What is the 1D Fourier transform of the function e^{-\lambda x^{2}}?

[Aquinox]

Homework Statement

Let A be a real, symmetric positively definite nxn - matrix.
$$f:\mathbb{R}^{n}\rightarrow\mathbb{R}\; s.t\;\vec{x}\rightarrow e^{-\frac{1}{2}<\vec{x},A\vec{x}>}$$
Show that the FT of f is given by:
$$\hat{f}(\vec{k})=\frac{1}{\sqrt{\det A}}e^{-\frac{1}{2}<\vec{k},A^{-1}\vec{k}>}$$

Homework Equations

If I'm not very much mistaken:
$$\hat{f}(\vec{k})=\int_{\mathbb{R}^{n}}f(\vec{x})e^{-2\pi i<\vec{k},\vec{x}>}d^{n}x$$

The Attempt at a Solution

Quite honestly I have no idea anymore. I suppose I'm missing sth. quite trivial.
I've tried to change <x,Ax> to $$x^tAx$$ and doing the same with <k,x> and then multiplying from right by A^-1*x and some more but kept running in circles.
I'm terrible with this matrix-stuff and on the solution of this task depends the solution of another one

 

[hunt_mat]

Isn't a diagonalisable which would reduce the problem somewhat?

Mat

 

[Aquinox]

Yep.
I've got that hint by a friend yesterday night which led me to:
$$\int e^{-\frac{1}{2}<x,T^{-1}TAT^{-1}Tx>-i<k,T^{-1}Tx>}\Rightarrow\int e^{-\frac{1}{2}<Tx,(TAT^{-1}=D)Tx>-i<Tk,Tx>}\Rightarrow\int e^{-\frac{1}{2}<y,Dy>-i<k',y>}d^{n}y$$

by using that TAT^-1 = D , with D-diagonal. defining y:=Tx, k':=Tk
using dy/dx=|T|=1

then using the representation of Exp as a series
$$\int e^{-\frac{1}{2}\sum_{l=0}^{n}d_{ll}y_{l}^{2}}e^{-i<k',y>}d^{n}y\Rightarrow\int e^{-\frac{1}{2}\sum_{l=0}^{n}d_{ll}y_{l}^{2}}\sum_{l=0}^{\infty}\frac{1}{l!}(-i<k',y>)^{l}d^{n}y$$

And here I'm stuck again not knowing anymore how it would be possible to simplify this.

 

[hunt_mat]

You're almost there, the n-dim integral can be then split up into a product of n 1-dim integrals, as
$$<y,Dy>=\sum_{i=1}^{n}\lambda_{i}y_{i}^{2}\quad <k',y>=\sum_{i=1}^{n}k_{i}'y_{i}$$
Where the $$\lambda_{i}$$ are the eigenvalues and the integral splits as:
$$\int e^{-\frac{1}{2}<y,Dy>-i<k',y>}d^{n}y=\prod_{i=1}^{n}\int_{-\infty}^{\infty} e^{-\frac{1}{2}\lambda_{i}y_{i}^{2}-ik_{i}'y_{i}}dy_{i}$$

 

[Dick]

You are making this WAY too complicated. Like Mat said, you can work in a basis where A is diagonal. That means f(x) is just a sum of squares of each coordinate times the corresponding diagonal entry. Now you can integrate one coordinates at a time. The only part of the problem that takes a little work is the one-dimensional case where f(x)=exp((-a/2)*x^2). You just have to complete the square in the exponent of the Fourier transform.

 

[Aquinox]

Why though? The way known to me to diagonalize a symmetric matrix is TAT^-1 = D
with orthogonal matrix T.
Following is the way I think I've solved it.
Taking off from hunt's equation:

$$\frac{1}{\sqrt{2\pi}^{n}}\prod_{s=1}^{n}\int_{\mathbb{R}^{1}}e^{-\frac{1}{2}d_{ss}y_{s}^{2}}e^{-ik'_{s}y_{s}}dy=\prod\sum_{l=0}^{\infty}\frac{(-i)^{l}}{l!}k'_{s}^{l}\frac{1}{\sqrt{2\pi}}\int e^{-\frac{1}{2}d_{ss}y_{s}^{2}}y_{s}^{l}dy$$

where I used Mathematica to cheat and check whether the d_ss will come up in the end in the way I guessed and the notes from the lectures giving this (2m) stuff (more human-readable than mathematica's output with Erf and Gamma).

$$=\prod\sum_{m=0}^{\infty}\frac{(-i)^{2m}}{2m!}k'_{s}^{2m}\frac{1}{\sqrt{2\pi}}\int e^{-\frac{1}{2}d_{ss}y_{s}^{2}}y_{s}^{2m}dy=\prod\sum_{m=0}^{\infty}\frac{(-i)^{2m}}{(2m)!}k'_{s}^{2m}\left(\frac{(2m)!}{d_{ss}^{m}2^{m}\sqrt{d_{ss}}m!}\right)=&\prod_{s=1}^{n}\sum_{m=0}^{\infty}\left(\frac{-k_{s}^{2}}{2d_{ss}}\right)^{m}\frac{1}{\sqrt{d_{ss}}m!}=\prod\frac{1}{\sqrt{d_{ss}}}e^{-\frac{1}{2}d_{ss}^{-1}k_{s}'^{2}}$$

$$=&\frac{1}{\sqrt{\prod d_{ss}}}e^{-\frac{1}{2}\sum_{s=0}^{n}k'_{s}d_{ss}^{-1}k'_{s}}=\frac{1}{\sqrt{\det D}}e^{-\frac{1}{2}<k',D^{-1}k'>}=\frac{1}{\sqrt{\det A}}e^{-\nicefrac{1}{2}<Tk,(TAT^{-1})^{-1}Tk>}=\frac{1}{\sqrt{\det A}}e^{-\nicefrac{1}{2}<k,A^{-1}k>}$$

 

[hunt_mat]

Again this is slightly more complicated than it needs to be. What is the 1D Fourier transform of the function:

$$f(x)=e^{-\lambda x^{2}}$$

This is a well known transform which can be found on wikipedia, use this and you're very close to getting the answer.