homework and exercises - QFT Propagator across spacelike separation

I have this general formula for the propagator.

$$D(x)=-i\iiint\dfrac{ d^3k}{\big( 2\pi \big)^32\omega_k}[e^{-i[\omega_kt-\vec k\cdot\vec x]}\Theta(t)+e^{i[\omega_kt-\vec k\cdot\vec x]}\Theta(-t)]$$

I am supposed to verify that it decays exponentially for spacelike separation. To impose this condition I use $t=0$ and $x>0$. The symbols $\vec k$ and $\vec x$ are 3-vectors and $\omega_k=+\sqrt{\vec k^2+m^2}$. Also, for the Heaviside step function $\Theta(0)=0.5$.

This is problem 1.3.1 in Zee's QFT in a Nutshell book, and the solution to this problem is in the back of the book. He writes that the first step should look like

$$D(x)=-i\iiint\dfrac{ d^3k}{\big( 2\pi \big)^32\sqrt{\vec k^2+m^2}}e^{-i\vec k\cdot\vec x}$$

but when I plug it in the values for $x$ and $\Theta$, I get

$$D(x)=-i\iiint\dfrac{ d^3k}{\big( 2\pi \big)^32\sqrt{\vec k^2+m^2}}\dfrac{1}{2}[e^{i\vec k\cdot\vec x}+e^{-i\vec k\cdot\vec x}]=-i\iiint\dfrac{ d^3k}{\big( 2\pi \big)^32\sqrt{\vec k^2+m^2}}\cos(\vec k\cdot\vec x)$$

What is the reasoning behind throwing out one of the exponentials instead of using both of them? I haven't done my integral yet, so I don't know if it still produces exponential decay, but it seems like won't, so I want to know what the discrepancy is before I start.