Would like help with a contour integral.

The second integral on the RHS is

$$i \rho \int_0^{-\pi} d\theta \, e^{i \theta} \frac{e^{-i t \rho e^{i \theta}}}{\rho e^{i \theta}+i\epsilon} = i \rho \int_0^{-\pi} d\theta \, e^{i \theta} \frac{e^{-i t \rho \cos{\theta}} e^{t \rho \sin{\theta}}}{\rho e^{i \theta}+i \epsilon}$$

The magnitude of the above integral is bounded by

$$\int_0^{-\pi} d\theta \, e^{t \rho \sin{\theta}} = \int_0^{\pi} d\theta \, e^{-t \rho \sin{\theta}}$$

Use symmetry and $\sin{\theta} \ge 2 \theta/\pi$, and the integral is further bounded by

$$ 2\int_0^{\pi} d\theta \, e^{-2 t \rho \theta/\pi} \le \frac{\pi}{t \rho}$$

As $\rho \to \infty$, the integral therefore vanishes as $\pi/(t \rho)$ when $t \gt 0$.

When $t \lt 0$, however, the above bounds do not apply; rather, we must close above the real axis. because there are no poles there, the integral is zero. Thus, the step function.