On can show that if a real function $t\mapsto f(t)$ is
Laplacetransformable (http://planetmath.org/LaplaceTransform), as well
is ${\int}_{0}^{t}}f(\tau )\mathit{d}\tau $. The latter is also
continuous^{} for $t>0$ and by the
Newton–Leibniz formula (http://planetmath.org/FundamentalTheoremOfCalculus),
has the derivative equal $f(t)$. Hence we may apply the
formula for Laplace transform of derivative, obtaining

$$F(s)=\mathcal{L}\{f(t)\}=s\mathcal{L}\left\{{\int}_{0}^{t}f(\tau )\mathit{d}\tau \right\}{\int}_{0}^{0}f(t)\mathit{d}t=s\mathcal{L}\left\{{\int}_{0}^{t}f(\tau )\mathit{d}\tau \right\},$$ 

i.e.

$\mathcal{L}\left\{{\displaystyle {\int}_{0}^{t}}f(\tau )\mathit{d}\tau \right\}={\displaystyle \frac{F(s)}{s}}.$ 

(1) 
Application. We start from the easily derivable rule

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

where the curved from the Laplacetransformed function^{} to the original function. The formula (1) thus yields successively

$$\frac{1}{{s}^{2}}\curvearrowright {\int}_{0}^{t}1\mathit{d}\tau =t,$$ 


$$\frac{1}{{s}^{3}}\curvearrowright {\int}_{0}^{t}\tau \mathit{d}\tau =\frac{{t}^{2}}{2!},$$ 


$$\frac{1}{{s}^{4}}\curvearrowright {\int}_{0}^{t}\frac{{\tau}^{2}}{2!}\mathit{d}\tau =\frac{{t}^{3}}{3!},$$ 

etc. Generally, one has

$\frac{1}{{s}^{n}}}\curvearrowright {\displaystyle \frac{{t}^{n1}}{(n1)!}}\mathit{\hspace{1em}}\forall n\in {\mathbb{Z}}_{+}.$ 

(2) 