Consider the integral $$k(\beta,s)=\int_0^\infty e^{-s u}\frac{e^{-\beta^2/4u}}{\sqrt{4\pi u}} \,dt,$$ the Laplace transform of $K(\beta,u)=\frac{1}{\sqrt{4\pi u}}e^{-\beta^2/4u}$. The significance of $K(\beta,u)$ is that it's the 1D heat kernel with position $\beta$ and time $u$. That is, it satisfies the heat equation $$\frac{\partial K}{\partial u}=\frac{\partial^2 K}{\partial \beta^2}$$ for all $u>0$ and converges to a Dirac delta $\delta(\beta)$ as $u\to 0$. (One may check that, for all $u>0$, $K(\beta,u)$ defines a Gaussian PDF with zero mean and variance $\sigma^2=u$.)
To make use of this fact, we take the Laplace transform of both sides of the heat equation:
\begin{align}
\int_0^\infty e^{-s u} \frac{\partial^2}{\partial\beta^2}K(\beta,u)\,du
&=\frac{\partial^2}{\partial\beta^2}\int_0^\infty e^{-s u} K(\beta,u)\,du=\frac{\partial^2 k}{\partial\beta^2}\\
&=\int_0^\infty e^{-s u} \frac{\partial}{\partial u}K(\beta,u)\,du\\
&=\left[e^{-s u}K(\beta,u)\right]^{\infty}_{u=0}+\int_0^\infty se^{-s u} K(\beta,u)\,du\\
&=-\delta(\beta)+s k(\beta,s)
\end{align}
where in the second-to-last line we have integrated by parts and subsequently used the boundary condition $K(\beta,u)\to \delta(\beta)$ as $u\to 0$. Regarding $s$ now as a parameter, we have the ODE $k''(\beta)=s\, k(\beta)-\delta(\beta)$. This may be solved to give $k(\beta,s)=\frac{1}{2}e^{-|\beta|\sqrt{s}}$. (Briefly: For $\beta\neq 0$, the equation is just $k''(\beta)=sk(\beta)$ with $k(\beta)$ vanishing at infinity; to account for the delta function, integrate both sides of the ODE from $-\epsilon$ to $\epsilon$ and take $\epsilon\to 0^+$ to obtain $k'(0^+)-k'(0^-)=1$.) Taking $s\to 1$ and assuming $\beta>0$, we obtain the announced result $2k(\beta,1)=e^{-\beta}$.
