Function orthogonal to powers of $1/\left(1+x^2\right)$ Does there exist any continuous function $f:\mathbb{R}\to\mathbb{R}$, $f(x)/(1+x^2)\in L^1(\mathbb R)$, such that $f(0)=1$ and
$$\int_{-\infty}^{\infty}\frac{f(x)}{\left(1+x^2\right)^p}dx=0$$ for every $1\leq p\leq 2$?
 A: The connection between this question and uniqueness sets for holomorphic functions has already been mentioned but I think that it could benefit from a more systematic treatment. Let me begin with Müntz’  result concerning when the span of the sequence $(x^{\lambda_n})$ is dense in $C[0,1]$. If one considers the  holomorphic function $F(z)=\int x^z dx$ then it can be dualised to the question of sets or sequences of uniqueness for bounded analytic functions in  the right half plane.  These are completely understood (e.g. by using a Möbius tranformation to reduce to functions on the unit disc and then Blaschke products). This provides a proof of the Müntz result in the special case where the exponents are real and go to infinity but also in many other cases (convergence to zero, complex exponentials which converge slowly to the imaginary small corrections—-axes).
In the case in question here one uses the function $F(z)=\int f(x)(1+x^2)^{-z} dx$ which is analytic on a suitable region of the plane.  Then sets of uniqueness provide results of this type—-in the case of the original posting the rather crude one of a real interval.
A: I think no such function exists. Assuming $\displaystyle{f(x)\over 1+x^2}\in L^1(\mathbb{R})$ the integral $\displaystyle\int_\mathbb{R}{f(x)\over (1+x^2)^p}dx$ is analytic wrto $p>1$: indeed, for any $p>1$ and $|t|<p-1$, expanding $(1+x^2)^{t}$ in powers of $t$ we have, by Tonelli's theorem (w.rto the product measure space $\mathbb{R}\times\mathbb{N}$)
$$\begin{align}\int_\mathbb{R}\sum_m \bigg|{f(x)\over (1+x^2)^p}\big( \log(1+x^2)\big)^m {t^m\over m!}\bigg|dx&=\\=\int_\mathbb{R} {|f(x)|\over (1+x^2)^{p-|t|}}\,dx&\le\int_\mathbb{R} {|f(x)|\over 1+x^2} \,dx<+\infty\end{align}$$
and then by Fubini's
$$\int_\mathbb{R}{f(x)\over (1+x^2)^{p+t}}dx=\sum_m \bigg(\int_\mathbb{R}{f(x)\over (1+x^2)^p}\big( \log(1+x^2)\big)^m\,dx\bigg) {(-t)^m\over m!}, $$
that is the power series expansion for the integral at $p$.
Therefore $\displaystyle\int_\mathbb{R}{f(x)\over (1+x^2)^p}dx=0$ for all $p\ge1$. Next, to simplify a little we may replace $f$ with its even part, and assume equivalently that the integral on $\mathbb{R}_+$ vanishes. Changing variable with $\displaystyle u={1\over 1+x^2}\in[0,1]$  we have
$$\int_0^1f\Big(\sqrt{{1\over u}-1}\Big)(1-u)^{-3/2}u^{p-1/2}du=0,\qquad \text{ for all } p\ge 1$$
that is, the function $g(u):=f\Big(\sqrt{{1\over u}-1}\Big)(1-u)^{-3/2}u^{1/2}$ has $\int_o^1g(u)u^pdu=0$ for all $p\in\mathbb{N}$, so it must be identically zero, and so must $f$. 
