Laplace transform of a gaussian function? 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? 
 A: 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

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

A: 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, 
$$L(f(t))=L(tg(t))=-\frac{d}{ds}(L(g)(s))
=-\frac{d}{ds}\left(\sqrt{\frac{\pi}{s}}e^{-2\sqrt{s}}\right)
\\=
-\sqrt{\pi}\left(-\frac{s^{-3/2}e^{-2\sqrt{s}}}{2}+s^{-1/2}\cdot\frac{-2e^{-2\sqrt{s}}}{2\sqrt{s}})\right)
=\frac{\sqrt{\pi} e^{-2 \sqrt{s}}}{2} \left (s^{-3/2} +2 s^{-1}  \right ).$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{}}$

\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}
