Let $f,g:\mathbb{R}\to\mathbb{R}$ be continuous functions with the following properties:


1. $f(x)$ and ${g(x)}/x$ are bounded;
2. ${g(x)}/{\left(1+x^2\right)}\in L^1\left(\mathbb{R}\right)$;
3. $\lim\_{x\to0}f(x)/x^2=1$;


and also
$$\int\_{-\infty}^\infty\frac{f(x)}{\left(1+x^2\right)^p}dx=p\int\_{-\infty}^{\infty}\frac{g(x)}{\left(1+x^2\right)^p}dx$$
for every real number $p\geq1$.


How special is the pair $\left(f,g\right)$?



Functions orthogonal to powers of $1/{\left(1+x^2\right)}$
Let $\tilde{f}(x) = f(x)+f(-x)$, and $\tilde{g} (x) = g(x)+g(-x)$, and let $F(x) = \int\_{0}^{x} \tilde{f}(t)dt$. Then your condition can be rewritten as (after integration by parts in the left hand side) 
$$
\int\_{0}^{\infty} \frac{2xF(x)-\tilde{g}(x) (1+x^{2})}{1+x^{2}} (1+x^{2})^{-p}dx =0
$$
for all $p\geq 1$. 


Now you can use the same density argument from here [Function orthogonal to powers of $1/\left(1+x^2\right)$](https://mathoverflow.net/questions/336892/function-orthogonal-to-powers-of-1-left1x2-right)


(notice that $\frac{2xF(x)-\tilde{g}(x) (1+x^{2})}{1+x^{2}} \cdot \frac{1}{1+x^{2}}\in L^{1}([0, \infty))$ to conclude that $\frac{2xF(x)}{1+x^{2}} = \tilde{g}(x)$ on $[0,\infty)$. 


