You cannot close the contour in the lower half-plane, because the exp in the numerator is large in the lower half-plane. You must close the contour in the upper half-plane. Then your integral is 0.

Carlo Beenakker's answer is right. When t>0, you cannot close the contour in the lower half-plane, because the exp in the numerator is large in the lower half-plane. You must close the contour in the upper half-plane. Then your integral is 0, for $t>0$. When $t<0$ your series computed with residues converges.

First of all, the singularity at the origin is not removable, it is a simple pole, if you count carefully. Second, you cannot close the contour in the lower half-plane, because the exp in the numerator is large in the lower half-plane. You must close the contour in the upper half-plane. Then your integral is easily computed using this residue at 0.

You cannot close the contour in the lower half-plane, because the exp in the numerator is large in the lower half-plane. You must close the contour in the upper half-plane. Then your integral is 0.