# Laplace transform of power function

In the defining integral (http://planetmath.org/ImproperIntegral)

 $\mathcal{L}\left\{t^{r}\right\}\;=\;\int_{0}^{\infty}\!e^{-st}t^{r}\,dt$

of the Laplace transform of the power function$t\mapsto t^{r}$,  we make the substitution (http://planetmath.org/SubstitutionForIntegration)  $u:=st$:

 $\mathcal{L}\left\{t^{r}\right\}\;=\;\int_{0}^{\infty}\!e^{-u}\left(\frac{u}{s}% \right)^{r}\frac{du}{s}\;=\;\frac{1}{s^{n+1}}\!\int_{0}^{\infty}\!e^{-u}u^{r+1% -1}\,du$

Here we have assumed that  $r>-1$  and $s>0$.  According to the definition of the gamma function, the last integral is equal to $\Gamma(r\!+\!1)$.  Thus we obtain

 $\displaystyle\mathcal{L}\left\{t^{r}\right\}\;=\;\frac{\Gamma(r\!+\!1)}{s^{r+1% }}.$ (1)

The special case  $r=-\frac{1}{2}$  gives the result

 $\displaystyle\mathcal{L}\left\{\frac{1}{\sqrt{t}}\right\}\;=\;\sqrt{\frac{\pi}% {s}}.$ (2)
Title Laplace transform of power function LaplaceTransformOfPowerFunction 2013-03-22 18:17:42 2013-03-22 18:17:42 pahio (2872) pahio (2872) 6 pahio (2872) Derivation msc 44A10 EvaluatingTheGammaFunctionAt12