# Fourier Transform of this function

I am trying to find if the Fourier transform of the following function is positive of not. $$f(x) = \frac{1}{2-e^{-x^2}}$$ How can I find the Fourier transform? I started with the classical definition but I could not solve the integral.

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Unless I'm missing something, I would comment that the function $f$ is not integrable ($f \notin L^p(\mathbb{R})$ for any $1 \leq p \leq 2$), so the Fourier Transform of $f(x)$ is likely not defined. – JavaMan Feb 24 '13 at 19:22
@JavaMan: You are right, thanks – neticin Feb 25 '13 at 9:54

On one hand, JavaMan is right: since the function $f$ is not in $L^1+L^2$ on $\mathbb R$, the classical definitions of the transform do not apply. On the other, the Fourier transform is positive in the sense of distributions; that is, it is a positive measure on $\mathbb R$. A nice measure, actually: smooth function plus a single point mass at $0$. Indeed, we have a series $$f(x)= \frac{1}{2}\frac{1}{1-2^{-1}e^{-x^2}} =\frac12\left(\sum_{n=0}^\infty 2^{-n}e^{-nx^2}\right) \tag1$$ which converges in almost every sense imaginable. The terms with $n\ge 1$ are Gaussians which transform to other Gaussians (positive functions). The term with $n=0$ is $\frac12$ which is of course not integrable, but its distributional Fourier transform is a positive point mass at $0$.