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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.

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up vote 9 down vote accepted

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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Thanks - I think I need to spend some time studying special functions. I don't suppose you'd be able to recommend an introductory text on the subject? – Skatche Dec 10 '11 at 17:12
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@Skatche: For Bessels only, you'll want to look into Watson. For special functions in general, either of Carlson, Temme, or Andrews/Askey/Roy should do. – J. M. Dec 10 '11 at 18:05
    
@Skatche this could prove a useful quick reference as well! – Brightsun Jan 20 at 0:04
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@Brightsun thanks for that. Adjusted – Sasha Jan 21 at 13:42
    
@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... – Brightsun Jan 21 at 13:45

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, consider a of circle with centre at $1$, radius $R$ and arc going from $R$ to $1-iR$; 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 $$ \int_{1}^{+\infty}\frac{\sqrt{\rho^2-1}}{\rho^2}e^{-imt\rho}d\rho = \int_0^{+\infty}\frac{\sqrt{y^2+i2y}}{y^2+2iy-1}e^{-mt(y+i)}dy. $$ 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{s^2(mt)^{-1}+i2s}\,e^{-s}ds\\ &\approx e^{-imt}\sqrt{m}t^{-3/2},\text{ as }t\to\infty. \end{align}

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