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I know that the result of the inverse laplace transform of the following is

$$\mathcal{L}^{-1}\left\{ \frac{1}{s^4}\right\} = \frac{1}{6}\cdot t^3$$

However I just can't seem to figure out where the fraction comes from. $$\frac{1}{6}$$

Can anyone explain to me in short? I got the feeling that I'm missing out on something really stupid.. Thanks in advance.

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$\mathcal{L}(t^3)=\frac{3!}{s^4}=\frac{6}{s^4}$ however we just have a $\frac{1}{s^4}$ so we need to get rid of that 6, to do this we divide by 6. So therefore $\mathcal{L}^{-1}\left(\frac{1}{s^4}\right )=\frac{t^3}{6}$

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Hint. One may observe that by differentiating $$ \int_0^\infty e^{-st}dt=\frac1s,\qquad s>0, $$ $n$ times one has $$ \left(\int_0^\infty e^{-st}dt\right)^{(n)}=\int_0^\infty (-1)^n t^ne^{-st}dt,\qquad s>0, $$ and $$ \left(\frac1s\right)^{(n)}=\left(s^{-1}\right)^{(n)}=(-1)^n \cdot n\cdot(n-1) \cdots 1 \cdot s^{-(n+1)}=(-1)^n\cdot n!\cdot s^{-(n+1)},\qquad s>0, $$ giving

$$ \int_0^\infty \frac{t^n}{\color{red}{n!}}\:e^{-st}dt=\frac{1}{s^{\color{red}{n+1}}},\quad s>0,\quad n=0,1,2,\cdots. $$

The OP question corresponds to $s=1$, $n=3$.

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    $\begingroup$ Great answer! (+1) $\endgroup$ – R.W Mar 29 '17 at 22:31
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}{\expo{ts} \over s^{4}}\,{\dd s \over 2\pi\ic} = \bracks{s^{3}}\expo{ts} = {t^{3} \over 3!} \end{align} It is equivalent to $$ {1 \over \pars{4 - 1}!}\,\lim_{s \to 0}\partiald[4 - 1]{\expo{ts}}{s} = {1 \over 3!}\,\lim_{s \to 0}\partiald[3]{\expo{ts}}{s} = {t^{3} \over 3!} $$

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