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Let $f:\mathbb{R}^d \rightarrow \mathbb{R}^d$ be the function with $f(x) = \exp(-\frac{1}{2}\mid x \mid^2)$. Show that the fourier transform of $f$ is given by $\hat f = (\sqrt{2 \pi})^d f$.

The fourier transform is given by $\hat f (\xi ) := \displaystyle\int_{\mathbb{R}^d} f(x) \exp(-i\xi \cdot x)\ \mu(dx)$.

I would like to start with looking at the case $d = 1$ but I dont know how to proceed.

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It's a Gaussian integral---complete the square up in the exponential to get $(2\pi)^{d/2} e^{-\|\xi\|^2/2}$. Standard result on multivariate Gaussians (and no need to worry about Lebesgue).

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    $\begingroup$ How do you know that $\int_{-\infty}^\infty e^{- (x-i \xi)^2/2}dx =\int_{-\infty}^\infty e^{- x^2/2}dx$ $\endgroup$ – reuns Nov 19 '18 at 17:39
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For $d=1$, we have $f(x)=e^{-x^2/2}$ and therefore (remember that $(i\xi+x)^2=-\xi^2+2i\xi x+x^2$) \begin{equation*} \begin{split} \hat{f}(\xi) & = \int_{-\infty}^{\infty} e^{-1/2(2i\xi x + x^2)} {\rm d}\mu(x) \\ & = \int_{-\infty}^{\infty} e^{-1/2((i\xi+x)^2+\xi^2)} {\rm d}\mu(x) \\ & = f(\xi) \int_{-\infty}^{\infty} e^{-(i\xi+x)^2/2} {\rm d}\mu(x) \\ & =f(\xi) \int_{-\infty}^{\infty} e^{-y^2/2} {\rm d}\mu(y) \\ & = \sqrt{2\pi} f(\xi). \end{split} \end{equation*}

It's pretty similar for $d>1$.

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  • $\begingroup$ How do you know that $\int_{-\infty}^\infty e^{- (x-i \xi)^2/2}dx =\int_{-\infty}^\infty e^{- x^2/2}dx$ $\endgroup$ – reuns Nov 19 '18 at 17:39
  • $\begingroup$ Change variable and the contribution from the ends of the contour (which gets shifted up or down) is negligible. $\endgroup$ – Richard Martin Nov 19 '18 at 17:49
  • $\begingroup$ How do I continue for $d=2$ and so on? $\endgroup$ – Arjihad Nov 20 '18 at 15:40
  • $\begingroup$ For any $d\geq 1$ you will get $$\hat{f}(\xi)=f(\xi)\int_{\mathbb{R}^d} e^{-|y|^2/2} {\rm d}\mu(y)$$ which is just $\hat{f}(\xi)=(2\pi)^{d/2} f(\xi)$. $\endgroup$ – Rodrigo Dias Nov 20 '18 at 23:50
  • $\begingroup$ Actually the fourth equality requires complex analysis which is not allowed in this task. $\endgroup$ – Arjihad Nov 21 '18 at 15:53

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