New answers tagged

0 votes

Laplace transform of derivative

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 $\mathcal L \{x\}$...
user avatar

Top 50 recent answers are included