I was playing around with the Faddeeva function $$\omega(z) = e^{-z^2} \left( 1 + \dfrac{2i}{\sqrt{\pi}} \int\limits_{0}^{z} e^{-t^2} dt \right)$$ and noticed numerically that for $z = x+iy$, and $x>0,y>0$, the function is always positive, see the picture below:
One can inspect an explicit integral representation of it's imaginary part: $$\text{Im} \:\omega(x+iy) = \dfrac{1}{\sqrt{\pi}} \int\limits_{0}^{\infty} e^{-t^2/4} e^{-yt}\sin(xt)dt$$, and can vaguely explain it by the fact that the dominant contribution comes from small $t$, where $\sin(xt)$ is positive for $x>0$, but is there any strict way to prove that?
Sorry, turned out to be quite easy.
One can split the whole integration interval over $t$ into: $$ \int\limits_{0}^{\infty} \ldots dt = \sum\limits_{j=0}^{\infty} \int\limits_{j \frac{\pi}{x}}^{(j+1) \frac{\pi}{x}} \ldots dt, $$ and by direct comparison of neighboring pairs one can see (for the first pair $j=0$, and $j=1$):
$$ I_{1} = \int\limits_{0}^{\frac{\pi}{x}} e^{-t^2/4} e^{-ty} \sin(xt) dt, \\ I_{2} = \int\limits_{\frac{\pi}{x}}^{2\frac{\pi}{x}} e^{-t^2/4} e^{-ty} \sin(xt) dt = -\int\limits_{0}^{\frac{\pi}{x}} e^{-(t + \pi/x)^2/4} e^{-(t+\pi/x)y} \sin(xt) dt.$$
It is clear that $|I_1|>|I_2|$ for $x>0,y>0$, while $I_1>0, I_2<0$. Therefore $I_1+I_2>0$, and the same argument holds for all such pairs, so, the overall integral is positive.