Translations:Dictionary:Hilbert transform/69/en

Hence, when ${\mbox{Im}}\;t<0,$ for the case when the integration contour closes in the upper-half-plane of $t,$ we must have $\operatorname {sgn}(\omega )=+1$ and when ${\mbox{Im}}\;t>0,$ when the contour closes in the lower-half-plane of $t,$ we must have $\operatorname {sgn}(\omega )=-1$ for the integral to exist.