# 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.

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ $$\ds{\bbox[5px,#ffd]{}}$$ 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}
• @GabrielePrivitera Yes, $\alpha$ is the one you wrote in your comment. I didn’t see any time $t$ in your question. Thanks. Commented Oct 6, 2020 at 21:01