During solving of one task I faced to new problem, that is connected with Inverse Fourier transform.

Let $ f(x) : R^n \to R $, Fourier transform of $f(x)$ exist, let $ F[f(x)] = \hat{f}(\xi) $. I'm working with $ e^{-\alpha |\xi|^2} \hat{f}(\xi), \alpha > 0 $. I would like to get $F^{-1} [e^{-\alpha |\xi|^2} \hat{f}(\xi)]$ expressed in terms of $f(x)$ or just simplify this expression.


1 Answer 1


This is a "standard trick" used (for example) when you solve a linear PDE using Fourier transform. I will assume that the Fourier transform of $f$ is given by $$\hat{f}(\xi)=\frac{1}{(2\pi)^{\frac{n}{2}}} \int_{\mathbb{R}^{n}}e^{-i x \cdot \xi}f(x)dx.$$ With this definition, we have the following relation: if $\hat{u}(\xi)=\hat{f}(\xi) \hat{g}(\xi)$, then $$u=\frac{1}{(2\pi)^{\frac{n}{2}}}f*g,$$ where $f*g$ denotes the convolution of $f$ with $g$. Then, using that $\mathcal{F}^{-1}[e^{-\alpha |\xi|^{2}}]=\left( \frac{1}{2 \alpha} \right)^{\frac{n}{2}}e^{-\frac{|x|^{2}}{4 \alpha}}$, you obtain that $$\mathcal{F}^{-1}[e^{-\alpha|\xi|^{2}}\hat{f}]=\left( \frac{1}{4 \pi \alpha} \right)^{\frac{n}{2}}f*e^{-\frac{|x|^{2}}{4 \alpha}}.$$

EDIT: Proof of $\mathcal{F}^{-1}[e^{-\alpha |\xi|^{2}}]=\left( \frac{1}{2 \alpha} \right)^{\frac{n}{2}}e^{-\frac{|\xi|^{2}}{4 \alpha}}$.

Step 1: Set $g(t)=(2\pi)^{-1/2}e^{-t^{2}/2}$ and $G(x)=(2\pi)^{-n/2}e^{-\frac{|x|^{2}}{2}}$. Note that $G(x)=g(x_{1})\cdots g(x_{n})$. Thus \begin{align*} \hat{G}(\xi)&=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}} e^{-i x \cdot \xi }G(x)dx \\ &=\int_{\mathbb{R}^{n}} \prod_{k=1}^{n}\frac{1}{\sqrt{2\pi}}e^{-ix_{k}\xi_{k}}g(x_{k}) dx \\ &=\prod_{k=1}^{n} \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}^{n}} \frac{1}{\sqrt{2\pi}}e^{-ix_{k}\xi_{k}}g(x_{k}) dx_{k} \\ &=\prod_{k=1}^{n}\hat{g}(\xi_{k}). \end{align*} Observe that the Fourier transform on the LHS is on $\mathbb{R}^{n}$, while the Fourier transform on (last part of) the RHS is on $\mathbb{R}$. Since $g$ satisfies $g'(t)=-tg(t)$, applying the Fourier transform to this ODE we find $$\xi_{k} \hat{g}(\xi_{k})=-\hat{g}(\xi_{k}). $$ Thus $\hat{g}(\xi_{k})=ce^{-\xi_{k}^{2}}$. Note that $$ c=\hat{g}(0)=\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \frac{1}{\sqrt{2\pi}}e^{-x_{k}^{2}/2}e^{-ix_{k} 0}dx_{k}=\frac{1}{\sqrt{2\pi}}.$$ Then, $\hat{g}(\xi_{k})=(2\pi)^{-1/2}e^{-\xi_{k}^{2}}=g(\xi_{k})$. Therefore $$\hat{G}(\xi)=\prod_{k=1}^{n}\hat{g}(\xi_{k})=\prod_{k=1}^{n}g(\xi_{k})=G(\xi).$$

Step 2: Define $G_{a}(x)=e^{-a|x|^{2}}$.

Remark: Step 1 says that $\mathcal{F}\left[G_{\frac{1}{2}} \right]=G_{\frac{1}{2}}$.

Using the Remark we can compute as follows \begin{align*} \mathcal{F}\left[G_{a} \right](\xi) &=\frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^{n}} e^{-i x \cdot \xi} e^{-a|x|^{2}}dx \\ & \underbrace{=}_{x=(2a)^{-1/2}y}\frac{1}{(2a )^{n/2}} \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^{n}} e^{-iy \cdot (2a)^{-1/2} \xi} e^{-|y|^{2}/2} dy \\ &=\frac{1}{(2a )^{n/2}} \mathcal{F} \left[G_{\frac{1}{2}} \right]((2a)^{-1/2}\xi) \\ &=\frac{1}{(2a )^{n/2}} G_{\frac{1}{2}}((2a)^{-1/2}\xi) \\ &=\frac{1}{(2a )^{n/2}} e^{-|(2a)^{-1/2}\xi|^{2}/2} \\ &=\frac{1}{(2a)^{n/2}} e^{-|\xi|^{2}/4a} \\ &=\frac{1}{(2a)^{n/2}} G_{\frac{1}{4a}} (\xi). \end{align*} Now set $\alpha=\frac{1}{4a}$. So far we have $$G_{\alpha}=\left( \frac{1}{2\alpha} \right)^{\frac{n}{2}} \mathcal{F}\left[G_{\frac{1}{4\alpha}} \right].$$ Applying the inverse Fourier transform we get $$\mathcal{F}^{-1}[e^{-\alpha |\xi|^{2}}]=\mathcal{F}^{-1} \left[ G_{\alpha} \right]=\left( \frac{1}{2\alpha} \right)^{\frac{n}{2}} G_{\frac{1}{4\alpha}} =\left( \frac{1}{2\alpha} \right)^{\frac{n}{2}} e^{-\frac{|x|^{2}}{4\alpha}}.$$ This finishes the proof.

  • $\begingroup$ +1 Nicely done. $\endgroup$ Jul 30, 2021 at 3:06
  • $\begingroup$ Thanks, very much, @Sebathon ! How did you get inverse Fourier transform for $ e^{-\alpha |\xi|^2}$ ? You just used formula of inverse Fourier transform, didn't you ? $\endgroup$
    – Elan
    Jul 30, 2021 at 11:46
  • $\begingroup$ @Ilya I just made an edit explaining that step. I tried to put all the details. $\endgroup$
    – Sebathon
    Jul 30, 2021 at 13:04
  • $\begingroup$ @Sebathon, thanks very much for your very detailed answer! very nice solution! $\endgroup$
    – Elan
    Jul 30, 2021 at 13:23
  • $\begingroup$ Please take a look at Enforcement of Quality Standards. $\endgroup$ Jul 30, 2021 at 13:43

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