Functions that are their own Fourier transform In Stein's Fourier Analysis, there's the following exercise:

The function $e^{-\pi x^2}$ is its own Fourier transform.   Generate other functions [presumably in the Schwartz space $S(\mathbb{R})$] that, up to a constant multiple, are their own FTs.  What must the constant multiples be?  To decide this, prove that $F^4 = I$ where $F$ is the FT operator.

My problem is that I don't really understand what the question is asking.  Is it asking us to find the class of all functions such that $F(f) = cf$ for constant c and prove the above identity, or is it asking us to just find other examples?  Or, is the $F^4 = I$ identity for any $f \in S(\mathbb{R})$?
Finding every function would mean finding all $f$ such that
$$f(\xi) = \int^{\infty}_{-\infty} f(x)e^{-2\pi i x \xi}dx$$ which seems rather difficult.
I can come up with a specific example: if I can find a function such that $\hat{f}(\xi) = 1$, I can just take $g = f \ast K_{\delta}(x)$ where $K_{\delta}$ is the family of Gaussian functions as approximations of the identity.
 A: Hint: for any $f$, take a linear combination of $f$ and $F(f)$ and ...
Another hint for proving $F^4 = I$: Fourier inversion formula.
A: Another example: the function $\frac{1}{\cosh(\pi x)}$ is its own Fourier transform, ie: 
 $$\frac{1}{\cosh(\pi \xi)}=\int_{-\infty}^\infty e^{-2\pi i x \xi}\frac{1}{\cosh(\pi x)}dx$$
(Possibly off-topic, but since Job Bouwman posted an example I figured I'd do the same for future visitors!)
A: Because Fourier transform is a linear transformation in the Hilbert space of square-normalizable curves $L^2$, this is essentially an eigenvalue problem. You are looking for a function that doesn't change under Fourier transform. Beside the Gaussian function, which is very famously an eigenfunction of the Fourier transform, there is a whole class of them that also satisfy this eigenvalue condition of $Ff=cf$. They are essentially a Gaussian times a Hermite polynomial:
http://en.wikipedia.org/wiki/Hermite_functions#Hermite_functions_as_eigenfunctions_of_the_Fourier_transform
On the other hand, $F^4=I$ is universal. If you observe the Fourier transform integral, you can see that the Fourier transform and inverse Fourier transform have an opposite sign in the imaginary exponent. This is why $F$ is not exactly $F^{-1}$ but it's close. $F^2$ flips the function (reverses time) because of this mismatch in sign, but reversing time twice brings you back again, which is why $F^4=I$ works for any function, not just the eigenfunctions. This is related to $i^4=1$: $F^4=I$ tells you that the only eigenvalues of $F$ are $c\in\{\pm 1, \pm i\}$ (try $F^4 f=F^3 (cf)=c^4 f=f$ which gives you $c^4=1$). Of course, this means, that the Fourier space is highly degenerate. A linear combination of any number of eigenfunctions with the same eigenvalue is also an eigenfunction. You can use the Hermite functions (they sequentially cycle through eigenvalues $i^n$, so take every fourth) as a basis spanning each of the four subspaces: you get an infinite family of eigenfunctions for each eigenvalue.
Be warned that this depends on the normalization of the Fourier transform. You need the unitary version. If you take the regular definition (the one used in physics and engineering), you get additional factors of $\sqrt{2\pi}^n$, which the transform blows up or extinguishes the amplitude of the function, and is not cyclic anymore.

You will notice that you can split any function into 4 components with eigenvalues $\{1,i,-1,-i\}$ by doing this:
$$\frac{1}{4}(1+F+F^2+F^3)f=f_1$$
$$\frac{1}{4}(1-iF-F^2+iF^3)f=f_i$$
$$\frac{1}{4}(1-F+F^2-F^3)f=f_{-1}$$
$$\frac{1}{4}(1+iF-F^2-iF^3)f=f_{-i}$$
where
$$Ff_1=f_1\quad Ff_i=if_i\quad Ff_{-1}=-f_{-1}\quad Ff_{-i}=-if_{-i}$$
Each of these components has a very specific symmetry. In particular, the $\pm i$ are odd and $\pm 1$ are even.
You can take any arbitrary function and use it to generate a function that is preserved under Fourier transform. This I think is the most general answer I can give you.
A: Here's another example that's hard to come by:
$$ \int_{-\infty}^{\infty} \left(\psi(1+|t|)-\log|t| \right) e^{-2\pi i xt} \, dt = \psi(1+|x|)-\log|x| $$
where $\psi$ is the digamma function and $x$ is real and non-zero. I learned of it from this paper. The formula was first found by Ramanujan, but had also been derived by others since.
A: The function $ f(\bf x) = \frac{1}{\sqrt{|x|}}$ is also its own Fourier Transform:
$$\mathcal{F}(f(x)) = \int^{\infty}_{-\infty} \frac{1}{\sqrt{|x|}}e^{-2\pi i x \xi}dx = \frac{1}{\sqrt{|\xi|}} = f(\xi) $$  
Although $f(x)$:


