For $\tau > 0$ define $\theta_{\tau}(x) = e^{\tau(x-x^{2})}$. I am curious about the asymptotics of $\widehat{\theta}_{\tau}(\tau)$, that is

$\int_{\mathbb{R}} e^{\tau(x - x^{2})}e^{-2\pi i \tau\cdot x}dx\ \sim\ ?\ \ \ \ \ \ \ \ (\tau \to +\infty)$

But I don't know how to get anything from the oscillation. Bringing absolute values inside the integral we have

$|\widehat{\theta}_{\tau}(\tau) |\leq \int_{\mathbb{R}} e^{\tau(x - x^{2})}dx \sim \sqrt{\frac{2\pi}{\tau}}e^{\tau/2}$

by Laplace's method. But shouldn't we be able to do much much better than this?

Thank you for any thoughts.