# Fourier transform of exp(exp(x))

I am interested in the Fourier transform of a function of the form $f(x) = \exp(g(\exp(x)))$, where g has a "simple" form, for example $g(y) = \frac{(y-1)^2}{y^2 - 1}$.

Has anyone a starting point for this? I think it would be a good start to see how to do the Fourier transform with $g=id$. I couldn't find anything about that, though. Any pointers would be helpful.

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Your "simple" $g$ is always rational? – J. M. Nov 10 '11 at 14:47
@J.M. yes, g is rational. Maybe that was my current definition of simple ;) – Andreas Mueller Nov 10 '11 at 14:59

The double exponential and in fact, the exponential function itself $f(x) = \exp (x)$ are not tempered distributions, and therefore they don't have Fourier transforms using classical techniques.

Basically, for the Fourier transform of a function $f$ to be defined, you need that the integral

$$\int f(x) \exp(i\xi x) dx = \lim_{R\to\infty}\int_{-R}^Rf(x)\exp(i\xi x) dx$$

to converge. (This requirement can be relaxed if you are willing to consider "generalised functions"; the techniques in the linked Wikipedia article can be useful.) In your case a necessary condition for such integrals to converge is that $\lim_{y\to\infty} g(y) = -\infty$ and $\lim_{y\to 0}g(y) = -\infty$ (and they are not sufficient to guarantee the integral converges). So for $g = id$ or $g = \ln$ the integrals fail to converge. Even for the case $g(y) = \frac{(y-1)^2}{y^2 - 1}$, by looking at the limits you see that the Fourier integral will fail to converge.

On the other hand, when $g(y) = \frac{(y-1)^2}{y^2-1} = \frac{y-1}{y+1}$, since $|g(y)| \leq 1$ for $y > 0$, you have that $f(x) = \exp\circ g\circ\exp(x)$ is a bounded (but not decaying) function, and so is a tempered distribution, which means that its Fourier transform will also exist as a tempered distribution. But I don't think it has any nice representations as "simple" functions.

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Thank you for the answer. It definitely makes me understand my question better ;) I will look at the wikipedia article on tempered distributions. Sampling from the corresponding distribution would actually be enough for my case. – Andreas Mueller Nov 10 '11 at 15:05
Those are different distributions. – Jonas Teuwen Nov 10 '11 at 15:19
Unfortunately, "distribution" is an overloaded word in mathematics. Probability distributions are quite different from Schwartz distributions (the one relevant to the above discussion) and also are quite different from tangent distributions from differential geometry... – Willie Wong Nov 10 '11 at 15:26