystyle \int_0^\pi e^{-R\sin\phi}d\phi=2\int_0^{\pi/2} e^{-R\sin\phi}d\phi$ when $R \to \infty$. You have:

$$\int_0^{\pi/2} e^{-R\sin\phi}d\phi=\int_0^{\arcsin \frac{1}{\sqrt{R}}} e^{-R\sin\phi}d\phi+\int_{\arcsin \frac{1}{\sqrt{R}}}^{\pi/2} e^{-R\sin\phi}d\phi$$

In first integral $e^{r\sin \phi} \leq 1$, so:

$$\int_0^{\arcsin \frac{1}{\sqrt{R}}}e^{-R\sin\phi}d\phi \leq \int_0^{\arcsin \frac{1}{\sqrt{R}}} 1 \; d\phi=\arcsin \frac{1}{\sqrt{R}} \to 0$$

when $R \to \infty$.Next consider second integral. $\sin$ on $[0,\pi/2]$ is monotonic increasing, so:

$$\int_{\arcsin \frac{1}{\sqrt{R}}}^{\pi/2} e^{-R\sin\phi}d\phi \leq \int_{\arcsin \frac{1}{\sqrt{R}}}^{\pi/2} e^{-R\sin(\arcsin \frac{1}{\sqrt{R}})}d\phi=\int_{\arcsin \frac{1}{\sqrt{R}}}^{\pi/2} e^{-\sqrt{R}}d\phi \to 0$$

when $R \to \infty$.