Suppose $F~$ is a probability distribution symmetrical about 0, for which all moments exist. Let $\mu_i~$be the $i$-th moment (of course $\mu_i=0$ if $i~$ is odd).

We know there are some conditions under which $\{\mu_i\}$ determines $F$, and in that case $\{\mu_i\}$ must determine the absolute moments of $F$.

Q1. How can we write the absolute moments directly in terms of $\{\mu_i\}$? (I have seen a formula of von Bahr (1965) quoted here, but I wonder if there is something simpler than an integral like that.)

Q2. In the case that $\{\mu_i\}$ doesn't determine $F$, does it still determine the absolute moments?

==========ADDITION==========

Here is the result of Bengt von Bahr, Ann. Math. Stat. 36 (1965) 808-818. I changed it in the fashion of Ushakov, Statistics & Probability Letters Volume 81, Issue 12, December 2011, Pages 2011-2015, which seems to correct a misplaced bracket in the original. All the integrals are over $(-\infty,\infty)$.

Let $H(x)$ be a function of bounded variation on $(-\infty,\infty)$ with finite absolute moment $ \int |x|^\nu |dH(x)|$ for $\nu>0$ not an even integer. Define the ordinary moments $\mu_j=\int x^j dH(x)$ for $j=0,1,\ldots,\lfloor \nu\rfloor$. Let $\phi(t)$ be the characteristic function of $H(x)$. Then $$ \int |x|^\nu dH(x) = (\Gamma(\nu+1)/\pi)\cos((\nu+1)\pi/2) \int \frac{\Re\phi(t) -\sum_{j=0}^m (-1)^j\mu_{2j}t^{2j}/(2j)!}{|t|^{\nu+1}} dt,$$ where $\Re$ stands for real part and $m=\lfloor \nu/2\rfloor$.