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I'm trying to workout the Fourier transform of $f(x) = x^2 \exp{(-x^2)}$. We know that \begin{equation} \tilde f(x) = \int_{-\infty}^{\infty} f(x)\exp{(-ikx)} \ \mathrm{d}x. \end{equation} I have managed to simplify this by subbing in the value of $f(x)$ into the integral to: \begin{equation} \tilde f(x) = \exp{(-\frac{1}{4} ik^2)} \int_{-\infty}^{\infty} x^2 \exp{(-(x+\frac{1}{2}ik)^2)} \ \mathrm{d}x \end{equation} by using the completing the square method. I'm stuck after this point as I'm struggling to integrate the term above. I've tried the substitution method where I sub $s=x+\frac{1}{2}ik$ but am still getting nowhere. I am missing something obvious here?

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Observe that $$ \int x^2 e^{-x^2} e^{-ikx} = - \frac{\partial^2}{\partial k^2} \int e^{-x^2} e^{-ikx} $$ and complete the square in the latter integral.

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  • $\begingroup$ So via completing the square, I have evaluated the whole integral as sqrt(Pi)*exp(-1/4 * k^2) but not sure how this helps. $\endgroup$ – tellap Mar 4 '15 at 15:03
  • $\begingroup$ Actually ignore my previous comment, I'm being stupid. So in the end we should get 1/4 exp(-1/4 k^2) sqrt(Pi)(2-k^2) $\endgroup$ – tellap Mar 4 '15 at 15:15
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We have $$\tilde{f}\left(\xi\right)=\int_{-\infty}^{\infty}x^{2}e^{-2\pi i\xi x-x^{2}}dx$$ and observing that $$x^{2}e^{-2\pi i\xi x-x^{2}}=\frac{1}{2}\left(-\frac{d}{dx}\left(xe^{-2\pi i\xi x-x^{2}}\right)+e^{-2\pi i\xi x-x^{2}}+2\pi i\xi xe^{-2\pi i\xi x-x^{2}}\right)$$ we have $$\tilde{f}\left(\xi\right)=\frac{1}{2}\left(-\int_{-\infty}^{\infty}\left(\frac{d}{dx}\left(xe^{-2\pi i\xi x-x^{2}}\right)\right)dx+\int_{-\infty}^{\infty}e^{-2\pi i\xi x-x^{2}}dx+2\pi i\xi\int_{-\infty}^{\infty}xe^{-2\pi i\xi x-x^{2}}dx\right)=$$ $$=-e^{-\pi^{2}\xi^{2}}\pi^{5/2}\xi^{2}+\frac{1}{2}\sqrt{\pi}-e^{-\pi^{2}\xi^{2}}$$ that can be rewritten as $$\frac{1}{4}e^{-\frac{1}{4}\omega^{2}}\sqrt{\pi}\left(2-\omega^{2}\right).$$

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  • $\begingroup$ How are you getting/why did you put 2Pi into the original problem? $\endgroup$ – tellap Mar 4 '15 at 14:39
  • $\begingroup$ @tellap the form of the integrands indicates to think a derivative of an exponential. About $2\pi$, the Fourier classical definition of Fourier transform have $2\pi$ (I studied with this definition). It's the definition of wikipedia, too en.wikipedia.org/wiki/Fourier_transform $\endgroup$ – Marco Cantarini Mar 4 '15 at 19:02

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