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  : Slater integrals and the spin-orbit interaction for Dy$^{3+}$ atoms in the Hartree-Fock approximation (in units of eV).

$\ast$In the RXS calculation, the above values of the anisotropic terms are reduced by multiplying a factor 0.8, while the values for $F^{(0)}(nl,n'l')$ are replaced by much smaller values, $F^{(0)}(4f,5d) = 3.0$, $F^{(0)}(4f,4f) = 7.0$, $F^{(0)}(2p,5d) = 4.0$, $F^{(0)}(2p,4f) = 12.0$. \[tab.slater\]

Before calculating the RXS spectra, we touch on the absorption coefficient. We calculate the absorption coefficient $A(\omega)$ in the $E_1$ process from the resolvent by using the relation, $$\begin{aligned}
 A(\omega) & \propto  \sum_{j} \sum_{\alpha}
   \langle\psi_j|x^\alpha(j)|m^ds^d\lambda\rangle 
\nonumber \\
&  \times
   \left(-\frac{1}{\pi}\right)
 {\rm Im}\left(\frac{1}{\hbar\omega - H_{\rm int} + i\delta}\right)_
    {m^ds^d\lambda;m'^ds'^d\lambda'} \nonumber\\
  & \times\langle m'^ds'^d\lambda'|x^\alpha(j)|\psi_j\rangle .
\label{eq.absorp}\end{aligned}$$ Here ${\rm Im}X$ indicates the imaginary part of the quantity $X$. Figure \[fig.absorp\] shows the calculated $A(\omega)$ in comparison with the fluorescence experiment.[@Hirota] We adjust the core hole energy such that the calculated peak coincides with the experimental one. The calculated curve reproduces the experimental one. Since the spectra is proportional to the $d$ DOS when the $5d$-$2p$ Coulomb interaction is neglected, the assumed DOS seems reasonable. The intensity seen in the high energy region of the experiment may come from the $d$ symmetric 