Deduce that $$\int_{-\pi}^{\pi} e^{-i\frac{\theta^2}{6t^2}} \frac{P(e^{-t}e^{i\theta})}{P(e^{-t})} \mathrm{d}\theta = O(t^{3/2}) \text{ when } t \text{ tends to } 0^+.$$
Deduce that
$$\int_{-\pi}^{\pi} e^{-i\frac{\theta^2}{6t^2}} \frac{P(e^{-t}e^{i\theta})}{P(e^{-t})} \mathrm{d}\theta = O(t^{3/2}) \text{ when } t \text{ tends to } 0^+.$$