Laplace transform of $t^{n}f(t)$

Let

$$F(s):=ℒ\{f(t)\}={\int }_0^{\mathrm{\infty }}{e}^{-st}f(t)𝑑t.$$

A differentiation under the integral sign with respect to $s$ yields

$$F^{\prime }(s)=-{\int }_0^{\mathrm{\infty }}{e}^{-st}tf(t)𝑑t=-ℒ\{tf(t)\}.$$

Differentiating again under the integral sign gives

$$F^{\prime \prime }(s)=+{\int }_0^{\mathrm{\infty }}{e}^{-st}{t}^{2}f(t)𝑑t=ℒ\{t^{2}f(t)\}.$$

One can continue similarly, and then we apparently have

$F^{(n)}(s)={(-1)}^n{\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-st}{t}^{n}f(t)𝑑t={(-1)}^nℒ\{t^{n}f(t)\}.$(1)

If this equation is multiplied by ${(-1)}^n$, it gives the

$ℒ\{t^{n}f(t)\}={(-1)}^n{F}^{(n)}(s)$(2)

which is true for  $n=1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}3},\mathrm{\dots }$

Application.  Evaluate the improper integral

$$I:={\int }_0^{\mathrm{\infty }}{t}^3{e}^{-t}\sin tdt.$$

By the parent entry (http://planetmath.org/LaplaceTransform), we have  $ℒ\{\sin t\}=\frac{1}{1+s^2}$.  Using this and (2), we may write

$${\int }_0^{\mathrm{\infty }}{t}^3{e}^{-st}\sin tdt=ℒ\{t^3\sin t\}={(-1)}^3\frac{d^3}{d{s}^3}\left(\frac{1}{s^2+1}\right)=\frac{24(s-s^3)}{{(1+s^2)}^4}.$$

The value of $I$ is obtained by substituting here  $s=1$:

$$I=\frac{24(1-1^3)}{{(1+1^2)}^4}=0.$$