# The Fourier transform of the Fourier transform

1. Feb 5, 2012

### Poopsilon

1. The problem statement, all variables and given/known data

Let f be a suitably regular function on ℝ. (whatever that means).

What function do we obtain when we take the Fourier transform of the Fourier transform of f?

2. Relevant equations

$$F(s) = \int_{x=-\infty}^{\infty}f(x)e^{-2\pi isx}dx$$

3. The attempt at a solution

$$\int_{x=-\infty}^{\infty}F(s)e^{-2\pi isx}dx = F(s)\int_{x=-\infty}^{\infty}e^{-2\pi isx}dx = F(s)[\frac{-e^{-2\pi isx}}{2\pi is}]_{x=-\infty}^{\infty} = \frac{F(s)i}{2\pi s}[e^{-2\pi isx}]_{x=-\infty}^{\infty}$$

But this simply oscillates indefinitely around a complex circle as x goes to negative infinity and infinity, where have I gone wrong?

Last edited: Feb 5, 2012
2. Feb 5, 2012

### sunjin09

when taking the Fourier transform of F(s), you must treat "s" as "x" when you substitute into the formula

3. Feb 5, 2012

### Poopsilon

Thanks sunjin09, could you also tell me what 'suitably regular' means?

4. Feb 5, 2012

### sunjin09

It means the Fourier transform exists, and the inverse transform gives back to the original function.

5. Feb 5, 2012

### Poopsilon

Ok, here's what I've got now:

$$\int_{x=-\infty}^{\infty}F(x)e^{-2\pi isx}dx = \int_{y=-\infty}^{\infty}(\int_{x=-\infty}^{\infty}f(x)e^{-2\pi ix^2}dx)e^{-2\pi isy}dy = \int_{x=-\infty}^{\infty}f(x)e^{-2\pi ix^2}(\int_{y=-\infty}^{\infty}e^{-2\pi isy}dy)dx = F(s)\int_{x=-\infty}^{\infty}f(x)e^{-2\pi ix^2}dx$$

Is this it, or can I simplify more?

Edit: Actually this isn't right, I think I may have made some illegal moves with the dummy variables.. Let's try this:

$$\int_{x=-\infty}^{\infty}F(x)e^{-2\pi isx}dx = \int_{y=-\infty}^{\infty}(\int_{x=-\infty}^{\infty}f(y)e^{-2\pi ixy}dx)e^{-2\pi isy}dy = \int_{x=-\infty}^{\infty}e^{-2\pi ixy}(\int_{y=-\infty}^{\infty}f(y)e^{-2\pi isy}dy)dx = F(s)\int_{x=-\infty}^{\infty}f(x)e^{-2\pi ixy}dx$$

Last edited: Feb 5, 2012
6. Feb 5, 2012

### sunjin09

When you write F(s) as F(x), you must write f(x) as say, f(u), don't use the same symbol for different quantities, the answer is very simple, just be careful with substitutions

7. Feb 5, 2012

### Poopsilon

Ok is the answer then, F(s)F(r)?

8. Feb 5, 2012

### Poopsilon

actually I think it's f(-x), is this correct?