For $t > 0$ let $\mu_t$ be the measure on $\R^+$ with density
$$
x\mapsto\frac{t\,\exp(-t^2/4x)}{\sqrt{4\pi x^3}}~.
$$
Show that the Laplace transform $\o_t(y)\colon=\int_0^\infty e^{-xy}\,\mu_t(dx)$ is given by $\o_t(y)=e^{-t\sqrt y}$. 2. Conclude that $\mu_s*\mu_t=\mu_{s+t}$. The measures $\mu_t$ are called $1/2$-stable probability measures on $\R^+$ ($1/2$ refers to the square root of $y$ in $\o_t$).
Put $a=t^2/4$ and $\a=\sqrt{a/y}$. By successive substitution $x\to\a s$, $s\to u^2$
and $u\to1/v$ we get:
\begin{eqnarray*}
\int_0^\infty x^{-3/2}e^{-a/x-yx}\,dx
&=&(y/a)^{1/4}\int_0^\infty s^{-3/2}e^{-\sqrt{ay}(1/s+s)}\,ds\\
&=&2e^{-2\sqrt{ay}}(y/a)^{1/4}
\int_0^\infty u^{-2}e^{-\sqrt{ay}(1/u-u)^2}\,du\\
&=&2e^{-2\sqrt{ay}}(y/a)^{1/4}
\int_0^\infty e^{-\sqrt{ay}(v-1/v)^2}\,dv\\
&=&2e^{-2\sqrt{ay}}(y/a)^{1/4}
\int_0^\infty e^{-\sqrt{ay}v^2}\,dv
=\sqrt{\pi/a}e^{-2\sqrt{ay}}
=\sqrt{4\pi}t^{-1}e^{-t\sqrt{y}}~.
\end{eqnarray*}
2. This follows from $\o_s(y)\o_t(y)=\o_{s+t}(y)$ and the facts that 1. the Laplace transform uniquely determines finite Borel measures on $\R^+$ and 2. the Laplace transform of the convolution is the product of the transforms.