to @simonreed, section 17, vol 4, for background on operator determinants. We stress the fact that all these formulas are finite dimensional.

From the last display, the strategy to show convergence in law in either Theorem \[ThJ\] or \[ThJextended\] is clear: fix $s_0$, show that under the relevant assumptions, $\mathbf{P}(l_{1,N}\leq s_0){\rightarrow}W_1(s_0)$, and use the fact that $W_1$ is continuous to conclude.

Complex case
------------

We just saw that to find the asymptotic behavior of $l_1$ is equivalent to showing the convergence of the determinant of a certain operator. This task can be reduced to showing convergence in trace class norm of this operator (see @simonreed for background on this, e.g, Lemma XIII.17.4 (p.323)). Through work from @widom99, @imj exhibits an integral representation formula for his operator, and the original problem is essentially transformed into showing that certain integrals have a predetermined limit. In somewhat more detail, if we call $\alpha=n-N$, and $L_k^{\alpha}$ the $k$-th Laguerre polynomial associated with $\alpha$ (as in @szego, p.100), let $$\phi_k(x)=\sqrt{\frac{k!}{(k+\alpha)!}}x^{\alpha/2}{\mathrm{e}}^{-x/2}L_k^{\alpha}(x) \; ,$$ $\xi_k(x)=\phi_k(x)/x$, $a_N=\sqrt{Nn}$, and finally $$\left\{
\begin{array}{cll}
\phi(x)&=(-1)^N\sqrt{\frac{a_N}{2}} (\sqrt{n}\xi_N(x)-\sqrt{N}\xi_{N-1}(x))  \; ,\\
\psi(x)&=(-1)^N\sqrt{\frac{a_N}{2}}(\sqrt{N}\xi_N(x)-\sqrt{n}\xi_{N-1}(x)) \; .
\end{array}
\right.$$ We note two things: first, there is a slight abuse of notation since $\phi$ and $\psi$ obviously depend 