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. 
 A: 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,
$$L^{-1}\{s^n\}(t) =  \delta^{(n)}(t)$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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}
