$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
A 'qualitative approach':
\begin{align} &{1 \over 4\pi^{2}}\int_{m}^{\infty}\root{E^{2} - m^{2}}\expo{-\ic Et}\,\dd E \,\,\,\stackrel{E\ =\ m + \varepsilon^{2}}{=}\,\,\, {\expo{-\ic mt} \over 2\pi^{2}}\int_{0}^{\infty}\root{\varepsilon^{2} + 2m}\expo{-\ic \varepsilon^{2} t}\varepsilon^{2}\,\dd\varepsilon \\[5mm] \stackrel{\varepsilon/\root{2m}\ \mapsto\ \varepsilon}{=}\,\,\, & \expo{-\ic mt}\,{2m^{2} \over \pi^{2}} \int_{0}^{\infty}\root{\varepsilon^{2} + 1} \exp\pars{-2mt\varepsilon^{2}\,\ic}\varepsilon^{2}\,\dd\varepsilon \end{align}
As $\ds{t \to \infty}$, the 'main contribution' arises ftom values of $\ds{\varepsilon \lesssim 1/\pars{2mt}^{1/2}}$ such that
\begin{align} &{1 \over 4\pi^{2}}\int_{m}^{\infty}\root{E^{2} - m^{2}}\expo{-\ic Et}\,\dd E \sim \expo{-\ic mt}\,{2m^{2} \over \pi^{2}} \int_{0}^{1/\pars{2mt}^{1/2}}\varepsilon^{2}\,\dd\varepsilon = \expo{-\ic mt}\,{2m^{2} \over \pi^{2}}\bracks{\pars{2mt}^{-3/2} \over 3} \\[5mm] = &\ \color{#f00}{{\root{2} \over 6\pi^{2}}}\,\root{m}t^{-3/2}\expo{-\ic mt} \end{align}
The $\color{#f00}{prefactors}$ can only be determined from the 'exact evaluation' ( see the @Brightsun fine answer ) but the qualitative evaluation yields the correct asymptotic behaviours $\bbx{\ds{\root{m}t^{-3/2}\expo{-\ic mt}}}$