# Tag Info

## New answers tagged laplace-transform

The existence of $\mathcal L\{x*\delta'\}=\mathcal L\{x\}\cdot\mathcal L\{\delta'\}$ requires the same subexponential behaviour from $x$ (if $\lim_{|t|\to\infty}xe^{-st} \ne 0$, then $\mathcal L \{x\}$...