Injectivity of an integral transform For a bounded function $F: \mathbb R_{\ge 0} \to \mathbb R$ (not necessarily non-negative),  is it true that
$$\int_0^\infty \frac{x^ks}{(s^2+x^2)^{(k+3)/2}} F(x) dx = 0 \text{ for all $s >0$} \iff F \equiv 0$$
where $k \in \mathbb{N}$ is a positive constant? Of course, one implication ($\leftarrow$) is true. What about the other one?
 A: Think that $|x| F(|x|)$ ($x \in \mathbb R^{k+2}$) is the boundary value of a harmonic function $u$ in the half-space $\mathbb R^{k+2} \times (0, \infty)$, given by an appropriate Poisson integral:
$$ u(x,s) = c_k \int_{\mathbb R^{k+2}} |y| F(|y|) \, \frac{s}{(|x - y|^2 + s^2)^{(k+3)/2}} \, dy $$
with $c_k$ the normalisation constant. Then
$$ \begin{aligned} u(0,s) & = c_k \int_{\mathbb R^{k+2}} |y| F(|y|) \, \frac{s}{(|y|^2 + s^2)^{(k+3)/2}} \, dy \\ & = c_k' \int_0^\infty r^k F(r) \, \frac{s}{(r^2 + s^2)^{(k+3)/2}} dr . \end{aligned} $$
Thus, your equation says that $u(0,s) = 0$ for every $s > 0$.
Note that $u(x, s)$ is a radial function of $x$ for every $s > 0$. Since $u$ is also real-analytic, its power series expansion at $(0, s)$ with respect to $x$ is of the form
$$ u(x, s) = \sum_{n=1}^\infty a_n(s) |x|^{2n} ,$$
and it can be differentiated under the sum. Since $\Delta_{x,s} u = 0$, we have
$$ 0 = \sum_{n=1}^\infty a_n(s) 2 n (2 n + k) |x|^{2n - 2} + \sum_{n=1}^\infty a_n''(s) |x|^{2n} ,$$
which shows that $a_n(s) = 0$ for every $n$. It follows that $u$ is identically zero, and therefore $F(x) = 0$ for almost all $x$.

Other proofs also seem feasible. For example:

*

*The equation says that the Mellin convolution of $F$ and another kernel, $x^k / (1 + x^2)^{(k+3)/2}$, is identically zero. Furthermore, the Mellin transform of the kernel $x^k / (1 + x^2)^{(k+3)/2}$ is everywhere non-zero (because by a substitution $x^2 = (1-t) / t$, it reduces to a beta integral). This means that the Mellin transform of $F$ (in the sense of distributions if $F$ is not nicely integrable) is necessarily equal to zero, and consequently $F$ is almost everywhere zero.


*After a substitution $x^2 = t$, the equation tells us that the generalised Cauchy–Stieltjes transform of $t^{k/2-1/2} F(t^{1/2})$ is zero along the half-line $(-\infty, 0)$. By appropriate inversion theorems for the generalised Cauchy–Stieltjes transform, one should be able to figure out that $F$ is zero almost everywhere.
