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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.

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Stationary phase? –  Qiaochu Yuan Dec 20 '12 at 20:12
3  
I'm too lazy to work out the answer right now, but you can evaluate the Fourier transform explicitly here since the Fourier transform of a Gaussian is another Gaussian. You can change variables ("complete the square") to remove the linear term. –  Matt Young Dec 20 '12 at 21:10

1 Answer 1

up vote 3 down vote accepted

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

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Amazing...so the oscillation converts what would be exponential growth to exponential decay?!? Can you help me get some intuition for why this happens? <boggles> –  Michael Tinker Dec 20 '12 at 22:48
1  
One slogan about the decay of that FT is ... the very-fast flattening results in very-fast fall-off of higher-frequency components. –  paul garrett Dec 20 '12 at 23:08

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