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Let $\displaystyle{K(x)= e^{- \pi |x|^2} \quad ,x \in \mathbb R^n}$ be the Gaussian kernel on $\mathbb R^n$. Prove that its Fourier transform is $$ \hat{K} (\xi) = e^{- \pi |\xi|^2} $$

I can prove this on $\mathbb R$ using the fact $\displaystyle{ \int_{- \infty}^{\infty} e^{ - \pi x^2} =1}$, but I do not know how to prove it on $\mathbb R^n$

I did a search, but all the things I found was for the $n=1$ case.

Any help?

passenger
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    You have $\exp\left(-\pi\sum (x_k + i 2y_k)x_k\right) = \prod \exp\left(-\pi(x_k-i2y_k)x_k\right)$. The $n$-dimensional Fourier-integral splits into a product of $n$ one-dimensional Fourier-integrals for $K$. Happiness ensues. – Daniel Fischer Nov 13 '13 at 21:23
  • Thank you very much! It feel so silly...I don't even think of it... thank's again! – passenger Nov 13 '13 at 21:29

1 Answers1

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Since $e^{a+b} = e^a e^b$, the Fourier integral

$$\begin{align} \int_{\mathbb{R}^n} \exp \left(-\pi\sum_{k=1}^n (x_k + 2i y_k)x_k\right)\, dx &= \int_{\mathbb{R}^n} \prod_{k=1}^n \exp \left(-\pi (x_k + 2iy_k)x_k\right)\,dx\\ &= \prod_{k=1}^n \int_{\mathbb{R}} e^{-\pi x_k^2}e^{-2\pi i x_ky_k}\, dx_k \end{align}$$

splits into a product of $n$ one-dimensional Fourier integrals of Gaussians. Since you know how to deal with the one-dimensional case, the remaining part is easy.

Daniel Fischer
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