Inverse Laplace Transform of $ \left(\frac{1-s^{1/2}}{s^2}\right)^2$ 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. 
 A: $\newcommand{\+}{^{\dagger}}
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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 !!!.

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