<s>[INST] 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? 
 [/INST] 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

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

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