# Inverse Laplace Transform of $s^n$

I want to calculate Inverse Laplace Transform of $s^n$. I have an idea, but I do not know if it works?

We have a formula to compute inverse laplace transforms of functions as below,

$$\mathcal{L}^{-1} [ F(s) ] = -\frac{\mathcal{L}^{-1} [ F^{\prime}(s) ]}{t}.$$

So from the given formula, we can obtain

$$\mathcal{L}^{-1} [ s ]= -\frac{\mathcal{L}^{-1} [ 1 ]}{t}= -\frac{\delta (t)}{t}.$$ and as a result, $$\mathcal{L}^{-1} [ s^n ] = (-1)^n\frac{n!\delta (t)}{t^n}$$ Is it right? In fact, I want to know the necessary conditions to use the given formula.

Intuitively, the derivative of the Dirac delta function $$\delta'$$ has Laplace transform $$s$$. The derivative of the Dirac delta is a generalized function that pulls out the derivative of the function with a change of sign: for any interval $$[a,b]$$ where $$a < 0 < b$$,

$$\int_a^b \delta'(t)f(t) \ dt = \left[ \delta(t) f(t) \right]_a^b - \int_a^b \delta(t) f'(t) \ dt = -f'(0)$$

Applying that procedure inductively,

$$L^{-1}\{s^n\}(t) = \delta^{(n)}(t)$$

• This answer is not correct. The inverse Laplace transform of $s^n$, for non-negative integer values of $n$ is given by $$\mathscr{L}^{-1}\{s^n\}=\delta^{(n)}(t)$$There is no factor of $(-1)^n$. Note that for $f(t)=e^{-st}$, $(-1)^n \frac{d^n e^{-st}}{dt^n}=s^ne^{-st}$. I'll edit the sign error accordingly. – Mark Viola Sep 30 '20 at 16:40


• I feel that you should give some more insights (especially regarding the difference to the answer of Simon S). I guess your result is only valid for $n<0$? – Fabian Mar 6 '18 at 11:17