Let $a<b$. When there is a smooth, strictly positive function $f$ on $[a,b]$ such that for any $n\in\mathbb Z, n\geq 0$ we have:

$$m_n(d\mu):=\int_{a}^bx^nd\mu \in\mathbb Z,$$

where $d\mu=f(x)dx$.

When $a=-1, b=1$ this is clearly impossible as $0<m_n(d\mu)$ and $m_n(d\mu)\to0$.

When $a=-2, b=2$ there is non-trivial example

\begin{equation} \dfrac 1{2\pi}\int\limits_{-2}^2\sqrt{4-x^2}x^k= \begin{cases} C_{k/2}, & \text{if}\ k\ \text{even} \\ 0, & \text{otherwise} \end{cases} \end{equation} Here $C_k$ is Catalan numbers.

When $a=0, b=+\infty$ we can take $f(x)=e^{-x}$.

Can we construct such a measure when $b-a < 4$?

  • $\begingroup$ Is the key point of this that you want all the $m_n$ to be integers? $\endgroup$
    – Henry
    Sep 18 at 12:55
  • $\begingroup$ Yes, you are correct. $\endgroup$ Sep 18 at 13:12
  • 2
    $\begingroup$ For what it is worth, stackoverflow.com/questions/65975300/… has an example with $a=0,b=4$ (in fact all the moments are Catalan numbers) and there may be simpler examples $\endgroup$
    – Henry
    Sep 18 at 13:16
  • $\begingroup$ Uniform distribution doesn't work! Because integral of $x^k$ has $k$ as denominator. $\endgroup$ Sep 18 at 19:58


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