Laplace transform of integral

On can show that if a real function  $t\mapsto f(t)$  isLaplace-transformable (http://planetmath.org/LaplaceTransform), as wellis ${\displaystyle {\int }_0^t}f(\tau )𝑑\tau$.  The latter is alsocontinuous for  $t>0$  and by theNewton–Leibniz formula (http://planetmath.org/FundamentalTheoremOfCalculus),has the derivative equal $f(t)$.  Hence we may apply theformula for Laplace transform of derivative, obtaining

$$F(s)=ℒ\{f(t)\}=sℒ\left\{{\int }_0^{t}f(\tau )𝑑\tau \right\}-{\int }_0^{0}f(t)𝑑t=sℒ\left\{{\int }_0^{t}f(\tau )𝑑\tau \right\},$$

i.e.

$ℒ\left\{{\displaystyle {\int }_0^t}f(\tau )𝑑\tau \right\}={\displaystyle \frac{F(s)}{s}}.$(1)

Application.  We start from the easily derivable rule

$$\frac{1}{s}\curvearrowright \mathrm{\hspace{0.33em}1},$$

where the curved from the Laplace-transformed function to the original function.  The formula (1) thus yields successively

$$\frac{1}{s^2}\curvearrowright {\int }_0^{t}1𝑑\tau =t,$$

$$\frac{1}{s^3}\curvearrowright {\int }_0^t\tau 𝑑\tau =\frac{t^2}{2!},$$

$$\frac{1}{s^4}\curvearrowright {\int }_0^t\frac{{\tau }^2}{2!}𝑑\tau =\frac{t^3}{3!},$$

etc.  Generally, one has

${\displaystyle \frac{1}{s^n}}\curvearrowright {\displaystyle \frac{t^{n-1}}{(n-1)!}}\mathit{\hspace{1em}}\forall n\in {\mathbb{Z}}_{+}.$(2)