Inverse Laplace transform of $s^k$ I want to calculate Inverse Laplace Transform of $s^k$. $0<k<1$
 I have an idea, but I do not know if it works?
We have a formula ,
$$\mathcal{L}^{-1} [ F(s) ] = -\frac{\mathcal{L}^{-1} [ F^{\prime}(s) ]}{t}.$$
So from the given formula, 
$$\mathcal{L}^{-1} [ s^k ] = -\frac{\mathcal{L}^{-1} [k s^{k-1}]}{t}= -\frac{k t^{-k-1}}{\Gamma(1-k)}$$
 Is it right?   What is the result of $$\mathcal{L}^{-1} [ s^k ]$$
Thank you very much.
 I also want to know the necessary conditions to use the given formula. 
 A: If $f(t)$ is piecewise continouns for $t\geq 0$ and $\left| f(t)\right| \leq M \exp (c t)$ as,$t\to +\infty$ then
$$\mathcal{L}_t[-t f(t)](s)=F'(s)$$
for $s>c$.Equivalently:
$$\color{Blue}{f(t)=\mathcal{L}_s^{-1}[F(s)](t)=-\frac{\mathcal{L}_s^{-1}\left[F'(s)\right](t)}{t}}$$
If $f(t)$ is piecewise continouns for $t\geq 0$ and $\underset{t\to 0}{\text{lim}}\frac{f(t)}{t}=\text{exist and is finite}$,and that
$\left| f(t)\right| \leq M \exp (c t)$ as,$t\to +\infty$ then
$$\mathcal{L}_t\left[\frac{f(t)}{t}\right](s)=\int_s^{\infty } F(a) \, da$$
for $s>c$.Equivalently:
$$\color{Blue}{f(t)=\mathcal{L}_s^{-1}[F(s)](t)=t \left(\mathcal{L}_s^{-1}\left[\int_s^{\infty } F(a) \, da\right](t)\right)}$$
For yours example:
$$\color{red}{\mathcal{L}_s^{-1}\left[s^k\right](t)}=t \left(\mathcal{L}_s^{-1}\left[\int_s^{\infty } a^k \, da\right](t)\right)=t
   \left(\mathcal{L}_s^{-1}\left[-\frac{s^{1+k}}{1+k}\right](t)\right)=\frac{t \left(-t^{-2-k}\right)}{(1+k) \Gamma (-1-k)}=\frac{t^{-1-k}}{\Gamma
   (-k)}=\color{red}{-\frac{k t^{-1-k}}{\Gamma (1-k)}}$$
If you go back:
$$\color{red}{\mathcal{L}_t\left[-\frac{k t^{-1-k}}{\Gamma (1-k)}\right](s)}=-\frac{k \int_0^{\infty } t^{-1-k} \exp (-s t) \, dt}{\Gamma (1-k)}=-\frac{k s^k
   \Gamma (-k)}{\Gamma (1-k)}=\color{red}{s^k}$$
A: There isn't a locally integrable function whose Laplace transform is $s^k, \;0 < k < 1$. In terms of distributions, if $t_+^k$ is the singular functional defined as
$$(t_+^k, \phi) = \int_0^\infty t^k (\phi(t) - \phi(0)) dt, \quad -2 < k < -1,$$
then its Laplace transform is $\Gamma(k + 1)s^{-k - 1}$, and $0 < -k - 1 < 1$.
