In some example, I saw the following bound: $$\int_1^\infty e^{-2t s^2} ds \leq e^{-2t}\sqrt{\pi/(2t)}.$$
I would like to prove this for my own understanding but ran into some trouble obtaining the factor $e^{-2t}$. What I know how to show is that \begin{align*} \int_1^\infty e^{-2ts^2}ds & \leq \int_{-\infty}^\infty e^{-2t s^2} ds\\ & = \sqrt{2\pi/(4t)},\\ & = \sqrt{\pi/(2t)}, \end{align*} using facts about the normalization constant of a Gaussian PDF. But it is not immediate to me how to get the, presumably better, bound including $e^{-2t}$. Any hints?
Hint: Try the substitution $s= \sqrt{1+u^2}$ and make use of the bound $\frac{ds}{du} \leq 1$