Let $$\displaystyle F(s) \,:=\, \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t)\,dt.$$

A differentiation under the integral sign with respect to $s$ yields $$F'(s) = -\int_0^\infty e^{-st}tf(t)\,dt = -\mathcal{L}\{tf(t)\}.$$ Differentiating again under the integral sign gives $$F''(s) = +\int_0^\infty e^{-st}t^2f(t)\,dt = \mathcal{L}\{t^2f(t)\}.$$ One can continue similarly, and then we apparently have

(1)

(2)

Application. Evaluate the improper integral $$I := \int_0^\infty t^3e^{-t}\sin{t}\,dt.$$ By the parent entry, we have $\mathcal{L}\{\sin{t}\} = \frac{1}{1+s^2}$ . Using this and (2), we may write $$\int_0^\infty t^3e^{-st}\sin{t}\,dt = \mathcal{L}\{t^3\sin{t}\} = (-1)^3\frac{d^3}{ds^3}\!\left(\frac{1}{s^2+1}\right) = \frac{24(s-s^3)}{(1+s^2)^4}.$$ The value of $I$ is obtained by substituting here $s = 1$ : $$I = \frac{24(1-1^3)}{(1+1^2)^4} = 0.$$