Inverse Laplace Transform of $ \left(\frac{1-s^{1/2}}{s^2}\right)^2$

Question

I found this question in my N.P Bali's Engineering Mathematics 7th Edition. I could not find any solved questions related to this.

How can I find the Inverse Laplace Transform of :

$\left({1-s^{1/2} \over s^{2}}\right)^2$

I know only to find the inverse Laplace of $s$ when the power is a whole number.

The answer to the question is given as : $\dfrac{t^3}{6} + \dfrac{t^2}{2} - \dfrac{16t^\frac{5}{2}}{15 \pi^{\frac{1}{2}}}$

I want to know how to get it.

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ With $\ds{\gamma > 0}$: \begin{align} &\int_{\gamma - \ic\infty}^{\gamma - \ic\infty} \pars{1 - {s^{1/2} \over s^{2}}}^{2}\expo{st}\,{\dd s \over 2\pi\ic} \\[3mm]&=\overbrace{\int_{\gamma - \ic\infty}^{\gamma - \ic\infty} \expo{st}\,{\dd s \over 2\pi\ic}}^{\ds{0}}\ -\ 2\color{#c00000}{\int_{\gamma - \ic\infty}^{\gamma - \ic\infty} {\expo{st} \over s^{3/2}}\,{\dd s \over 2\pi\ic}}\ +\ \overbrace{\int_{\gamma - \ic\infty}^{\gamma - \ic\infty} {\expo{st} \over s^{3}}\,{\dd x \over 2\pi\ic}}^{\ds{\half\,t^{2}}} \end{align}

\begin{align} &\color{#c00000}{\int_{\gamma - \ic\infty}^{\gamma - \ic\infty} {\expo{st} \over s^{3/2}}\,{\dd s \over 2\pi\ic}} \\[3mm]&=\lim_{\epsilon \to 0^{+}}\left\lbrack% -\int_{-\infty}^{-\epsilon}{\expo{xt} \over \pars{-x}^{3/2}\expo{3\pi\ic/2}}\, {\dd x \over 2\pi\ic} -\int_{\pi}^{-\pi}{\exp\pars{t\epsilon\expo{\ic\theta}} \over \epsilon^{3/2}\expo{3\ic\theta/2}}\,{\epsilon\expo{\ic\theta}\ic \,\dd\theta \over 2\pi\ic} \right. \\[3mm]&\phantom{\lim_{\epsilon \to 0^{+}}\bracks{.}}\left. -\int^{-\infty}_{-\epsilon}{\expo{xt} \over \pars{-x}^{3/2}\expo{-3\pi\ic/2}}\, {\dd x \over 2\pi\ic} \right\rbrack \\[3mm]&=\lim_{\epsilon \to 0^{+}}\bracks{% -\,{1 \over 2\pi}\int_{\epsilon}^{\infty}x^{-3/2}\expo{-xt}\,\dd x -{2 \over \pi}\,\epsilon^{-1/2} -\,{1 \over 2\pi}\int_{\epsilon}^{\infty}x^{-3/2}\expo{-xt}\,\dd x} \\[3mm]&=-\,{1 \over \pi}\lim_{\epsilon \to 0^{+}}\bracks{% \int_{\epsilon}^{\infty}x^{-3/2}\expo{-xt}\,\dd x - 2\epsilon^{-1/2}} \\[3mm]&=-\,{1 \over \pi}\lim_{\epsilon \to 0^{+}}\bracks{% 2\epsilon^{-1/2} - 2t\int_{\epsilon}^{\infty}x^{-1/2}\expo{-xt}\,\dd x - 2\epsilon^{-1/2}} ={2t \over \pi}\,{1 \over t^{1/2}}\int_{0}^{\infty}x^{-1/2}\expo{-x}\,\dd x \\[3mm]&={2 \over \pi}\,t^{1/2}\Gamma\pars{\half}={2 \over \pi}\,t^{1/2}\root{\pi} =\color{#c00000}{{2 \over \root{\pi}}\,t^{1/2}} \end{align} $\ds{\Gamma\pars{z}}$ is the Gamma Function and $\ds{\Gamma\pars{\half} = \root{\pi}}$

$$\color{#00f}{\large% \int_{\gamma - \ic\infty}^{\gamma - \ic\infty} \pars{1 - {s^{1/2} \over s^{2}}}^{2}\expo{st}\,{\dd s \over 2\pi\ic} =\half\,t^{2} + {2 \over \root{\pi}}\,t^{1/2}} $$

The OP changed the question after I solved it !!!.

Answer 2

The partial fraction expansion yields;

$$\left(\dfrac{1-s^{1/2}}{s^{2}}\right)^2 = -\dfrac{2}{s^{7/2}}+\dfrac{1}{s^4}+\dfrac{1}{s^3} $$

Using this table of Laplace Transforms (item $6$ and item $3$ (twice)) yields:

$$\mathscr{L}^{-1} \left(-\dfrac{2}{s^{7/2}}+\dfrac{1}{s^4}+\dfrac{1}{s^3}\right) = -\dfrac{2 \times 2^3}{1 \times 3 \times 5 ~ \sqrt{\pi}}t^{5/2} +\dfrac{t^3}{6} + \dfrac{t^2}{2} = -\dfrac{16}{15~ \sqrt{\pi}}t^{5/2} +\dfrac{t^3}{6} + \dfrac{t^2}{2}$$