# Inverse Fourier transform of $e^{-\alpha |\xi|^2} \hat{f}(x)$ [closed]

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.

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.

• +1 Nicely done. Jul 30, 2021 at 3:06
• 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 ?
– Elan
Jul 30, 2021 at 11:46
• @Ilya I just made an edit explaining that step. I tried to put all the details. Jul 30, 2021 at 13:04
• @Sebathon, thanks very much for your very detailed answer! very nice solution!
– Elan
Jul 30, 2021 at 13:23
• Please take a look at Enforcement of Quality Standards. Jul 30, 2021 at 13:43