for Fredholm determinants (using eqs.(\[DFr\],\[DN0\])): $$\label{Ffr}
\e^{+\mi \varphi/4} \Delta_N^+ (\e^{-\mi \varphi} \lambda) \Delta_N^- (\lambda)
-\e^{-\mi \varphi/4} \Delta_N^+ (\lambda) \Delta_N^- (\e^{-\mi \varphi} \lambda)
\equiv 2 \mi \sin \, \varphi/4 .$$ In the latter formula, holding for all $\varphi = 4\pi /(N+2)$, the expansions of both sides in powers of $\varphi \to 0$ should then be identified order by order. Eq.(\[Ffr\]) having the form of a “quantum Wronskian" identity for finite $\varphi$ [@BLZ; @DT], it is not surprising that the identification to the leading order $O(\varphi)$ discloses a “classical" Wronskian structure: $$\label{frinf}
\Delta^+ \Bigl(\lambda {\d \over \d\lambda} \Delta^- \Bigr)
- \Bigl(\lambda {\d \over \d\lambda} \Delta^+ \Bigr) \Delta^-
\equiv {1 \over 2} (1- \Delta^+ \Delta^-) .$$ However, we know neither how to interpret the right-hand side, nor how to solve this functional relation directly (and identification at the next order in $\varphi \propto 1/N$ in eq.(\[Ffr\]) does not appear to yield any new information either): this constitutes an interesting open problem, since the finite-$N$ equation admits of constructive solutions by an exact quantization method using eq.(\[EQN\]); this method itself however seems totally singular in the $N \to +\infty$ limit. At the same time, the Fredholm determinants of the infinite square well are known, 