This is probably going to be a stupid question ( I don't feel great today) but I can't get around this problem.
$$I = \int_\mathbb R \frac 1 {(3x-2i)^2} dx $$
I thought that using contour integration this was going to be a piece of cake.First note that the integral vanishes on every circle or semicircle in the complex plane.
If we integrate over the positive semicircle, we get $I = 2 \pi i \text{ Res }(f, \frac 23i) = 2\pi i /9$
If we integrate over the negative semicircle, we get $I = 0$ (which is the value wolfram alpha reports).
What is going on?
The integral is zero. Your trouble stems from the fact that you have a double pole. The residue at the pole is clearly zero, so the integral is zero.