\frac{\pi}{L}x \right)\psi_\beta^0(x)
\label{gegenbauer}$$ with the coefficient $${\cal N}^\ell_\beta = \left( -\frac{\pi}{L} \right)^\ell 
\frac{\ell!\, (2\beta+2\ell-1)!\, (\beta-1)! }
{2^\ell\, (2\beta+\ell-1)!\, (\beta+\ell-1)!},$$ where $C_\ell^\beta (z)$ denotes the Gegenbauer polynomial of the $\ell$th order [@tableofintegral; @Kuramoto]. The coefficient ${\cal N}^\ell_\beta$ can be obtained by demanding that $A_\beta \psi^\ell_\beta (x) = \left( \frac{\pi}{L} \right)^2 \ell (\ell+2\beta) \psi^{\ell-1}_{\beta+1} (x)$ and the following recursion relation for the Gegenbauer polynomials are consistent: $$\frac{d}{dz} C^\beta_\ell (z) = 2\beta\, C^{\beta+1}_{\ell-1} (z).$$ Then, the corresponding eigenvalue of $H_\beta$ is obtained as $$\mu= \left( \frac{\pi}{L} \right)^2(\beta+\ell)^2,
\label{chem_p}$$ for $\ell=0,1,2, \cdots$. Thus, if the sum of $\beta$ and $\ell$ is constant, say $\beta+\ell=q$, then $\mu$ takes the same value among $\psi^0_q (x), \psi^1_{q-1}(x), \cdots, \psi^{q-\beta}_{\beta}(x), \cdots, \psi^q_0 (x)$.

Here, we list in Table I the first few eigenfunctions of $H_\beta$ with $\beta=0, 1, ..., 5$, from which the relation between the $1/\sin^2$-potential problem and the SSD problem can be seen clearly. On the one hand, one finds the eigenfunctions of $H_\beta$ in the column labeled by $\beta$. On the other hand, the functions in the row labeled by $q$ give the solutions of the inverse problem (up to unimportant overall factors). For a particular chemical potential parameterized by $\mu= (\frac{\pi}{L})^2 q^2$, we obtain 