Fourier transform of a rapidly decreasing function

Posted February 20th, 2008 by Structure

Let

$\widehat{f}(\xi)=\int_{-\infty}^{+\infty}f(x)e^{-ix\xi}dx$

be the Fourier transform of a function in Lesbegue space $\mathcal{L}^1(R)$
For function $f(x)=e^{-x^2}$ we have

$\widehat{f}(\xi)=\int_{-\infty}^{+\infty}e^{-x^2}e^{-ix\xi}dx=e^{-\frac{\xi^2}{4}}\int_{-\infty}^{+\infty}e^{-x^2-ix\xi-\frac{i^2\xi^2}{4}}dx<br /> =e^{-\frac{\xi^2}{4}}\int_{-\infty}^{+\infty}e^{-(x+i\frac{\xi}{2})^2}dx=\sqrt{\pi}e^{-\frac{\xi^2}{4}}$

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Fourier transform of a rapidly decreasing function

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