most recent 30 from http://mathoverflow.net 2013-05-19T04:59:28Z http://mathoverflow.net/feeds/question/116901 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116901/asymptotics-of-a-one-parameter-family-of-schwartz-functions Michael Tinker 2012-12-20T19:58:59Z 2012-12-20T22:30:36Z

<p>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</p> <p>$\int_{\mathbb{R}} e^{\tau(x - x^{2})}e^{-2\pi i \tau\cdot x}dx\ \sim\ ?\ \ \ \ \ \ \ \ (\tau \to +\infty)$</p> <p>But I don't know how to get anything from the oscillation. Bringing absolute values inside the integral we have</p> <p><code>$|\widehat{\theta}_{\tau}(\tau) |\leq \int_{\mathbb{R}} e^{\tau(x - x^{2})}dx \sim \sqrt{\frac{2\pi}{\tau}}e^{\tau/2}$</code></p> <p>by Laplace's method. But shouldn't we be able to do much much better than this?</p> <p>Thank you for any thoughts.</p>

http://mathoverflow.net/questions/116901/asymptotics-of-a-one-parameter-family-of-schwartz-functions/116930#116930 Bazin 2012-12-20T22:30:36Z 2012-12-20T22:30:36Z

<p>$$ I(\tau)=\int_{\mathbb R}e^{-\pi\frac{\tau}{\pi} x^2}e^{-2i\pi \tau x (1-\frac{1}{2i\pi})}dx= (\frac{\pi}{\tau})^{1/2}e^{-\pi\frac{\pi}{\tau} \tau^2 (1-\frac{1}{2i\pi})^2}= (\frac{\pi}{\tau})^{1/2}e^{-{\pi^2\tau} (1-\frac{1}{2i\pi})^2}, $$ so that $$ I(\tau)=(\frac{\pi}{\tau})^{1/2}e^{-{\pi^2\tau} (1-\frac{1}{4\pi^2})}e^{-i\pi\tau}. $$</p>