Any hint for this :$\int_{-\infty}^\infty \frac{e^{ 3ix}} {x-i}\ dx $?

Question

I have tried to get the closed form of this integral:

$$\int_{-\infty}^{\infty} \frac{e^{3ix}}{x-i} \, dx $$

using Euler formula $e^{ix} = \cos x + i\sin x$, but I don't succeed. However, that integral is converge as shown here in WolframAlpha which is

$$ \frac{\operatorname{Ei}(3i(x-i))}{e^3}. $$

Then any simple method for it?