# 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.

• Another useful formula:$\mathcal{L}_s^{-1}[F(s)](t)=-t \left(\mathcal{L}_s^{-1}[\int F(s) \, ds](t)\right)$ – Mariusz Iwaniuk Jan 30 '18 at 10:54
• Sorry, I don't know how to do it. Could you explain clearly, please?Thank you very much. – Caroline Jan 30 '18 at 11:18
• See this book:amazon.com/Differential-Equations-Boundary-Value-Problems/dp/… on pages: 469 and 471 – Mariusz Iwaniuk Jan 30 '18 at 11:27
• $$\mathcal{L}^{-1} [ s^k ] = -t\mathcal{L}^{-1} [∫ s^{k}ds]= -t\mathcal{L}^{-1} [\frac{s^{k+1}}{k+1}]$$????Sorry, I still can not get the result. It will be infinity. – Caroline Jan 30 '18 at 12:20

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:

$$\color{red}{\mathcal{L}_t\left[-\frac{k t^{-1-k}}{\Gamma (1-k)}\right](s)}=-\frac{k \int_0^{\infty } t^{-1-k} \exp (-s t) \, dt}{\Gamma (1-k)}=-\frac{k s^k \Gamma (-k)}{\Gamma (1-k)}=\color{red}{s^k}$$

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 $0 < -k - 1 < 1$.