For a bounded function $\operatorname{F}: \mathbb{R}\_{\,\ge\ 0} \to \mathbb{R}$ (*not necessarily non-negative*), if
$$
\int\_{0}^{\infty}\frac{x^{k}\,s}{(s^{2} + x^{2})^{\left(k + 3\right)/2}\,\,}\, \operatorname{F}\left(x\right)\,{\rm d}x = 0\quad
\forall s > 0 
$$
where $k \in \mathbb{N}$ is a positive constant,
is it true that
$$
\int\_{0}^{\infty}\left(\frac{x^{k}\,s}{(s^{2} + x^{2})^{\left(k + 3\right)/2}\,\,}\right)^{h} \, \operatorname{F}\left(x\right)\,{\rm d}x = 0\quad
\forall s > 0 
$$
where $h \in \mathbb{N}$?


This question is inspired by a comment to the answer in <https://math.stackexchange.com/questions/3996738/condition-for-an-integral-to-be-zero> (which required to check the assumptions of the Stone-Weierstrass theorem).


More generally, ignoring the question above, my main concern is this:


* Use Stone-Weierstrass to prove that
$$
\int\_{0}^{\infty}\frac{x^{k}\,s}{(s^{2} + x^{2})^{\left(k + 3\right)/2}\,\,}\, \operatorname{F}\left(x\right)\,{\rm d}x = 0\quad
\forall s > 0 
$$
if and only if $F \equiv 0$.



An integral transform and the Stone-Weierstrass theorem
Method 1: There is a clear harmonic-analytic interpretation: if
\[ u(x, s) = C\_k \int\_{\mathbb R^{k+2}} \frac{s}{(s^2 + |x - y|^2)^{(k+3)/2}} \times |y|^{-1} F(|y|) dy , \]
then $u$ a harmonic function in the half-space $x \in \mathbb R^{k+2}$, $s > 0$, and $u(0, s) = 0$ for all $s > 0$. We claim that $u$ is identically zero, and hence $F = 0$ almost everywhere. This follows by standard tools: we have $\partial\_s^n u(0, 1) = 0$ for all $n$, and so $\Delta\_x^n u(0, 1) = 0$ for all $n$, and consequently all terms in the Taylor expansion of $u$ about $(0, 1)$ vanish.




---


Method 2 *(Edit: This is the same as Georgio Metafune's answer, which I did not realise in the beginning. Sorry!)* : Up to some powers of $s$, the integral is equal to the Mellin convolution of $F$ and the kernel
\[ x \mapsto \frac{x^k}{(1 + x^2)^{(k+3)/2}} . \]
The Mellin transform of that kernel is
\[ t \mapsto \frac{\Gamma(\tfrac{k-t}{2}) \Gamma(\tfrac{k+t}{2})}{2 \Gamma(k)} , \]
which has no zeros in the strip $0 \leqslant \Re t \leqslant k$. By the Mellin convolution theorem, we have $F = 0$.


The last step requires some care, since the Mellin transform of $F$ may fail to exist in the usual sense. One way around is to use distribution theory, as in Giorgio Metafune's answer. Another approach might be to split $F$ into two pieces ("small $x$" and "large $x$"), write the Mellin inversion formula for both of them, and deform the contour of integration.


