# Fourier transform and lebesgue integral

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.

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).

• How do you know that $\int_{-\infty}^\infty e^{- (x-i \xi)^2/2}dx =\int_{-\infty}^\infty e^{- x^2/2}dx$ – reuns Nov 19 '18 at 17:39

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 $$d>1$$.

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