The Fourier transform of the Fourier transform

[Poopsilon]

Homework Statement

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?

Homework Equations

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

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?

 

[sunjin09]

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

 

[Poopsilon]

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

 

[sunjin09]

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

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

 

[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:

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

 

[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

 

[Poopsilon]

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

 

[Poopsilon]

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