Use Fourier inversion with the stated Fourier transform convention:
For , close the contour integral in the lower half-plane, where decays. Jordan lemma eliminates the large semicircle. The contour is clockwise, so the residue theorem supplies times the residue at . That simple pole has
Consequently
The minus sign uses both the clockwise orientation and the residue ; it is consistent with the full odd function away from zero.

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