# Comparison of (square) of a function and its Fourier transform in an integral

I am completely stuck on a comparison between $f(t)^2$ and $\hat{f}(t)^2$ in an integral.

Considering $f(t)$ of rapid decrease at infinity such that near zero: $f(t) \sim_0 t^{-\frac{1}{2}- \alpha}+o(t)$ with $0<\alpha<\frac{1}{2}$ I would like to study the sign of the following integral:

$$I=\int\limits_{0}^\infty \frac{1}{x} \int\limits_{x}^{\infty} 2 \alpha (f(t)^2 + \hat{f}(t)^2 ) + (t f(t)^2 - t \hat{f}(t)^2 )' dt dx$$

Where $\hat{f}(t)$ is the Fourier transform of $f(|t|)$.

If $f(t)$ was in $L^2$ we would use the Parseval theorem for the second part of $I$:

$$\int\limits_{0}^\infty \frac{1}{x} \int\limits_{x}^{\infty} (t f(t)^2 - t \hat{f}(t)^2 )' dt dx = \int\limits_{0}^\infty f(t)^2 - \hat{f}(t)^2 dx =0$$

but this is not the case and $I$ cannot be easily simplified. Any reference about the behavior of $f^2$ versus $\hat{f}(t)^2$ ? Any clue on how to cope with such integral ?

Note that the property of Fourier transform gives that $\hat{f}(t) \sim_{\infty} t^{-\frac{1}{2}+ \alpha}$

Note also that we consider $f$ such that following integral is null (to allow existence of $I$) $$\int\limits_{0}^{\infty} 2 \alpha (f(t)^2 + \hat{f}(t)^2 ) + (t f(t)^2 - t \hat{f}(t)^2 )' dt dx$$