*

*is not continuous and defined at $ x = 0 $ 

*is not integrable over $<-\infty , +\infty>$   


Proof: 
Let:
$$g(\xi) = \mathcal{F}(f(x)) =  \int^{\infty}_{-\infty} \frac{1}{\sqrt{|x|}}e^{-2\pi i x \xi}dx $$
Since $f(x) = \frac{1}{\sqrt{|x|}}$ is even (and real), $g(\xi)$ must be even and real. So to avoid problems with the absolute value, I'll work with $x > 0$ and $\xi > 0 $ 
$$g(\xi) =  2\int^{\infty}_{0} \frac{1}{\sqrt{x}}\cos(2\pi x \xi)dx \,\,\,\,\,\, \text{   for   } \,\,\,\,\,\, \xi > 0 $$
Substitute: 
$$u =2\pi x \xi \implies   \sqrt{x}=\sqrt{\frac{u}{2\pi \xi}} \implies  dx = \frac{du}{2\pi  \xi}$$
to get: 
$$g(\xi) =  2\int^{\infty}_{0} \frac{\sqrt{2\pi\xi}}{\sqrt{u}}\cos(u)\frac{du}{2\pi \xi} $$
$$g(\xi) =  \frac{2}{\sqrt{2\pi \xi}}\int^{\infty}_{0} \frac{1}{\sqrt{u}}\cos(u)du $$
Substitute:
$$ u = t^{2} \implies \sqrt{u}=t\implies du=2tdt $$
$$g(\xi) =  \sqrt{\frac{2}{\pi}}\frac{1}{\sqrt{\xi}} \int^{\infty}_{0} \frac{1}{t}\cos(t^{2})2tdt $$
in order to get the $\color{blue}{\text{Fresnel integral}}$ : 
$$g(\xi) =  2\sqrt{\frac{2}{\pi}}\frac{1}{\sqrt{\xi}}\color{blue}{\int^{\infty}_{0} \cos(t^{2})dt} $$
$$g(\xi) =  \sqrt{\frac{8}{\pi}}\frac{1}{\sqrt{\xi}}\color{blue}{\sqrt{\frac{\pi}{8}}} = \frac{1}{\sqrt{\xi}}$$  
Now we use that $g(\xi)$ is even to expand its domain to $\xi \neq 0 $: 
$$g(\xi) =\frac{1}{\sqrt{|\xi|}} = f(\xi)\,\,\,\,\,\,\,\,\,\,   \blacksquare$$
A: Not answer, only a recommendation:
Don't look for
$$f(x) = \int^{\infty}_{-\infty} f(x)e^{-2\pi i x \xi}dx$$
but (firstly) for
$$1 = \int^{\infty}_{-\infty} \int^{\infty}_{-\infty} f(x)e^{-2\pi i y \xi}e^{-2\pi i x \xi}dydx$$
which would be $F^2 = I$ 
And I guess there are also f(x) with
$$-1 = \int^{\infty}_{-\infty} \int^{\infty}_{-\infty} f(x)e^{-2\pi i y \xi}e^{-2\pi i x \xi}dydx$$
which would result in $F^4 = I$
At least I don't see another reason to use $F^4 = I$ instead of $F^2 = I$
Have a try
A: Observe that, Fourier inversion formula gives us, $\widehat{(\hat{f})}(x) =f(-x)$; and  temporally put, $g(x)= f(-x)$; again by Fourier inversion formula, $\widehat{(\hat{g})}(x) =g(-x)$ and clearly, $g(-x)= f(x).$
A: Let $f(x) = e^{-\pi x^2}$. Note that multiplication is the differentiation in the transformed space, $$\mathcal{F}(2\pi ixf(x)) = \frac{d}{d\xi}F(f(\xi)) = -2\pi\xi e^{\pi\xi^2}.$$
This can be generalized to give a family of polynomials $\{G_n\}_{n\in\mathbb{N}}$ such that $\mathcal{F}(G_n\cdot f) = i^n (G_n\cdot f)$.
My guess is this was the intention as suggested by "generate" in the prompt.
