Oscillatory integral giving me the willies So now that my term's over, I've been brushing up on my quantum field theory, and I came across the following line in my textbook without any justification:
$$\frac{1}{4\pi^2}\int_m^{\infty}\sqrt{E^2-m^2}e^{-iEt}dE \sim e^{-imt}\text{ as }t\to\infty$$
Well, I can see intuitively that if most of the integral cancels out, the main contribution will be from the region $E\approx m$, since (under a coordinate transformation) that's the region of stationary phase.  But I'm a mathematician, dangit, not a physicist, and I want this to be rigourous.
The Riemann-Lebesgue lemma, if I'm not mistaken, doesn't apply since $\sqrt{E^2-m^2}$ is unbounded as $E\to\infty$, and it certainly isn't $L^1$.  And I guess I could shift the path of the integral off the real axis in the complex plane, but I don't see why that would be the right way to take the integral.  The whole thing is giving me the heebie-jeebies, and I was hoping one of you folks could assuage my fears.
 A: Considering the integrand as the Fourier transform of a tempered distribution, it makes sense then to write
$$
\int_{m}^{+\infty}\sqrt{E^2-m^2}e^{-iEt}dE = -\frac{\partial^2}{\partial t^2}\int_{1}^{+\infty}\frac{\sqrt{\rho^2-1}}{\rho^2}e^{-imt\rho}d\rho.
$$

Now, in the complex plane, let us consider the contour in the figure: a quarter of circle centered at $1$ of radius $R$ with arc going from $R$ to $1-iR$ and a small indent around $1$. Integrating 
$$
f(z)=\frac{\sqrt{z^2-1}}{z^2}e^{-imtz}
$$
along such a contour and choosing the branch cut from $-1$ to $+1$, we get a vanishing contribution (by Jordan's lemma) from the arc and hence
$$
 \int_{1}^{+\infty}\frac{\sqrt{\rho^2-1}}{\rho^2}e^{-imt\rho}d\rho =
\int_0^{+\infty}\frac{\sqrt{y^2+i2y}}{(1-iy)^2}e^{-imt(1-iy)}dy.
$$ 
Differentiating twice, by the above consideration, 
\begin{align}
\int_{m}^{+\infty}\sqrt{E^2-m^2}e^{-iEt}dE
&=m^2\int_0^{+\infty}\sqrt{y^2+i2y} \, e^{-mt(y+i)}dy
\end{align}
and rescaling $s=mty$
\begin{align}
\int_{m}^{+\infty}\sqrt{E^2-m^2}e^{-iEt}dE
&= e^{-imt}\sqrt{m}t^{-3/2} \int_0^{+\infty} \sqrt{\frac{s^2}{mt}+i2s}\,e^{-s}ds\\
&= e^{-imt}\sqrt{m}t^{-3/2} \left( \sqrt\frac{i\pi}{2} + O\left(t^{-1}\right) \right),
\end{align} 
asymptotically as $t\to\infty$.
A: Clearly the integral as stated diverges, so one needs to regularize it. To that end, consider $t$ complex with small negative imaginary part, to ensure that it converges at $E\to\infty$.
The integral directly matches the following integral representation of the Bessel function of the second kind:
$$
   K_{\nu }(z)=\frac{\sqrt{\pi } z^{\nu } }{2^{\nu } \, \Gamma \left(\nu
   +\frac{1}{2}\right)} \, \int_1^{\infty }
   \left(t^2-1\right)^{\nu -\frac{1}{2}} e^{-z t} \, \mathrm{d}t
$$
valid for $\mathfrak{Re}(\nu) > -\frac{1}{2}$ and $\mathfrak{Re}(z) > 0$.
Thus:
$$
  \frac{1}{4 \pi^2} \int_m^\infty \mathrm{e}^{-i t \mathcal{E}} \sqrt{ \mathcal{E}^2-m^2} \mathrm{d} \mathcal{E} = -i \frac{m}{4 \pi^2 t} K_1\left( i m t \right)
$$
By means of regularization we proclaim that integral equal to the rhs even for real $t$. Expanding:
$$
  -i \frac{m}{4 \pi^2 t} K_1\left( i m t \right) = -i \frac{m}{8 \pi t} H_1(m t) + i \frac{m}{4 \pi t} \operatorname{sign}(t) J_1(m t)  
$$
This allows to conclude that for $t \to +\infty$, the expression is proportional to $\mathrm{e}^{-i m t} \frac{\mathrm{e}^{-i 3 \pi/4}}{8} \sqrt{m} (\pi t)^{-3/2}$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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 $\bbx{\ds{\root{m}t^{-3/2}\expo{-\ic mt}}}$

