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Given $f:\Bbb R^n \to \Bbb R$. Let $\hat{f} = \mathcal{F}(f)$ be its Fourier transform. The inverse Fourier transform of $\hat{f}$ is defined as: $$ [\mathcal{F}^{-1}(\hat{f})](x) =\frac{1}{(2\pi)^{n/2}} \int_{\Bbb R^n}\hat{f}(\xi)\exp(i \xi \cdot x)\,d\xi $$

Further, if $f \in L^1(R^n)$ such that $\hat{f} \in L^1(\Bbb R^n)$ then $f=\mathcal{F}^{-1}(\hat{f})$ a.e. in $\Bbb R^n$.

Let $\hat{f}:\Bbb R^n \to \Bbb R$ be defined as: $$ \hat{f}(\xi)=\frac{1}{(2\pi)^{n/2}}\frac{1}{\lambda+|\xi|^2} $$ with $\lambda >0$.

Prove that: $$ f(x) = \mathcal{F}^{-1}(\hat{f}) = \int_0^{\infty}\exp(-\lambda t)(4\pi t)^{-n/2}\exp(-\frac{|x|^2}{4t})\,dt $$

Attempt: I applied the inverse Fourier transform to $\hat{f}$ but didn't know where the $t$ in the question came from.

Bernard
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A Slow Learner
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  • Well, it is the dumb variable in the integral, which goes from $0$ to $\infty$. – Bernard Nov 06 '20 at 19:49
  • @Bernard The inverse Fourier transform of $\hat{f}$ has no $t$ in it. Further, $t$ here is just a one-dimensional variable. So, I don't know how to get such $t$ in the result. – A Slow Learner Nov 06 '20 at 19:54
  • But $t$ is a dumb variable! You may name it as you please. There's no $t$ in the result. – Bernard Nov 06 '20 at 20:01
  • @Bernard I am still stuck at this problem. Do you think you can provide me a hint? – A Slow Learner Nov 06 '20 at 22:26
  • At the moment, unfortunately I can't (I'm no specialist of Fourier transforms). Note my comments were at a very general level. I'll look into it tomorrow (it's late here), but I think some specialist will answer you. – Bernard Nov 06 '20 at 22:32
  • Here's a hint: instead of taking the inverse Fourier transform of $\hat{f}$, try taking the Fourier transform of the integrally-defined function at the end, using Fubini's theorem (i.e. see that the resulting calculation matches the definition of $\hat{f}$). – Jason Nov 09 '20 at 03:37
  • @Jason I tried. However, all I could do was writing out the Fourier transform of $f$, but then I didn't know how to combine all the exponential functions in the integrands (was I having the correct idea?). – A Slow Learner Nov 11 '20 at 07:21
  • @ASlowLearner I'd say you don't need to combine all the exponential terms. After you change the order of integration from $dt , dx$ to $dx , dt$, you can first bring some of the factors outside of the inner integral. What's left in the inner integral becomes the Fourier transform of a Gaussian - I'll leave that for you to look up. – Jason Nov 11 '20 at 17:44

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