Question

Laplace transforms Help!!
computing integral from 0 to infinity of te^-t sin 2t using a laplace transform.

Answer 1

\[\int\limits_0^\infty te^{-t}\sin(2t)\,\mathrm dt\\=\mathcal L\big\{t\sin(2t)\big\}|_{s\to1}\]

Answer 2

\[\mathcal L\big\{tf(t)\big\}=-\frac{\mathrm d}{\mathrm ds}\mathcal L\big\{f(t)\big\}\]

Answer 3

use a table on this thing lol