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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.

3 Answers 3

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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 $\mathrm{e}^{-i m t} \frac{\mathrm{e}^{-i 3 \pi/4}}{8} \sqrt{m} (\pi t)^{-3/2}$

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    @Sasha Sorry! I deleted the comment as I thought it was unnecessary now. By the way, I am not entirely sure that the coefficient in front of my asymptotic expansion is precisely the same as yours...2016-01-21
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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. $ enter image description here 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 $t\to\infty$.

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