Laplace transform of a gaussian function?

Question

How do I find the Laplace transform of this Gaussian function $f(t)=\sqrt{t}e^{-\frac{1}{t}}$.
I have from a table that $L\Big[{\frac{1}{\sqrt{t}} e^{-\frac{a^2}{4t}}}\Big ]=\sqrt{\frac{\pi}{s}}e^{-a\sqrt{s}}$ for a Gaussian Normal Distribution.
If I set $a=2$ then the formula are alike apart for $\frac{1}{\sqrt{t}}$. How do I deal with that?

Answer 1

The LT you seek may be written as

$$\begin{align} \int_0^{\infty} dt \, \sqrt{t} \, e^{-1/t} e^{-s t} &= 2 \int_{0}^{\infty} du \, u^2 \, e^{-(s u^2+1/u^2)} \\ &= 2 e^{2 \sqrt{s}} \int_0^{\infty} du \, u^2 \, e^{-\left (\sqrt{s} u + 1/u \right )^2} \\ \end{align}$$

Sub $v=\sqrt{s} u+1/u$, which implies two branches of the integration region:

$$u = \frac1{2 \sqrt{s}} \left (v \pm \sqrt{v^2-4 \sqrt{s}} \right ) $$ $$du = \frac1{2 \sqrt{s}} \left (1 \pm \frac{v}{\sqrt{v^2-4 \sqrt{s}}} \right ) dv$$

Thus, the LT we seek may be written as

$$e^{2 \sqrt{s}} \int_{\infty}^{2 s^{1/4}} dv \frac1{\sqrt{s}} \left (1 - \frac{v}{\sqrt{v^2-4 \sqrt{s}}} \right ) \left (\frac{v - \sqrt{v^2-4 \sqrt{s}}}{2 \sqrt{s}} \right )^2 e^{-v^2} \\ + e^{2 \sqrt{s}} \int_{2 s^{1/4}}^{\infty} dv \frac1{\sqrt{s}} \left (1 + \frac{v}{\sqrt{v^2-4 \sqrt{s}}} \right ) \left (\frac{v + \sqrt{v^2-4 \sqrt{s}}}{2 \sqrt{s}} \right )^2 e^{-v^2} $$

This looks horrendous but it simplifies considerably. Further, if we sub $v=2 s^{1/4} \cosh{w}$ and perform a little algebra, we get for the LT:

$$2 s^{-3/4} e^{-2 \sqrt{s}} \int_0^{\infty} dw \, \cosh{w} (1+4 \sinh^2{w}) e^{-4 \sqrt{s} \sinh^2{w}}$$

which again simplifies considerably to

$$2 s^{-3/4} e^{-2 \sqrt{s}} \int_0^{\infty} dp \, (1+4 p^2) e^{-4 \sqrt{s} p^2} $$

And now we are at a standard integral for which no further elaboration should be needed. The result we seek is

$$\int_0^{\infty} dt \, \sqrt{t} \, e^{-1/t} \, e^{-s t} = \frac12 \sqrt{\pi} e^{-2 \sqrt{s}} \left (s^{-3/2} +2 s^{-1} \right ) $$

Answer 2

Let $g(t)=\frac{1}{\sqrt{t}} e^{-\frac{1}{t}}$. Then $f(t)=tg(t)$.

Now, by the frequency-domain derivative property of the Laplace transform, $$L(f(t))=L(tg(t))=-\frac{d}{ds}(L(g)(s)) =-\frac{d}{ds}\left(\sqrt{\frac{\pi}{s}}e^{-2\sqrt{s}}\right) \\= -\sqrt{\pi}\left(-\frac{s^{-3/2}e^{-2\sqrt{s}}}{2}+s^{-1/2}\cdot\frac{-2e^{-2\sqrt{s}}}{2\sqrt{s}})\right) =\frac{\sqrt{\pi} e^{-2 \sqrt{s}}}{2} \left (s^{-3/2} +2 s^{-1} \right ).$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$

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\begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}\root{t}\expo{-1/t} \expo{-st}\dd t} \\[5mm] \stackrel{st\ \mapsto\ t}{=}\,\,\,& {1 \over s^{3/2}}\int_{0}^{\infty} \exp\pars{-t - {\bracks{\color{red}{2\root{s}}}^{2} \over 4t}}\, {\dd t \over t^{\color{red}{-3/2} + 1}} \\[5mm] = &\ {1 \over s^{3/2}}\bracks{\on{K}_{\color{red}{-3/2}}\,\pars{\color{red}{2\root{s}}} \over \pars{1/2}\pars{\color{red}{2\root{s}}/2}^{\color{red}{-3/2}}} = {2 \over s^{3/4}}\on{K}_{3/2}\,\pars{2\root{s}} \end{align}

See this link. $\ds{\on{K}_{\nu}}$ is a Modified Bessel Function. Note that $\ds{\on{K}_{\nu}\pars{z} = \on{K}_{-\nu}\pars{z}}$

However, $\ds{K_{3/2}\pars{z} = \root{\pi \over 2}\expo{-z}\ {z + 1 \over z^{3/2}}}$. See ${\bf\color{black}{10.2.17}}$ in A & S Table.

Then, \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}\root{t}\expo{-1/t} \expo{-st}\dd t} = {2 \over s^{3/4}}\bracks{\root{\pi \over 2}\expo{-2\root{s}} {2\root{s} + 1 \over \pars{2\root{s}}^{3/2}}} \\[5mm] = & \bbx{{\root{\pi} \over 2}\pars{s^{-3/2} + 2s^{-1}}\expo{-2\root{s}}} \\ & \end{align}