# Laplace transform of integral

On can show that if a real function$t\mapsto f(t)$  is Laplace-transformable (http://planetmath.org/LaplaceTransform), as well is $\displaystyle\int_{0}^{t}f(\tau)\,d\tau$.  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

 $F(s)\;=\;\mathcal{L}\{f(t)\}\;=\;s\,\mathcal{L}\left\{\int_{0}^{t}\!f(\tau)\,d% \tau\right\}-\int_{0}^{0}\!f(t)\,dt\;=\;s\,\mathcal{L}\left\{\int_{0}^{t}\!f(% \tau)\,d\tau\right\},$

i.e.

 $\displaystyle\mathcal{L}\left\{\int_{0}^{t}\!f(\tau)\,d\tau\right\}\;=\;\frac{% F(s)}{s}.$ (1)

Application.  We start from the easily derivable rule

 $\frac{1}{s}\;\curvearrowright\;1,$

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

 $\frac{1}{s^{2}}\;\curvearrowright\;\int_{0}^{t}\!1\,d\tau\;=\;t,$
 $\frac{1}{s^{3}}\;\curvearrowright\;\int_{0}^{t}\!\tau\,d\tau\;=\;\frac{t^{2}}{% 2!},$
 $\frac{1}{s^{4}}\;\curvearrowright\;\int_{0}^{t}\!\frac{\tau^{2}}{2!}\,d\tau\;=% \;\frac{t^{3}}{3!},$

etc.  Generally, one has

 $\displaystyle\frac{1}{s^{n}}\;\curvearrowright\;\frac{t^{n-1}}{(n\!-\!1)!}% \quad\forall\,n\in\mathbb{Z}_{+}.$ (2)
Title Laplace transform of integral LaplaceTransformOfIntegral 2014-03-17 10:43:31 2014-03-17 10:43:31 pahio (2872) pahio (2872) 8 pahio (2872) Derivation msc 44A10