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.
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