Let's consider $g(x)$ a function in $C^{\infty}(\mathbb{R}^+ \to \mathbb{C})$ such that $g(x)$ is asymptotic to $x^{\alpha}$ for $x$ near $+\infty$ with $g(0)=0$, and $\alpha$ a complex number such that $Re(\alpha)<-\frac{1}{2}$.
If $g$ satisfy following condition:
$$\int\limits_{0}^{\infty} g(x) \overline{x^{\alpha} } dx=0 $$
then following integral is well defined:
$$ I=\int\limits_{0}^\infty \frac{1}{x} \int\limits_{x}^{\infty} g(y) \;\overline{y^{\alpha} }+4 \; \hat{g(y)} \;\overline{\hat{y^{\alpha} }} \; dy$$
where $\hat{f}(x)$ is the cosine transform ($\hat{f}(x)=\int\limits_{0}^\infty f(t) \cos(2\pi xt) dt)$
It is quite intuitive that $I$ is not zero for all $g$ functions satisfying given conditions above, but how to prove it ?
How to prove that there exists $g$ functions such that $I \ne 0$ ? Is there an elegant / simple way ? Which functions can be used for a simple proof ? I took linear combinations of functions of the form $x^b e^{-x^2}$ but calculation of $I$ is not so simple ! What type of functions can be good candidates for simple demonstration ? Any trick to show there is a contradiction if we assume all $g$ funcitons give $I=0$ ?