Recovering the spatial Fourier transform from the space-time Fourier transform

This CW question is aimed at developing some intuition (grokking) about a certain formula of Fourier analysis. Any kind of explanation (physical, geometrical, analytical ...) is welcome.

If we have a function $$\begin{array}{cc}\phi\colon \mathbb{R}\times\mathbb{R}^n\to \mathbb{C},&\phi=\phi(t, x),\end{array}$$ we can take the space-time Fourier transform $$\widetilde{\phi}(\tau, \xi)=\int_{\mathbb{R}\times\mathbb{R}^n}\phi(t,x) e^{-i(t\tau+x\cdot\xi)}\, dt dx$$ and the spatial Fourier transform (on the time slice $t=0$) $$\widehat{f}(0, \xi)=\int_{\mathbb{R}^n} \phi(0, x)e^{-i x \cdot \xi}\, dx.$$

From the (space-time) Fourier inversion formula it follows that $$\tag{1}\widehat{f}(0, \xi)=\int_{\mathbb{R}}\widetilde{\phi}(\tau,\xi)\, d\tau.$$

Can you give some explanation of formula (1) that allows us to grok it?

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Perhaps the following makes it more grokkable:

If $\phi$ allows separating the space and time parts, $\phi(t,x) = a(t)\cdot b(x)$, then

$$\tilde{\phi}(\tau,\,\xi) = \widehat{a}(\tau)\cdot\widehat{b}(\xi)$$

and $(1)$ is simply the Fourier inversion applied to the time part. The span of separated functions is dense, hence $(1)$ is valid for all $\phi$ by continuity.

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This one gives some insight. Very nice. –  Giuseppe Negro Jul 24 '13 at 17:18

I think you may be off by a factor of $2 \pi$, but here's how I see it:

$$\int_{\mathbb{R}} d\tau \, \widetilde{\phi}(\tau, \xi) = \int_{\mathbb{R}} d\tau \,\int_{\mathbb{R}\times\mathbb{R}^n} dt \, dx \,\phi(t,x) e^{-i(t\tau+x\cdot\xi)}$$

Changing the order of integration:

$$\int_{\mathbb{R}} d\tau \, \widetilde{\phi}(\tau, \xi) =\int_{\mathbb{R}\times\mathbb{R}^n} dt \, dx \,\phi(t,x) e^{-i x\cdot\xi} \,\int_{\mathbb{R}} d\tau \, e^{-i t\tau}$$

The innermost integral is equal to $2 \pi \delta(t)$. Using the sifting property of the delta function, we get

$$\int_{\mathbb{R}} d\tau \, \widetilde{\phi}(\tau, \xi) = 2 \pi \int_{\mathbb{R}^n} dx \,\phi(0,x)\, e^{-i x\cdot\xi} = 2 \pi \widehat{f}(0,\xi)$$

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I should have mentioned that I am "choosing units so that $2\pi=1$". :-) –  Giuseppe Negro Jul 24 '13 at 16:52
@GiuseppeNegro: ah, so $\pi=1/2$ and indeed we may square the circle after all. Brilliant! –  Ron Gordon Jul 24 '13 at 16:54

Here is a measure-theoretic interpretation. Let $$\mu(d\tau, d\xi)=\widetilde{\phi}(\tau, \xi)d\tau d\xi$$ be the absolutely continuous measure on $\mathbb{R}\times\mathbb{R}^n$ associated to the space-time Fourier transform. Its marginal distributions are then $$\tag{2}\mu_\xi(d\xi)=\left(\int_{\mathbb{R}}\widetilde{\phi}(\tau, \xi)\, d\tau\right)d\xi,$$ $$\tag{3} \mu_\tau(d\tau)=\left(\int_{\mathbb{R}^n}\widetilde{\phi}(\tau, \xi)\, d\xi\right)d\tau.$$ So the formula (1) in Question above tells us that, when viewed as an absolutely continuous measure, the spatial Fourier transform is a marginal distribution of the space-time Fourier transform.

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