Exponential decay of Feynman propagator outside the lightcone

Question

In Chapter three (I.3) of A. Zee's Quantum Field Theory in a Nutshell, the author derives the Feynman propagator for a scalar field: $$ \begin{aligned} D(x)&=\int \frac{\operatorname{d}^4 \mathbf{k}}{(2\pi)^4} \frac{e^{ikx}}{k^2-m^2+i\epsilon} \\ &=-i\int \frac{\operatorname{d}^3 \mathbf{k}}{(2\pi)^3 2\omega_k} \left[e^{-i(\omega_kt-\mathbf{k}\cdot \mathbf{x})}\theta(t)+ e^{i(\omega_kt-\mathbf{k}\cdot \mathbf{x})}\theta(-t)\right] \end{aligned} $$ where $\omega_k=\sqrt{\mathbf{k}^2+m^2}$.

Without working through the $\mathbf{k}$ integral, the behavior of the propagator for events inside and outside the light-cone can be roughtly analyzed (or so the text states): for time-like events in the future cone, e.g., $x=(t,\mathbf{x}=0)$, with $t>0$, the propagator is a sum of plane waves $$D(t,0)=-i\int \frac{\operatorname{d}^3 \mathbf{k}}{(2\pi)^3 2\omega_k} e^{-i\omega_kt}$$ Likewise, for time-like events in the past cone ($t<0$) the propagator is a sum of plane waves with the opposite phase.

Now, for space-like events, e.g., $x=(0,\mathbf{x})$, after interpreting $\theta(0)=\frac{1}{2}$ and observing the propagator allows for the exchange $\mathbf{k}\rightarrow -\mathbf{k}$, we obtain $$D(0,\mathbf{x})=-i\int \frac{\operatorname{d}^3 \mathbf{k}}{(2\pi)^3 2\sqrt{\mathbf{k}^2+m^2}} e^{-i\mathbf{k}\cdot \mathbf{x}}$$

The author then states that "...the square root cut starting at $\pm im$ leads to an exponential decay $\sim e^{-m|\mathbf{x}|}$, as we would expect." It is left to the reader to verify this as a later problem.

The question is: how can I see that the above is true, without going through the $\mathbf{k}$ integral?

Secondarily, what does "the square root cut starting at $\pm im$" mean? I know that one must supply the complex square root with a branch cut, but said branch cut must be a whole ray of the plane, not just a segment.

I have tried going through the integral; by rotating the $\mathbf{k}$ so that $\mathbf{x}$ points along the $k^3$ direction and switching to spherical coordinates ($k=|\mathbf{k}|, x=|\mathbf{x}|$) the integral becomes:

$$ \begin{aligned} D(0,\mathbf{x})&=-i\int_0^\infty \operatorname{d}k \int_0^\pi \operatorname{d}\theta \int_0^{2\pi} \operatorname{d}\varphi \left( \frac{k^2 \sin{\theta}e^{-i kx\cos{\theta}}}{(2\pi)^3 2\sqrt{k^2+m^2}} \right)\\ &=-i\int_0^\infty \operatorname{d}k \int_0^\pi \operatorname{d}\theta \left( \frac{k^2 \sin{\theta}e^{-i kx\cos{\theta}}}{(2\pi)^2 2\sqrt{k^2+m^2}} \right)\\ &=\frac{-i}{(2\pi)^2}\int_0^\infty \operatorname{d}k \left( \frac{k}{2ix\sqrt{k^2+m^2}}\right)e^{ikx}-e^{-ikx}\\ &=\frac{-i}{(2\pi)^2}\int_0^\infty \operatorname{d}k \frac{k\sin{kx}}{x\sqrt{k^2+m^2}}\sim \frac{1}{|\mathbf{x}|} \end{aligned} $$ Which is not the desired result.

Answer 1

See the Wikipedia article on the Feynman propagator. It's real-space form is: $$G_F(x,y) = \left\{\begin{array}{cc} -\frac{1}{4\pi}\delta(\tau^2) + \frac{m}{8\pi \tau} H_1^{(2)}(m\tau) & \tau^2 \ge 0 \\ -\frac{im}{4\pi^2 \sqrt{|\tau|}} K_1(m|\tau|) & \tau^2 < 0, \end{array}\right.$$ where $\tau^2 \equiv (x^0 - y^0)^2 - (\vec{x} - \vec{y})^2$, $H_1^{(2)}$ is a Hankel function and $K_1$ is a modified Bessel function of the second kind. The desired result follows directly from the asymptotic properties of $K_1$ for large arguments.

If the desired result is to explore the properties of the integral, then you may find an answer digging around in the integral representations of the modified Bessel functions.

If you are fine with a proof that the modified Bessel function is part of a propagator, though not particularly the Feynman propagator, and aren't concerned about the light cone delta function, it follows from analytic continuation of the 4-d Euclidean Green's function.

Answer 2

What I will do isn't mathematicaly valid, but I think it works a little like physical intuition: $$\oint \frac{e^{-ikx} \frac{k^2}{\sqrt{k-im}}}{\sqrt{k+im}} \approx \frac{-m^2}{\sqrt{-2im}} \oint \frac{e^{-ikx} }{\sqrt{k+im}}$$ Usin fractional calculus generalization for cauchy's integral formula: $$\frac{-m^2}{\sqrt{-2im}} \frac{2\pi i}{\Gamma(1/2)} \frac{\partial^{\frac{1}{2}} e^{-ikx}}{\partial k^{\frac{1}{2}}} \approx \frac{-m^2}{\sqrt{-2im}} \frac{2\pi i}{\sqrt{-i\pi x}} e^{-mx}$$... well... I got a $1/\sqrt{x}$ but it has the exponential...