Compute this Laplace and inverse Laplace transform

Question

I know that $$ \mathcal{L}(\frac{e^{-(x-\eta)^{2}/4t}}{2\sqrt{\pi t}}) =\frac{e^{-\sqrt{s}|x-\eta|}}{2\sqrt{s}}$$ I am interested in a proof via showing that $$ \int \frac{e^{-(x-\eta)^{2}/4t}}{2\sqrt{\pi t}}e^{st}dt=\frac{e^{-\sqrt{s}|x-\eta|}}{2\sqrt{s}}$$ and via $$ \int_{c-i\infty}^{c+i\infty}\frac{e^{-\sqrt{s}|x-\eta|}}{2\sqrt{s}}e^{st}ds=\frac{e^{-(x-\eta)^{2}/4t}}{2\sqrt{\pi t}} $$ Just some general indications would be enough.

Answer 1

Let us work on the following integral $$ \int_0^\infty e^{-a^2x^2-\frac{b^2}{x^2}}dx. $$ Under the change of variable $x=\gamma t$ ($\gamma>0$ will be determined later), one has \begin{eqnarray} \int_0^\infty e^{-a^2x^2-\frac{b^2}{x^2}}dx&=&\gamma\int_0^\infty e^{-a^2\gamma^2t^2-\frac{b^2}{\gamma^2t^2}}dt. \end{eqnarray} Letting $a^2\gamma^2=\frac{b^2}{\gamma^2}$ gives $\gamma=\sqrt{\frac{b}{a}}$ and hence \begin{eqnarray} \int_0^\infty e^{-a^2x^2-\frac{b^2}{x^2}}dx&=&\gamma\int_0^\infty e^{-a^2\gamma^2t^2-\frac{b^2}{\gamma^2t^2}}dx\\ &=&\gamma\int_0^\infty e^{-a^2\gamma^2t^2-\frac{b^2}{\gamma^2t^2}}dt\\ &=&\gamma\int_0^\infty e^{-ab(t^2+\frac{1}{t^2})}dt\\ &=&\gamma e^{-2ab}\int_0^\infty e^{-ab(t-\frac{1}{t})^2}dt\\ &=&\sqrt{\frac{b}{a}}e^{-2ab}\frac{\sqrt\pi}{2\sqrt{ab}}\\ &=&\frac{\sqrt\pi}{2a}e^{-2ab}. \end{eqnarray} Using this, one has \begin{eqnarray} \mathcal{L}(\frac{e^{-(x-\eta)^{2}/4t}}{2\sqrt{\pi t}}) &=&\int_0^\infty e^{-st}\frac{e^{-(x-\eta)^{2}/4t}}{2\sqrt{\pi t}}dt\\ &=&\frac{1}{\sqrt{\pi}}\int_0^\infty e^{-st^2-(x-\eta)^{2}/4t^2}dt\\ &=&\frac{e^{-\sqrt{s}|x-\eta|}}{2\sqrt{s}} \end{eqnarray}