How can the electric susceptibility be 0 in this case?

I'm given the frequency-dependent electric susceptibility from the Lorentz model:

$$\chi_e(\omega) = \frac{\omega_p^2}{\omega_0^2-\omega^2-i2\gamma\omega}$$

And being asked to find the temporal dependence through:

$$\tilde \chi_e(\tau)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\chi_e(\omega)\exp(-i\omega\tau)d\omega$$

Through Jordan's lemma (note that we would have to reverse the integration), we can decompose this integral into two others, one of the being a closed loop and the other being a semi-circle on the superior complex plane. This semi-circle integral is $$0$$ as through Jordan's lemma.

Now, $$\chi_e$$ has poles on the inferior side of the complex plane. So if we integrate using a path on the superior side of the complex plane, we have no poles and thus through the residue theorem, the other integral is also $$0$$. Meaning:

$$\tilde \chi_e(\tau)=0$$

But how can this be? Any ideas?

$$\chi(\tau)$$ is causal. It should be zero for $$\tau<0$$. It will be non-zero fot $$\tau>0$$ as Jordan's lemma requires us to close above or below depending on the sign of $$\tau$$.