# 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

$$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} &\left.\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ -\ \infty\ic}s^{n}\expo{ts} {\dd s \over 2\pi\ic}\,\right\vert_{\ t\ >\ 0} = \int_{0^{+}\ -\ \infty\ic}^{0^{+}\ -\ \infty\ic}s^{n}\expo{ts}{\dd s \over 2\pi\ic} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} &\ -\int_{-\infty}^{-\epsilon}\pars{-s}^{n}\expo{n\pi\ic}\expo{ts} \,{\dd s \over 2\pi\ic} - \int_{\pi}^{-\pi}\epsilon^{n}\expo{\ic n\theta}\epsilon\expo{\ic\theta}\ic\, \,{\dd s \over 2\pi\ic} - \int_{-\epsilon}^{-\infty}\pars{-s}^{n}\expo{-n\pi\ic}\expo{ts} \,{\dd s \over 2\pi\ic} \\[5mm] = &\ -\expo{n\pi\ic}\int_{\epsilon}^{\infty}s^{n}\expo{-ts}\,{\dd s \over 2\pi\ic} - {\epsilon^{n + 1}\sin\pars{n\pi} \over \pars{n + 1}\pi} + \expo{-n\pi\ic}\int_{\epsilon}^{\infty}s^{n}\expo{-ts}\,{\dd s \over 2\pi\ic} \\[5mm] = &\ -\,{\epsilon^{n + 1}\sin\pars{n\pi} \over \pars{n + 1}\pi} - {\sin\pars{n\pi} \over \pi} \int_{\epsilon}^{\infty}s^{n}\expo{-ts}\,\dd s \\[5mm] = &\ -\,{\epsilon^{n + 1}\sin\pars{n\pi} \over \pars{n + 1}\pi} - {\sin\pars{n\pi} \over \pi}\bracks{% -\,{\epsilon^{n + 1} \over n + 1}\,\expo{-t\epsilon} - \int_{\epsilon}^{\infty}{s^{n + 1} \over n + 1}\expo{-ts}\pars{-t}\,\dd s} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} &\ -\,{\sin\pars{n\pi} \over \pars{n + 1}\pi}\,t \int_{\epsilon}^{\infty}s^{n + 1}\expo{-ts}\,\dd s \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to} -\,{\sin\pars{n\pi} \over \pars{n + 1}\pi}\,t^{-n - 1}\,\Gamma\pars{n + 2} \\[5mm] = &\ -\,{\sin\pars{n\pi} \over \pars{n + 1}\pi}\,t^{-n - 1}\, {\pi \over \Gamma\pars{-1 - n}\sin\pars{\pi\bracks{n + 2}}} = -\,{1 \over n + 1}\,t^{-n - 1}\, {1 \over \Gamma\pars{-n}/\pars{-1 - n}} \\[5mm] = & \bbx{\ds{1 \over t^{n + 1}\Gamma\pars{-n}}} \\ & \end{align}

• 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