Compute the inverse Laplace transform of $\mathrm{e}^{\left( i- 1\right)\sqrt{ms/h}\ x}$ 
*

*I'm trying to solve the Schrödinger's equations using the method of Laplace transforms, but I can't figure out the inverse transform once I've reached this point.

*I'm a high school student, so I'm self-taught and I haven't much experience with the theory behind the Inverse Laplace transform.

*I'm sorry for the bad typing of my formulas but this is my first post.

 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
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 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$
Basically, you want to compute
$\ds{\bbox[5px,#ffd]{\left.\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}\!
\expo{\alpha x\!\root{s}}\!\!\expo{ts}
{\dd s \over 2\pi\ic}\,\right\vert_{\ t\ >\ 0}}}$
where
$\ds{\alpha \equiv \pars{\ic - 1}\root{m \over h} \mbox{is a}\ constant}$.

\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}\!
\expo{\alpha x\!\root{s}}\!\!\expo{ts}
{\dd s \over 2\pi\ic}\,\right\vert_{\ t\ >\ 0}}\\[5mm] = &\
-\int_{-\infty}^{0}\expo{\alpha x\root{-s}\ic}\expo{ts}
\,{\dd s \over 2\pi\ic} -
\int_{0}^{-\infty}\expo{\alpha x\root{-s}\pars{-\ic}}\expo{ts}
\,{\dd s \over 2\pi\ic}
\\[5mm] = &\
-\int_{0}^{\infty}\expo{\alpha x\root{s}\ic}\expo{-ts}
\,{\dd s \over 2\pi\ic} +
\int_{0}^{\infty}\expo{\alpha x\root{s}\pars{-\ic}}\expo{-ts}
\,{\dd s \over 2\pi\ic}
\\[5mm] = &\
-\,{1 \over \pi}\int_{0}^{\infty}\sin\pars{\alpha x\root{s}}
\expo{-ts}\,\dd s
\\[5mm] = &\ \stackrel{s\ \mapsto\ s^{2}}{=}\,\,\,
-\,{2 \over \pi}\int_{0}^{\infty}s\sin\pars{\alpha xs}
\expo{-ts^{2}}\,\dd s
\\[5mm] = &\
\bbx{-\,{\alpha \over 2\root{\pi}}
\,{x\exp\pars{-\alpha^{2}x^{2}/\bracks{4t}} \over t^{3/2}}\,,\qquad \Re\pars{x} > 0} \\ &
\end{align}
