Complicated Laplace Transform I have found the following Laplace Transform in a list
$$\int\limits_0^{\infty}e^{-st}\frac{e^{-u^2/4t}}{\sqrt{\pi t}}dt = \frac{e^{-u\sqrt{s}}}{\sqrt{s}}.$$
I am wondering how to prove this? I tried to do some substitutions for the integral, but nothing worked. Can someone explain it to me?
I appreciate any help!
 A: First, enforcing the substitution $t\to t^2$ in the integral of interest yields
$$\begin{align}
\int_0^\infty e^{-st}\,\frac{e^{-u^2/4t}}{\sqrt{\pi t}}\,dt&=\frac{2}{\sqrt\pi}\int_0^\infty e^{-s\left(t^2+u^2/4st^2\right)}\,dt\\\\
&=\frac{2e^{-u/\sqrt{s}}}{\sqrt\pi}\int_0^\infty e^{-s\left(t-u/2\sqrt{s}t\right)^2}\,dt \tag 1
\end{align}$$

Second, enforcing the substitution $t\to \sqrt{\frac{u}{2\sqrt{s}}}\,t$ into the right-hand side of $(1)$ and letting $\alpha =\sqrt{\frac{u}{2\sqrt{s}}}$ reveals
$$\frac{2e^{-u/\sqrt{s}}}{\sqrt\pi}\int_0^\infty e^{-s\left(t-u/2\sqrt{s}t\right)^2}\,dt=\frac{2e^{-u\sqrt{s}}}{\sqrt\pi}\alpha \int_0^\infty e^{-\alpha^2 s\left(t-1/t\right)^2}\,dt \tag 2$$

Third, enforcing the substitution $t\to 1/t$ in the integral on the right-hand side of $(3)$, we obtain
$$\frac{2e^{-u\sqrt{s}}}{\sqrt\pi}\alpha \int_0^\infty e^{-\alpha^2 s\left(t-1/t\right)^2}\,dt =\frac{2e^{-u\sqrt{s}}}{\sqrt\pi}\alpha \int_0^\infty e^{-\alpha^2 s\left(t-1/t\right)^2}\,\left(\frac{1}{t^2}\right)\,dt \tag 3$$

Next, adding $(2)$ and $(3)$ and dividing by $2$, we find that
$$\begin{align}
\frac{2e^{-u\sqrt{s}}}{\sqrt\pi}\alpha \int_0^\infty e^{-\alpha^2 s\left(t-1/t\right)^2}\,dt&=\frac{e^{-u\sqrt{s}}}{\sqrt\pi}\alpha \int_0^\infty e^{-\alpha^2 s\left(t-1/t\right)^2}\,\left(1+\frac{1}{t^2}\right)\,dt\\\\
&=\frac{e^{-u\sqrt{s}}}{\sqrt\pi}\alpha \int_{-\infty}^{\infty}e^{-\alpha^2 s u^2}\,du\\\\
&=\frac{e^{-u\sqrt{s}}}{\sqrt{s}} \tag 4
\end{align}$$

Finally, substituting $(4)$ into $(1)$ we find 
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty e^{-st}\,\frac{e^{-u^2/4t}}{\sqrt{\pi t}}\,dt=\frac{e^{-u\sqrt{s}}}{\sqrt{s}}}$$
as was to be shown!
A: It is probably easier to prove first that if $A,B\in\mathbb{R}^+$ we have:
$$ \int_{0}^{+\infty}\exp\left(-A^2 x^2-\frac{B^2}{x^2}\right)\,dx = \frac{\sqrt{\pi}}{2A} e^{-2AB}\tag{1} $$
that is the same as proving that for every $C\in\mathbb{R}^+$ we have:
$$ \int_{0}^{+\infty}\exp\left(-x^2-\frac{C^2}{x^2}\right)\,dx = \frac{\sqrt{\pi}}{2}e^{-2C}\tag{2} $$
or:
$$ \int_{0}^{+\infty}\exp\left[-\left(x-\frac{C}{x}\right)^2\right]\,dx = \frac{\sqrt{\pi}}{2}.\tag{3} $$
However, $(3)$ is a trivial consequence of Glasser's master theorem.
So you just have to use the substitution $t=v^2$, then exploit $(1)$.
